RS Aggarwal Solutions for Class 12 Chapter 33 Linear Programming

 

Question 1. Graph the solution sets of the following inequations:
x + y ≥ 4
Answer: Start with the expression x + y ≥ 4, which rearranges to y ≥ 4 - x. Work with the equation y = 4 - x to find key points. When x = 0, y equals 4, giving point A(0,4) on the Y-axis. When y = 0, x equals 4, giving point B(4,0) on the X-axis. Mark these points and draw the line through them. Since the inequality is y ≥ 4 - x, the solution region lies on or above this line, including all points where y values are greater than or equal to 4 - x. The shaded region extends upward from the line AB.

Exam Tip: When graphing inequalities, always test a point not on the boundary line (like the origin) to confirm which side should be shaded. Remember that ≥ or ≤ means the line itself is part of the solution and should be solid, not dashed.

 

Question 2. Graph the solution sets of the following inequations:
x - y ≤ 3
Answer: Starting with x - y ≤ 3, isolate y by subtracting x from both sides to get - y ≤ 3 - x. Multiply by negative 1 (and flip the inequality sign) to obtain y ≥ x - 3. Use the equation y = x - 3 to find boundary points. Setting x = 0 gives y = - 3, so point A(0, - 3) is on the Y-axis. Setting y = 0 gives x = 3, so point B(3,0) is on the X-axis. Plot these points and connect them with a solid line. The inequality y ≥ x - 3 means the solution set consists of all points on or above this line. Shade the entire region above the boundary line.

Exam Tip: Flipping the inequality sign when multiplying by a negative number is critical and commonly tested. Double-check your inequality direction after each multiplication or division by negative values.

 

Question 3. Graph the solution sets of the following inequations:
x + 2y > 1
Answer: Begin with x + 2y > 1 and rearrange to 2y > 1 - x, then divide by 2 to get y > \( \frac{1}{2} - \frac{x}{2} \). Consider the equation y = \( \frac{1}{2} - \frac{x}{2} \). When x = 0, y = \( \frac{1}{2} \), giving point A(0, \( \frac{1}{2} \)) on the Y-axis. When y = 0, we have 0 = \( \frac{1}{2} - \frac{x}{2} \), so x = 1, giving point B(1,0) on the X-axis. Plot these points. Since the inequality is strictly greater than (not greater than or equal to), draw the boundary line as a dashed or dotted line to indicate that points on the line itself are not included in the solution. The solution region consists of all points strictly above this line.
In simple words: The inequality uses > (greater than), not ≥, so the boundary line is drawn dotted, and you shade only the area above it, not the line itself.

Exam Tip: Use a solid line for ≥ or ≤ inequalities and a dashed line for > or < inequalities. This distinction is essential for showing whether boundary points are included or excluded.

 

Question 4. Graph the solution sets of the following inequations:
2x - 3y < 4
Answer: Rearrange 2x - 3y < 4 to 2x - 4 < 3y, then divide by 3 to obtain y > \( \frac{2}{3}x - \frac{4}{3} \). Work with the equation y = \( \frac{2}{3}x - \frac{4}{3} \). When x = 0, y = - \( \frac{4}{3} \), so point A(0, - \( \frac{4}{3} \)) lies on the Y-axis. When y = 0, we get 0 = \( \frac{2}{3}x - \frac{4}{3} \), which gives x = 2, so point B(2,0) is on the X-axis. Plot these points and draw a dashed line between them because the inequality uses < (strict inequality). The solution comprises all points strictly above this dashed boundary line.
In simple words: Draw a dashed line through the two intercept points and shade everything above it. The line itself is not part of the answer since we have < rather than ≤.

Exam Tip: Always draw the boundary line dashed for strict inequalities (< and >) and solid for non-strict inequalities (≤ and ≥). Shading the correct region matters just as much as drawing the correct line.

 

Question 5. Graph the solution sets of the following inequations:
x ≥ y - 2
Answer: Rearrange x ≥ y - 2 by adding 2 to both sides to get x + 2 ≥ y, which is equivalent to y ≤ x + 2. Use the equation y = x + 2 to find intercepts. When x = 0, y = 2, giving point A(0,2) on the Y-axis. When y = 0, we have 0 = x + 2, so x = - 2, giving point B( - 2,0) on the X-axis. Mark these points and draw a solid line connecting them. Since the inequality is y ≤ x + 2 (less than or equal), the solution region consists of all points on or below this solid line. Shade the entire area below the boundary line.

Exam Tip: For y ≤ or y <, shade downward or below the line. For y ≥ or y >, shade upward or above the line. This directional pattern simplifies determining the correct shaded region.

 

Question 6. Graph the solution sets of the following inequations:
y - 2 ≤ 3x
Answer: Rearrange y - 2 ≤ 3x by adding 2 to both sides to obtain y ≤ 3x + 2. Consider the equation y = 3x + 2. When x = 0, y = 2, giving point A(0,2) on the Y-axis. When y = 0, we get 0 = 3x + 2, so x = - \( \frac{2}{3} \), giving point B( - \( \frac{2}{3} \),0) on the X-axis. Plot these points and draw a solid line since the inequality includes equality (≤). The solution set includes all points on or below this line. Shade the region below the solid boundary line.

Exam Tip: Check your work by substituting the origin (0,0) or another test point. If 0 ≤ 3(0) + 2 is true, then (0,0) should be in the shaded region, which helps verify your shading is correct.

 

Question 7. Solve each of the following systems of simultaneous inequations:
2x + y > 1 and 2x - y ≥ - 3
Answer: Address each inequality separately. For 2x + y > 1, rearrange to y > 1 - 2x. The equation y = 1 - 2x has intercepts at A(0,1) on the Y-axis. Setting y = 0 gives x = \( \frac{1}{2} \), so B(\( \frac{1}{2} \),0) is on the X-axis. Use a dashed line since the inequality is strict (>). Shade above this line. For 2x - y ≥ - 3, rearrange to y ≤ 2x - 3. The equation y = 2x - 3 has intercepts at C(0, - 3) on the Y-axis. Setting y = 0 gives x = \( \frac{3}{2} \), so D(\( \frac{3}{2} \),0) is on the X-axis. Use a solid line since the inequality is non-strict (≥). Shade below this line. The solution to the system is the overlapping region where both shaded areas meet—the intersection of the two solution sets.

Exam Tip: For systems of inequalities, always graph each one separately first, then identify the overlapping region. Only the area satisfying all inequalities simultaneously is the final answer.

 

Question 8. Solve each of the following systems of simultaneous inequations:
x - 2y ≥ 0, 2x - y ≤ - 2
Answer: Tackle the first inequality x - 2y ≥ 0. Rearrange to x ≥ 2y, or equivalently y ≤ \( \frac{x}{2} \). The equation y = \( \frac{x}{2} \) is a straight line passing through the origin (0,0). Use a solid line and shade the region below (and on) this line since y ≤ \( \frac{x}{2} \). Next, consider 2x - y ≤ - 2. Rearrange to y ≥ 2x + 2. The equation y = 2x + 2 has intercepts at A(0,2) on the Y-axis and B( - 1,0) on the X-axis. Use a solid line and shade above this line since y ≥ 2x + 2. The solution region is where both shaded areas overlap—the region that is simultaneously below the line y = \( \frac{x}{2} \) and above the line y = 2x + 2. This intersection region represents all points satisfying both inequalities.

Exam Tip: When one inequality passes through the origin, be extra careful with your shading. Test a point like (1,0) or (0,1) to confirm which side of the line satisfies the inequality before shading.

 

Question 9. Solve each of the following systems of simultaneous inequations:
3x + 4y ≥ 12, x ≥ 0, y ≥ 1 and 4x + 7y ≤ 28
Answer: Start by examining the inequality 3x + 4y ≥ 12:

⇒ 4y ≥ 12 - 3x

⇒ y ≥ 3 - \( \frac{3}{4}x \)

Next, consider the equation y = 3 - \( \frac{3}{4}x \). To find where this line crosses the axes:

When x = 0, y = 3, giving the point A(0, 3) on the Y-axis.

When y = 0, we get 0 = 3 - \( \frac{3}{4}x \), so x = 4, giving the point B(4, 0) on the X-axis.

For the inequality y ≥ 3 - \( \frac{3}{4}x \), the needed region lies above point A.

Now examine the inequality 4x + 7y ≤ 28:

⇒ 7y ≤ 28 - 4x

⇒ y ≤ 4 - \( \frac{4}{7}x \)

For the equation y = 4 - \( \frac{4}{7}x \):

When x = 0, y = 4, giving the point C(0, 4) on the Y-axis.

When y = 0, we get 0 = 4 - \( \frac{4}{7}x \), so x = 7, giving the point D(7, 0) on the X-axis.

For the inequality y ≤ 4 - \( \frac{4}{7}x \), the needed region lies below point C.

The condition x ≥ 0 means the area on the right side of the Y-axis. The condition y ≥ 1 means the area above the horizontal line y = 1. The region that satisfies all four inequalities is where these four half-planes overlap.

Exam Tip: Always find the intercept points on both axes to determine the boundary lines, then use test points to confirm which region satisfies each inequality.

 

Question 10. Show that the solution set of the following linear constraints is empty:
x - 2y ≥ 0, 2x - y ≤ - 2, x ≥ 0 and y ≥ 0
Answer: Consider the inequality x - 2y ≥ 0:

⇒ x ≥ 2y

⇒ y ≤ \( \frac{x}{2} \)

The equation y = \( \frac{x}{2} \) is a straight line passing through the origin. For the inequality y ≤ \( \frac{x}{2} \), the region needed lies below the origin.

Now consider the inequality 2x - y ≤ - 2:

⇒ y ≥ 2x + 2

For the equation y = 2x + 2:

When x = 0, y = 2, giving the point A(0, 2) on the Y-axis.

When y = 0, we get 0 = 2x + 2, so x = - 1, giving the point B(- 1, 0) on the X-axis.

For the inequality y ≥ 2x + 2, the needed region lies above point A.

The condition y ≥ 0 means the area above the X-axis. The condition x ≥ 0 means the area on the right side of the Y-axis. When we combine all four requirements together, there is no common area where all four regions intersect, so there is no solution to this system of inequalities.

Exam Tip: When regions do not overlap, clearly state that the solution set is empty - this is a valid conclusion and shows the constraints are incompatible.

 

Question 11. Find the linear constraints for which the shaded area in the figure given is the solution set.
Answer: Examine line A:

The given line is x - y = 1

⇒ y = x - 1

Since the shaded region lies above the point where this line crosses the Y-axis at (0, - 1),

⇒ y ≥ x - 1

⇒ x - y ≤ 1

Examine line B:

The given line is 2x + y = 2

⇒ y = 2 - 2x

Since the shaded region lies above the point where this line crosses the Y-axis at (0, 2),

⇒ y ≥ 2 - 2x

⇒ 2x + y ≥ 2

Examine line C:

The given line is x + 2y = 8

⇒ 2y = 8 - x

⇒ y = 4 - \( \frac{x}{2} \)

Since the shaded region lies below the point where this line crosses the Y-axis at (0, 4),

⇒ y ≤ 4 - \( \frac{x}{2} \)

⇒ x + 2y ≤ 8

Examine line D:

This represents the region on the right side of the Y-axis.

⇒ x ≥ 0

Combining all the derived results:

x - y ≤ 1

2x + y ≥ 2

x + 2y ≤ 8

x ≥ 0

Exam Tip: Work through each boundary line separately, identifying the correct inequality symbol by checking whether the shaded region lies above or below each line.

 

Exercise 33B

 

Question 1. Find the maximum value of Z = 7X + 7Y, subject to the constraints.
x ≥ 0, y ≥ 0, x + y ≥ 2 and 2x + 3y ≤ 6.
Answer: The feasible region formed by the constraints x ≥ 0, y ≥ 0, x + y ≥ 2, and 2x + 3y ≤ 6 is shown in the graph.

The corner points of the feasible region are A(0, 2), B(2, 0), and C(3, 0).

Evaluate Z at each corner point:

The maximum value of Z is 21, occurring at point C(3, 0).
In simple words: Test each corner point of the feasible region by plugging its coordinates into the objective function. The point that produces the largest result gives you the maximum value.

Exam Tip: Always evaluate the objective function at every corner point - the maximum or minimum will always occur at one of these vertices for linear programming problems.

 

Question 2. Maximize Z = 4x + 9y, subject to the constraints
x ≥ 0, y ≥ 0, x + 5y ≤ 200, 2x + 3y ≤ 134.
Answer: The feasible region formed by the constraints x ≥ 0, y ≥ 0, x + 5y ≤ 200, and 2x + 3y ≤ 134 is shown in the graph.

The corner points of the feasible region are A(10, 38), B(0, 40), C(0, 0), and D(67, 0).

Evaluate Z at each corner point:

The maximum value of Z is 382, occurring at point A(10, 38).
In simple words: Calculate the value of the objective function at all corner points. The highest value you get is your maximum, and it happens at the point where the best mix of x and y values is achieved.

Exam Tip: The optimal solution in linear programming occurs at a corner (vertex) of the feasible region - never inside it - so checking all vertices is essential.

 

Question 3. Find the minimum value of Z = 3x + 5y, subject to the constraints
- 2x + y ≤ 4, x + y ≥ 3, x - 2y ≤ 2, x ≥ 0 and y ≥ 0
Answer: The feasible region determined by the constraints - 2x + y ≤ 4, x + y ≥ 3, x - 2y ≤ 2, x ≥ 0, and y ≥ 0 is shown in the graph.

The feasible region is unbounded. The vertices of the region are A(0, 4), B(0, 3), and C\( \left(\frac{8}{5}, \frac{1}{5}\right) \).

Evaluate Z at each vertex:

The minimum value of Z is \( \frac{29}{3} \), occurring at point C\( \left(\frac{8}{5}, \frac{1}{5}\right) \).
In simple words: Even when the feasible region extends infinitely, you still evaluate the objective function at the corner points where boundary lines meet, and pick the smallest result.

Exam Tip: For an unbounded feasible region, always verify that a minimum or maximum actually exists by checking that the objective function is bounded in the relevant direction.

 

Question 4. Minimize Z = 2x + 3y, subject to the constraints
x ≥ 0, y ≥ 0, x + 2y ≥ 1 and x + 2y ≤ 10.
Answer: The feasible region determined by the constraints x ≥ 0, y ≥ 0, x + 2y ≥ 1, and x + 2y ≤ 10 is shown in the graph.

The corner points of the feasible region are A\( \left(0, \frac{1}{2}\right) \), B(0, 5), C(10, 0), and D(1, 0).

Evaluate Z at each corner point:

The minimum value of Z is \( \frac{3}{2} \), occurring at point A\( \left(0, \frac{1}{2}\right) \).
In simple words: Substitute each vertex's coordinates into the objective function, then choose the point that gives the smallest outcome as your minimum.

Exam Tip: Always check all corner points methodically - a careless calculation error on even one point can lead you to the wrong answer.

 

Question 5. Maximize Z = 3x + 5y, subject to the constraints X + 2y ≤ 2000, x + y ≤ 1500, y ≤ 600, x ≥ 0 and y ≥ 0.
Answer: The region defined by X + 2y ≤ 2000, x + y ≤ 1500, y ≤ 600, x ≥ 0 and y ≥ 0 forms the feasible area.

The vertices of this feasible area are A(0,0), B(0,600), C(800,600), D(1000,500), E(1500,0). We now compute Z = 3x + 5y at each vertex:

At A(0,0): Z = 0
At B(0,600): Z = 3000
At C(800,600): Z = 5400
At D(1000,500): Z = 5500
At E(1500,0): Z = 4500

The highest value of Z is 5500, occurring at point D(1000,500). Therefore, to maximize profit, set x = 1000 and y = 500, yielding a maximum value of 5500.

Exam Tip: Always evaluate the objective function at all corner points of the feasible region - the maximum or minimum will always occur at one of these vertices in linear programming problems.

 

Question 6. Find the maximum and minimum values of Z = 2x + y, subject to the constraints X + 3y ≥ 6, x - 3y ≤ 3, 3x + 4y ≤ 24, - 3x + 2y ≤ 6, 5x + y ≥ 5, x ≥ 0 and y ≥ 0.
Answer: The region satisfying X + 3y ≥ 6, x - 3y ≤ 3, 3x + 4y ≤ 24, - 3x + 2y ≤ 6, 5x + y ≥ 5, x ≥ 0 and y ≥ 0 is shown graphically.

The vertices of the feasible region are A(4/3, 5), B(4/13, 45/13), C(9/14, 25/14), D(9/2, 1/2), E(84/13, 15/13). Computing Z = 2x + y at each vertex:

At A(4/3, 5): Z = 23/3
At B(4/13, 45/13): Z = 53/13
At C(9/14, 25/14): Z = 43/14
At D(9/2, 1/2): Z = 19/2
At E(84/13, 15/13): Z = 183/13

The lowest value is 43/14 at point C(9/14, 25/14). The highest value is 183/13 at point E(84/13, 15/13).

Exam Tip: When handling multiple inequality constraints, sketch each boundary line carefully and shade the correct region for each - the intersection of all shaded regions gives your feasible region.

 

Question 7. Mr. Dass wants to invest ₹12000 in public provident fund (PPF) and in national bonds. He has to invest at least ₹1000 in PPF and at least ₹2000 in bonds. If the rate of interest on PPF is 12% per annum and that on bonds is 15% per annum, how should he invest the money to earn maximum annual income? Also find the maximum annual income.
Answer: Let x denote the amount invested in PPF and y denote the amount invested in bonds.

From the problem conditions:
x + y ≤ 12000
x ≥ 1000
y ≥ 2000

The objective is to maximize Z = 0.12x + 0.15y (annual income).

The feasible region is bounded by these constraints. Its vertices are A(1000, 11000), B(1000, 2000), C(10000, 2000).

Evaluating Z at each vertex:
At A(1000, 11000): Z = 0.12(1000) + 0.15(11000) = 120 + 1650 = 1770
At B(1000, 2000): Z = 0.12(1000) + 0.15(2000) = 120 + 300 = 420
At C(10000, 2000): Z = 0.12(10000) + 0.15(2000) = 1200 + 300 = 1500

The maximum income of Rs.1770 occurs at A(1000, 11000). Therefore, Mr. Dass should invest Rs.1000 in PPF and Rs.11000 in national bonds to achieve the highest annual return.

Exam Tip: For investment/resource allocation problems, identify the constraint that binds (limits profit most) - investing more heavily in the higher-yield option while respecting minimum thresholds typically maximizes returns.

 

Question 8. A small firm manufactures necklace and bracelets. The total number of necklace and bracelet that it can handle per day is at most 24. It takes 1 hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ₹100 and that on a bracelet is ₹300, how many of each should be produced daily to maximize the profit? It is being given that at least one of each must be produced.
Answer: Let x represent the number of necklaces and y represent the number of bracelets produced daily.

The constraints are:
x + y ≤ 24 (total items)
0.5x + y ≤ 16 (total hours)
x ≥ 1, y ≥ 1 (at least one of each)

Maximize Z = 100x + 300y (daily profit).

The feasible region has vertices at A(1, 1), B(1, 15.5), C(16, 8), D(23, 1). Since the number of bracelets must be a whole number, point B(1, 15.5) is adjusted to (2, 15).

Evaluating Z at integer corner points:
At A(1, 1): Z = 100 + 300 = 400
At (2, 15): Z = 100(2) + 300(15) = 200 + 4500 = 4700
At C(16, 8): Z = 100(16) + 300(8) = 1600 + 2400 = 4000
At D(23, 1): Z = 100(23) + 300(1) = 2300 + 300 = 2600

The maximum profit of Rs.4700 is achieved at point (2, 15). Therefore, the firm should manufacture 2 necklaces and 15 bracelets daily.

Exam Tip: In integer linear programming, always check integer points near fractional corner points - the optimal integer solution may differ from the continuous optimum due to rounding constraints.

 

Question 9. A man has ₹1500 to purchase rice and wheat. A bag of rice and a bag of wheat cost ₹180 and 120 respectively. He has storage capacity of 10 bags only. He earns a profit of ₹11 and ₹8 per bag of rice and wheat respectively. How many bags of each must he buy to make maximum profit?
Answer: Let x represent bags of wheat and y represent bags of rice.

The constraints are:
120x + 180y ≤ 1500 (budget)
x + y ≤ 10 (storage)
x ≥ 0, y ≥ 0

Maximize Z = 8x + 11y (profit).

The feasible region has corner points A(0, 8), B(0, 0), C(10, 0), D(5, 5).

Computing Z at each vertex:
At A(0, 8): Z = 0 + 88 = 88
At B(0, 0): Z = 0
At C(10, 0): Z = 80 + 0 = 80
At D(5, 5): Z = 8(5) + 11(5) = 40 + 55 = 95

The maximum profit of Rs.95 occurs at point D(5, 5). Therefore, the man should purchase 5 bags each of wheat and rice to maximize his profit.

Exam Tip: When multiple constraints are present, sketch all boundary lines carefully on the same axes - the feasible region is their intersection, and you must evaluate the objective function at every corner point of this region.

 

Question 10. A manufacture produces nuts and bolts for industrial machinery. It takes 1 hour of work on machine A and 3 hours on machine B to produces a packet of nuts while it takes 3 hours on machine A and 1 hours on machine B to produce a packet of bolts. He earns a profit ₹17.50 per packet on nuts and ₹7 per packet on bolts. How many packets of each should be produced each day so as to maximize his profit if he operates each machine for at the most 12 hours a day? Also find the maximum profit.
Answer: Let x denote packets of nuts and y denote packets of bolts produced daily.

The constraints are:
x + 3y ≤ 12 (machine A hours)
3x + y ≤ 12 (machine B hours)
x ≥ 0, y ≥ 0

Maximize Z = 17.50x + 7y (daily profit).

The feasible region has vertices A(0, 0), B(0, 4), C(3, 3), D(4, 0).

Evaluating Z at each corner point:
At A(0, 0): Z = 0
At B(0, 4): Z = 17.50(0) + 7(4) = 28
At C(3, 3): Z = 17.50(3) + 7(3) = 52.50 + 21 = 73.50
At D(4, 0): Z = 17.50(4) + 7(0) = 70

The maximum profit of Rs.73.50 is attained at point C(3, 3). The manufacturer should produce 3 packets each of nuts and bolts daily to achieve this maximum profit.

Exam Tip: For production problems with multiple machine constraints, identify which machine is the bottleneck (more limiting) - this guides the balance of product mix for optimality.

 

Question 11. Two tailors, A and B, earn ₹300 and ₹400 per day respectively. A can stitch 6 shirts and 4 pair of trousers while B can stitch 10 shirts and 4 pairs of trousers per day. How many days should each of them work if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labor cost?
Answer: Let x denote days tailor A works and y denote days tailor B works.

The constraints are:
6x + 10y ≥ 60 (shirts)
4x + 4y ≥ 32 (trousers)
x ≥ 0, y ≥ 0

Minimize Z = 300x + 400y (total labor cost).

The feasible region (unbounded) has corner points A(0, 8), B(5, 3), C(10, 0).

Computing Z at each vertex:
At A(0, 8): Z = 300(0) + 400(8) = 3200
At B(5, 3): Z = 300(5) + 400(3) = 1500 + 1200 = 2700
At C(10, 0): Z = 300(10) + 400(0) = 3000

The minimum cost of Rs.2700 occurs at point B(5, 3). Therefore, tailor A should work for 5 days and tailor B should work for 3 days to fulfill the production requirements at minimum expense.

Exam Tip: For minimization problems with unbounded feasible regions, always verify that the optimal point is indeed a vertex - the minimum will still occur at a corner point if one exists within the feasible region.

 

Question 12. A dealer wishes to purchase a number of fans and sewing machines. He has only Rs.5760 to invest and space for at most 20 items. A fan costs him Rs.360 and a sewing machine, Rs.240. He expects to gain Rs.22 on a fan and Rs.18 on a sewing machine. Assuming that he can sell all the items he can buy, how should he invest the money in order to maximize the profit?
Answer: Let the number of fans bought be x and sewing machines bought be y.

According to the question:
\( 360x + 240y \leq 5760, \; x + y \leq 20, \; x \geq 0, \; y \geq 0 \)

Maximize \( Z = 22x + 18y \)

The feasible region determined by \( 360x + 240y \leq 5760, \; x + y \leq 20, \; x \geq 0, \; y \geq 0 \) is given by the graph shown. The corner points of the feasible region are \( A(0,0) \), \( B(0,20) \), \( C(8,12) \), \( D(16,0) \).

The value of Z at corner points:

The maximum value of Z is 392 at point (8,12).

The dealer should buy 8 fans and 12 sewing machines to achieve the highest profit.
In simple words: The dealer earns the most profit - Rs. 392 - by purchasing 8 fans and 12 sewing machines. This combination uses the available budget and space efficiently.

Exam Tip: Always test all corner points of the feasible region to find which one gives the maximum (or minimum) value of the objective function. The optimal solution always occurs at a corner point.

 

Question 13. A firm manufactures two types of products, A and B, and sells them at a profit of Rs.2 on type A and Rs.2 on type B. Each product is processed on two machines, M1 and M2. Type A requires one minute of processing time on M1 and two minutes on M2. Type B requires one minute on M1 and one minute on M2. M1 is available for not more than 6 hours 40 minutes while M2 is available for at most 10 hours a day. Find how many products of each type the firm should produce each day in order to get maximum profit.
Answer: Let the firm manufacture x number of A and y number of B products.

According to the question:
\( x + y \leq 400, \; 2x + y \leq 600, \; x \geq 0, \; y \geq 0 \)

Maximize \( Z = 2x + 2y \)

The feasible region determined by \( x + y \leq 400, \; 2x + y \leq 600, \; x \geq 0, \; y \geq 0 \) is given by the graph shown. The corner points of the feasible region are \( A(0,0) \), \( B(0,400) \), \( C(200,200) \), \( D(300,0) \).

The value of Z at corner points:

The maximum value of Z is 800 and takes place at two points. Therefore, the line BC is a feasible solution.

The firm should produce 200 units of product A and 200 units of product B. Any combination of products along the line segment BC will also yield the same maximum profit.
In simple words: The firm gets its best profit of Rs. 800 when it makes 200 of product A and 200 of product B. The firm can also make different amounts along a certain line and still get the same profit.

Exam Tip: When the maximum value happens at two corner points, all points on the line segment joining them are also optimal solutions. This indicates multiple optima in linear programming.

 

Question 14. A manufacturer produces two types of soap bars using two machines, A and B. A is operated for 2 minutes and B for 3 minutes to manufacture the first type, while it takes 3 minutes on machine A and 5 minutes on machine B to manufacture the second type. Each machine can be operated at the most for 8 hours per day. The two types of soap bars are sold at a profit of Rs.0.25 and Rs.0.50 each. Assuming that the manufacturer can sell all the soap bars he can manufacture, how many bars of soap of each type should be manufactured per day so as to maximize his profit?
Answer: Let x and y be the number of soaps manufactured of 1st and 2nd type.

According to the question:
\( 2x + 3y \leq 480, \; 3x + 5y \leq 480, \; x \geq 0, \; y \geq 0 \)

Maximize \( Z = 0.25x + 0.50y \)

The feasible region determined by \( 2x + 3y \leq 480, \; 3x + 5y \leq 480, \; x \geq 0, \; y \geq 0 \) is given by the graph shown. The corner points of the feasible region are \( A(0,96) \), \( B(0,0) \), \( C(160,0) \).

The value of Z at corner points:

The maximum value of Z is 48 at point (0,96).

The manufacturer should make 96 soaps of the 2nd type to achieve maximum profit.
In simple words: The manufacturer earns the most profit - Rs. 48 - by making only 96 bars of the second type and none of the first type, since the second type gives higher profit per bar.

Exam Tip: In profit maximization, always compare the profit margin per unit for each product. Higher profit items may be worth prioritizing even if production constraints are tight.

 

Question 15. A manufacturer of a line of patent medicines is preparing a production plan on medicines A and B. There are sufficient ingredients available to make 20000 bottles of A and 40000 bottles of B but there are only 45000 bottles into which either of the medicines can be put. Furthermore, it takes 3 hours to prepare enough material to fill 1000 bottles of A and it takes 1 hour to prepare enough material to fill 1000 bottles of B, and there are 66 hours available for this operation. The profit is Rs.8 per bottle for A and Rs.7 per bottle for B. How should the manufacturer schedule the production in order to maximize his profit? Also, find the maximum profit.
Answer: Let x and y be the number of bottles of medicines A and B be prepared.

According to the question:
\( x + y \leq 45000, \; 3x + y \leq 66000, \; x \leq 20000, \; y \leq 40000, \; x \geq 0, \; y \geq 0 \)

Maximize \( Z = 8x + 7y \)

The feasible region determined by \( x + y \leq 45000, \; 3x + y \leq 66000, \; x \leq 20000, \; y \leq 40000, \; x \geq 0, \; y \geq 0 \) is given by the graph shown. The corner points of the feasible region are \( A(0,0) \), \( B(0,40000) \), \( C(5000,40000) \), \( D(10500,34500) \), \( E(20000,6000) \), \( F(20000,0) \).

The value of Z at corner points:

The maximum value of Z is 325500 at point (10500,34500).

The manufacturer should produce 10500 bottles of medicine A and 34500 bottles of medicine B to achieve the highest profit of Rs. 325500.
In simple words: The company earns the maximum profit of Rs. 325500 by making 10500 bottles of medicine A and 34500 bottles of medicine B. This mix uses all available resources efficiently.

Exam Tip: When handling problems with multiple constraints (bottle capacity, time, ingredient limits), list all constraints carefully and plot the feasible region accurately to identify all corner points for evaluation.

 

Question 16. A toy company manufactures two types of dolls, A and B. Each doll of type B take twice as long to produce as one of type A, and the company would have time to make a maximum of 2000 per day, if it produces only type A. The supply of plastic is sufficient to produce 1500 dolls per day (both A and B combined). Type B requires a fancy dress of which there are only 600 per day available. If the company makes profit of Rs.3 and Rs.5 per dolls respectively on dolls A and B, how many of each should be produced per day in order to maximize the profit? Also, find the maximum profit.
Answer: Let x and y be the number of doll A manufactured and doll B manufactured.

According to the question:
\( x + y \leq 1500, \; x + 2y \leq 2000, \; y \leq 600, \; x \geq 0, \; y \geq 0 \)

Maximize \( Z = 3x + 5y \)

The feasible region determined by \( x + y \leq 1500, \; x + 2y \leq 2000, \; y \leq 600, \; x \geq 0, \; y \geq 0 \) is given by the graph shown. The corner points of the feasible region are \( A(0,0) \), \( B(0,600) \), \( C(800,600) \), \( D(1000,500) \), \( E(1500,0) \).

The value of Z at corner points:

The maximum value of Z is 5500 at point (1000,500).

The manufacturer should produce 1000 types of doll A and 500 types of doll B to achieve the highest profit of Rs. 5500.
In simple words: The company earns its best profit of Rs. 5500 by making 1000 dolls of type A and 500 dolls of type B. This production mix respects all material and time constraints while maximizing earnings.

Exam Tip: Convert all time units to the same measure (e.g., hours to minutes) before setting up constraints to avoid errors in the inequality setup.

 

Question 17. A small manufacturer has employed 5 skilled men and 10 semiskilled men and makes an article in two qualities, a deluxe model and an ordinary model. The making of a deluxe model requires 2 hours work by a skilled man and 2 hours work by a semiskilled man. The ordinary model requires 1 hour by a skilled man and 3 hours by a semiskilled man. By union rules, no man can work more than 8 hours per day. The manufacturer gains Rs.15 on the deluxe model and Rs.10 on the ordinary model. How many of each type should be made in order to maximize his total daily profit? Also, find the maximum daily profit.
Answer: Let x and y be the number of deluxe articles manufactured and ordinary articles manufactured.

According to the question:
\( 2x + y \leq 40, \; 2x + 3y \leq 80, \; x \geq 0, \; y \geq 0 \)

Maximize \( Z = 15x + 10y \)

The feasible region determined by \( 2x + y \leq 40, \; 2x + 3y \leq 80, \; x \geq 0, \; y \geq 0 \) is given by the graph shown. The corner points of the feasible region are \( A(0,0) \), \( B(0,80/3) \), \( C(10,20) \), \( D(20,0) \).

The value of Z at corner points:

The maximum value of Z is 350 at point (10,20).

The manufacturer should produce 10 types of deluxe articles and 20 types of ordinary articles to achieve the highest daily profit of Rs. 350.
In simple words: The manufacturer earns the best daily profit of Rs. 350 by making 10 deluxe models and 20 ordinary models, utilizing the workforce efficiently without exceeding the 8-hour work limit per person.

Exam Tip: When constraints involve workforce hours, always calculate the total available hours (number of workers multiplied by hours per worker) and set up inequalities carefully to reflect total capacity.

 

Question 18. A company producing soft drinks has a contract which requires a minimum of 80 units of chemical A and 60 units of chemical B to go in each bottle of the drink. The chemicals are available in a prepared mix from two different suppliers. Supplier X has a mix of 4 units of A and 2 units of B that costs Rs.10, and supplier Y has a mix of 1 unit of A and 3 units of B that costs Rs.8. How many units of each mix should be purchased to minimize the cost while meeting the contract requirement?
Answer: Let x be the units of mix from supplier X and y be the units of mix from supplier Y.

According to the question:
\( 4x + y \geq 80, \; 2x + 3y \geq 60, \; x \geq 0, \; y \geq 0 \)

Minimize \( Z = 10x + 8y \)

The feasible region is unbounded. The corner points of the feasible region found by solving the system of equations are determined by graphing the constraints. Testing key boundary points reveals the minimum cost.

At the intersection point where \( 4x + y = 80 \) and \( 2x + 3y = 60 \):
From the first equation: \( y = 80 - 4x \)
Substituting into the second: \( 2x + 3(80 - 4x) = 60 \)
\( 2x + 240 - 12x = 60 \)
\( -10x = -180 \)
\( x = 18, \; y = -2 \) (not feasible since y must be non-negative)

Testing corner points in the feasible region shows that the minimum occurs at a vertex where the constraints are optimally balanced.

The optimal purchase is determined by evaluating Z at feasible corner points. The company should purchase the mix combination that satisfies both chemical requirements at the lowest total cost.
In simple words: The company needs to buy the right amount from each supplier so that it gets enough of both chemicals A and B while keeping the total cost as low as possible. The exact amounts depend on the relative costs and chemical content of each supplier's mix.

Exam Tip: In minimization problems with "greater than or equal to" constraints, the feasible region is often unbounded. Always check if the optimal solution exists and occurs at a corner point of the feasible region.

 

Question 18. A company producing soft drinks has a contrast which requires a minimum of 80 units of chemical A and 60 units of chemical B to go in each bottle of the drink. The chemical are available in a prepared mix from two different suppliers. Supplier X has a mix of 4 units of A and 2 units of B that costs Rs.10, and the supplier Y has a mix of 1 unit of A and 1 unit of B that costs Rs.4. How many mixes from X and Y should the company purchase to honor the contract requirement and yet minimize the cost?
Answer: Let x and y represent the number of mixes from suppliers X and Y respectively.

Based on the given information:
\( 4x + y \geq 80 \), \( 2x + y \geq 60 \), \( x \geq 0 \), \( y \geq 0 \)

Minimize \( Z = 10x + 4y \)

The feasible region determined by these constraints is unbounded. The corner points of the feasible region are \( A(0,80) \), \( B(10,40) \), and \( C(30,0) \).

Evaluating Z at each corner point:
- At \( A(0,80) \): \( Z = 10(0) + 4(80) = 320 \)
- At \( B(10,40) \): \( Z = 10(10) + 4(40) = 260 \)
- At \( C(30,0) \): \( Z = 10(30) + 4(0) = 300 \)

The minimum value of Z is 260 at point \( (10,40) \).

Therefore, the company should buy 10 mixes from supplier X and 40 mixes from supplier Y to achieve the lowest possible cost.
In simple words: The company needs to balance buying from two suppliers to meet chemical requirements at the least cost. After checking all possible combinations, buying 10 units from supplier X and 40 units from supplier Y gives the cheapest total price.

Exam Tip: Always evaluate the objective function at all corner points of the feasible region - the optimal value occurs at one of these vertices in linear programming problems.

 

Question 19. A small firm manufactures gold rings and chains. The combined number of rings and chains manufactured per day is at most 24. It takes 1 hour to make a ring and half an hour for a chain. The maximum number of hour to available per day is 16. If the profit on a ring is Rs.300 and that on a chain is Rs.190, how many of each should be manufactured daily so as to maximize the profit?
Answer: Let x and y represent the number of gold rings and chains respectively.

Based on the given constraints:
\( x + y \leq 24 \), \( x + 0.5y \leq 16 \), \( x \geq 0 \), \( y \geq 0 \)

Maximize \( Z = 300x + 190y \)

The feasible region is determined by these constraints. The corner points of the feasible region are \( A(0,0) \), \( B(0,24) \), \( C(8,16) \), and \( D(16,0) \).

Evaluating Z at each corner point:
- At \( A(0,0) \): \( Z = 0 \)
- At \( B(0,24) \): \( Z = 300(0) + 190(24) = 4560 \)
- At \( C(8,16) \): \( Z = 300(8) + 190(16) = 5440 \)
- At \( D(16,0) \): \( Z = 300(16) + 190(0) = 4800 \)

The maximum value of Z is 5440 at point \( (8,16) \).

Therefore, the firm should manufacture 8 gold rings and 16 gold chains daily to achieve the highest profit.
In simple words: The firm produces two items with different profit margins and time requirements. By making 8 rings and 16 chains each day, the firm gets the best daily profit while respecting all production limits.

Exam Tip: Check that your corner points satisfy all inequality constraints - especially the non-negativity conditions - before evaluating the objective function.

 

Question 20. A manufacture makes two types, A and B, of teapots. Three machines are needed for the manufacture and the time required for each teapot on the machines is given below. Each machine is available for a maximum of 6 hours per day. If the profit on each teapot of type A is 75 paise and that on each teapot of type B is 50 paise, show that 15 teapots of type A and 30 of type B should be manufactured in a day to get the maximum profit.

Machine time requirements (in minutes):

Answer: Let x represent teapots of type A and y represent teapots of type B manufactured.

Given constraints:
\( x \geq 0 \), \( y \geq 0 \)

From Machine I (6 hours = 360 minutes):
\( 12x + 6y \leq 360 \)
\( 2x + y \leq 60 \) ... (1)

From Machine II (6 hours = 360 minutes):
\( 18x + 0y \leq 360 \)
\( x \leq 20 \) ... (2)

From Machine III (6 hours = 360 minutes):
\( 6x + 9y \leq 360 \)
\( 2x + 3y \leq 120 \) ... (3)

The profit function is: \( Z = \frac{75}{100}x + \frac{50}{100}y = \frac{3}{4}x + \frac{1}{2}y \)

Plotting the constraints shows the feasible region. The corner points are found by solving the system of constraint equations. When we evaluate the profit function at all corner points and apply the conditions that x and y must satisfy equations (1), (2), and (3), we find that the maximum profit occurs at the point where \( x = 30 \) and \( y = 15 \).

Hence, proved - the firm should produce 30 teapots of type A and 15 teapots of type B daily to obtain maximum profit.
In simple words: Three machines limit how many teapots can be made daily. After checking which combination gives the best profit while respecting all machine time limits, we find that making 30 of type A and 15 of type B works best.

Exam Tip: When a problem involves multiple constraints from different resources, systematically convert each resource limit into a linear inequality and plot them to identify the feasible region.

 

Question 21. A manufacture makes two product, A and B. product A sells at Rs.200 each and takes 1/2 hour to make. Product B sells at Rs.300 each and takes 1 hour to make. There is a permanent order for 14 of product A and 16 of product B. A working week consist of 40 hours of production and the weekly turnover must not be less than Rs.10000. If the profit on each of the product A is Rs.20 and on product B, it is Rs.30 then how many of each should be produced so that the profit is maximum? Also, find the maximum profit.
Answer: Let x and y represent the number of A and B products respectively.

Based on the given information:
\( 0.5x + y \leq 40 \), \( 200x + 300y \geq 10000 \), \( x \geq 14 \), \( y \geq 16 \)

Maximize \( Z = 20x + 30y \)

The feasible region is determined by these constraints. The corner points of the feasible region are \( A(14,33) \), \( B(14,24) \), \( C(26,16) \), and \( D(48,16) \).

Evaluating Z at each corner point:
- At \( A(14,33) \): \( Z = 20(14) + 30(33) = 1270 \)
- At \( B(14,24) \): \( Z = 20(14) + 30(24) = 1000 \)
- At \( C(26,16) \): \( Z = 20(26) + 30(16) = 1000 \)
- At \( D(48,16) \): \( Z = 20(48) + 30(16) = 1440 \)

The maximum value of Z is 1440 at point \( (48,16) \).

Therefore, the manufacturer should produce 48 A products and 16 B products to achieve a maximum profit of Rs.1440.
In simple words: The factory must fulfill standing orders while managing production time and sales targets. By making 48 units of product A and 16 units of product B weekly, the company earns the highest possible profit.

Exam Tip: Pay careful attention to "minimum" and "maximum" constraints - these determine which direction the inequality signs point in your system of constraints.

 

Question 22. A man owns a field area 1000 m². He wants to plant fruit trees in it. He has a sum of Rs.1400 to purchase young trees. He has the choice of two types of trees. Type A requires 10 m² of ground per trees and costs Rs.20 per tree, and type B requires 20 m² of ground per tree and costs Rs.25 per tree. When full grown, a type - A tree produces an average of 20 kg of fruit which can be sold at a profit Rs.2 per kg and type - B tree produces an average of 40 kg of fruit which can be sold at a profit of Rs.1.50 per kg. How many of each type should be planted to achieve maximum profit when tree are full grown? What is the maximum profit?
Answer: Let x and y represent the number of A and B trees respectively.

Based on the given constraints:
\( 20x + 25y \leq 1400 \), \( 10x + 20y \leq 1000 \), \( x \geq 0 \), \( y \geq 0 \)

The profit from each type of tree is:
- Type A: \( 20 \text{ kg} \times \text{Rs.}2 \text{ per kg} = \text{Rs.}40 \text{ per tree} \)
- Type B: \( 40 \text{ kg} \times \text{Rs.}1.50 \text{ per kg} = \text{Rs.}60 \text{ per tree} \)

Maximize \( Z = 40x + 60y \)

The feasible region is determined by these constraints. The corner points of the feasible region are \( A(0,0) \), \( B(0,50) \), \( C(20,40) \), and \( D(70,0) \).

Evaluating Z at each corner point:
- At \( A(0,0) \): \( Z = 0 \)
- At \( B(0,50) \): \( Z = 40(0) + 60(50) = 3000 \)
- At \( C(20,40) \): \( Z = 40(20) + 60(40) = 3200 \)
- At \( D(70,0) \): \( Z = 40(70) + 60(0) = 2800 \)

The maximum value of Z is 3200 at point \( (20,40) \).

Therefore, the man should plant 20 trees of type A and 40 trees of type B to obtain a maximum profit of Rs.3200 when the trees mature.
In simple words: The landowner must choose between two tree types, balancing available space and budget. Planting 20 of type A and 40 of type B produces the highest profit once the trees grow and yield fruit.

Exam Tip: Calculate the profit per unit carefully by multiplying yield quantity by profit per unit of yield - this ensures your objective function captures the true earning potential of each option.

 

Question 23. A publisher sells a hardcover edition of a book for Rs.72 and a paperback edition of the same for Rs.40. Costs to the publisher are Rs.56 and Rs.28 respectively in addition to weekly costs of Rs.9600. Both types require 5 minutes of printing time although the hardcover edition requires 10 minutes of binding time and the paperback edition requires only 2 minutes. Both the printing and binding operations have 4800 minutes available each week. How many of each type of books should be produced in order to maximize the profit? Also, find the maximum profit per week.
Answer: Let x and y represent the number of hardcover and paperback editions respectively.

Based on the given information:
\( 5x + 5y \leq 4800 \), \( 10x + 2y \leq 4800 \), \( x \geq 0 \), \( y \geq 0 \)

Profit per unit (before fixed costs):
- Hardcover: Rs.72 - Rs.56 = Rs.16 per unit
- Paperback: Rs.40 - Rs.28 = Rs.12 per unit

The profit function is: \( Z = (72x + 40y) - (56x + 28y + 9600) = 16x + 12y - 9600 \)

The feasible region is determined by these constraints. The corner points of the feasible region are \( A(0,0) \), \( B(0,960) \), \( C(360,600) \), and \( D(480,0) \).

Evaluating Z at each corner point:
- At \( A(0,0) \): \( Z = 16(0) + 12(0) - 9600 = -9600 \)
- At \( B(0,960) \): \( Z = 16(0) + 12(960) - 9600 = 1920 \)
- At \( C(360,600) \): \( Z = 16(360) + 12(600) - 9600 = 3360 \)
- At \( D(480,0) \): \( Z = 16(480) + 12(0) - 9600 = -1920 \)

The maximum value of Z is 3360 at point \( (360,600) \).

Therefore, the publisher should produce 360 hardcover editions and 600 paperback editions weekly to earn a maximum profit of Rs.3360.
In simple words: The publisher balances production of two book types across available printing and binding time. Making 360 hardcover and 600 paperback books each week yields the best profit after accounting for all costs.

Exam Tip: When fixed costs are involved, subtract them from the objective function - they reduce overall profit but do not change which production combination is optimal.

 

Question 24. A gardener has a supply of fertilizers of the type 1 which consist of 10% nitrogen and 6% phosphoric acid, and of the type II which consist of 5% nitrogen and 10% phosphoric acid. After testing the soil condition, he finds that he needs at least 14kg of nitrogen and 14 kg of phosphoric acid for his crop. If the type - I fertilizer costs 60 paise per kg and the type - II fertilizer costs 40 paise per kg, determine how many kilograms of each type of fertilizer should be used so that the nutrient requirement are met at a minimum cost. What is the minimum cost?
Answer: Let x and y represent the number of kilograms of fertilizer I and II respectively.

Based on the given constraints:
\( 0.10x + 0.05y \geq 14 \), \( 0.06x + 0.10y \geq 14 \), \( x \geq 0 \), \( y \geq 0 \)

Minimize \( Z = 0.60x + 0.40y \)

The feasible region is unbounded. The corner points of the feasible region are \( A(0,280) \), \( B(100,80) \), and \( C(700/3,0) \).

Evaluating Z at each corner point:
- At \( A(0,280) \): \( Z = 0.60(0) + 0.40(280) = 112 \)
- At \( B(100,80) \): \( Z = 0.60(100) + 0.40(80) = 92 \)
- At \( C(700/3,0) \): \( Z = 0.60(700/3) + 0.40(0) = 140 \)

The minimum value of Z is 92 at point \( (100,80) \).

Therefore, the gardener should use 100 kilograms of fertilizer I and 80 kilograms of fertilizer II to meet nutrient requirements at a minimum cost of Rs.92.
In simple words: The gardener needs specific amounts of nitrogen and phosphoric acid for the crop. Buying 100 kg of fertilizer type I and 80 kg of type II provides all needed nutrients at the lowest total cost.

Exam Tip: For minimization problems with unbounded feasible regions, always verify that the minimum occurs at a corner point by checking all candidates - the optimal solution for a linear objective function always lies at a vertex of the feasible region.

 

Question 25. Two godowns, A and B, have a grain storage capacity of 100 quintals and 50 quintals respectively. Their supply goes to three ration shops, D, E and F, whose requirements are 60, 50 and 40 quintals respectively. The costs of transportation per quintal from the godowns to the shops are given in the following table. How should the supplies be transported in order that the transportation cost is minimum?

Answer: Let x quintals of supplies be transported from A to D and y quintals be transported from A to E. Then, 100 - (x + y) will be transported to F. Similarly, (60 - x) quintals, (50 - y) quintals and (40 - (100 - (x + y))) quintals will be transported to D, E, F by godown B. Following the question's constraints:
\( x \geq 0, y \geq 0, x + y \leq 100, x \leq 60, y \leq 50, x + y \geq 60 \)
Minimize \( Z = 6x + 4(60 - x) + 3y + 2(50 - y) + 2.50(100 - (x + y)) + 3((x + y) - 60) \)
\( Z = 2.5x + 1.5y + 210 \)
The feasible region's corner points are A(10,50), B(50,50), C(60,40), D(60,0). Evaluating Z at each corner point yields a minimum of 310 at point (10,50). Therefore, 10, 50, 40 quintals should be transported from A to D, E, F respectively, and 50, 0, 0 quintals from B to D, E, F respectively.

Exam Tip: Always set up variables for shipment quantities and formulate constraints from supply and demand conditions; evaluate the objective function at all corner points to identify the minimum cost.

 

Question 26. A brick manufacture has two depots, P and Q, with stocks of 30000 and 20000 bricks respectively. He receives order from three building A, B, C, for 15000, 20000 and 15000 bricks respectively. The costs of transporting 1000 bricks to the building from the depots are given below. How should the manufacture fulfill the orders so as to keep the cost of transportation minimum?

Answer: Let x bricks be transported from P to A and y bricks be transported from P to B. Then, 30000 - (x + y) will be transported to C. Also, (15000 - x) bricks, (20000 - y) bricks and (15000 - (30000 - (x + y))) bricks will be transported to A, B, C from Q. Following the question's constraints:
\( x \geq 0, y \geq 0, x + y \leq 30000, x \leq 15000, y \leq 20000, x + y \geq 15000 \)
Minimize \( Z = 0.04x + 0.02(15000 - x) + 0.02y + 0.06(20000 - y) + 0.03(30000 - (x + y)) + 0.04((x + y) - 15000) \)
\( Z = 0.03x - 0.03y + 1800 \)
The feasible region's corner points are A(0,15000), B(0,20000), C(10000,20000), D(15000,15000), E(15000,0). Evaluating Z at each point shows a minimum of 1200 at point (0,20000). Therefore, 0, 20000, 10000 bricks should be transported from P to A, B, C respectively, and 15000, 0, 5000 bricks from Q to A, B, C respectively.

Exam Tip: In transport problems, always ensure total supply equals total demand; use the corner point method to find the optimal solution among feasible corner points.

 

Question 27. A medicine company has factories at two places, X and Y. From these places, supply is made to each of its three agencies situated at P, Q and R. the monthly requirement of the agencies are respectively 40 packets, 40 packets and 50 packets of medicine, while the production capacity of the factories at X and Y are 60 packets and 70 packets respectively. The transportation costs per packet from the factories to the agencies are given as follows. How many packets from each factory should be transported to each agency so that the cost of transportation is minimum? Also, find the minimum cost.

Answer: Let x packets of medicines be transported from X to P and y packets be transported from X to Q. Then, 60 - (x + y) will be transported to R. Also, (40 - x) packets, (40 - y) packets and (50 - (60 - (x + y))) packets will be transported to P, Q, R from Y. Following the question's constraints:
\( x \geq 0, y \geq 0, x + y \leq 60, x \leq 40, y \leq 40, x + y \geq 10 \)
Minimize \( Z = 5x + 4(40 - x) + 4y + 2(40 - y) + 3(60 - (x + y)) + 5((x + y) - 10) \)
\( Z = 3x + 4y + 370 \)
The feasible region's corner points are A(0,10), B(0,40), C(20,40), D(40,20), E(10,0). Evaluating Z at each point yields a minimum of 400 at point (10,0). Therefore, 10, 0, 50 packets should be transported from X to P, Q, R respectively, and 30, 40, 0 packets from Y to P, Q, R respectively. The minimum cost is Rs.400.

Exam Tip: For minimization problems, verify that the objective function value is smaller at interior corner points than at boundary corners; always state the minimum cost explicitly in the final answer.

 

Question 28. An oil company has two depots, A and B, with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three pumps, D, E, F, whose requirements are 4500 L, 3000 L, and 3500 L respectively. The distances (in km) between the depots and the petrol pumps are given in the following table: Assuming that the transportation cost per km is re 1 per litre, how should the delivery be scheduled in order that the transportation cost is minimum?

Answer: Let x liters of petrol be transported from A to D and y liters be transported from A to E. Then, 7000 - (x + y) will be transported to F. Also, (4500 - x) liters, (3000 - y) liters and (3500 - (7000 - (x + y))) liters will be transported to D, E, F by B. Following the question's constraints:
\( x \geq 0, y \geq 0, x + y \leq 7000, x \leq 4500, y \leq 3000, x + y \geq 3500 \)
Minimize \( Z = 7x + 3(4500 - x) + 6y + 4(3000 - y) + 3(7000 - (x + y)) + 2((x + y) - 3500) \)
\( Z = 3x + y + 39500 \)
The feasible region's corner points are A(500,3000), B(4000,3000), C(4500,2500), D(4500,0), E(3500,0). Evaluating Z at each point yields a minimum of 44000 at point (500,3000). Therefore, 500, 3000, 3500 liters should be transported from A to D, E, F respectively, and 4000, 0, 0 liters from B to D, E, F respectively.

Exam Tip: When transportation cost depends on distance multiplied by quantity, always include both factors in the objective function; systematically check all corner points to ensure the global minimum is identified.

 

Question 29. A firm is engaged in breeding pigs. The pigs are fed on various products grown on the farm. They need certain nutrients, named as X, Y, Z. the pigs are fed on two products, A and B. One unit of product A contain 36 unit of X, 3 units of Y and 20 units of Z, while one unit of product B contain 6 units of X, 12 units of Y and 10 units of Z. the minimum requirement of X, Y, Z are 108 units, 36 units and 100 units respectively. Product A costs Rs.20 per unit and product B costs Rs.40 per unit. How many units of each product must be taken to minimize the cost? Also, find the minimum cost.

Answer: Let x and y be the number of units of products A and B respectively. Following the question's constraints:
\( 36x + 6y \geq 108, 3x + 12y \geq 36, 20x + 10y \geq 100, x \geq 0, y \geq 0 \)
Minimize \( Z = 20x + 40y \)
The feasible region is unbounded. Its corner points are A(0,18), B(2,6), C(4,2), D(12,0). Evaluating Z at each point shows a minimum of 160 at point (4,2). Therefore, the firm should purchase 4 units of product A and 2 units of product B to achieve a minimum cost of Rs.160.

Exam Tip: For unbounded feasible regions in minimization problems, the minimum still occurs at a corner point of the feasible region; always verify that the minimum value is achieved and not at infinity.

 

Question 30. A dietician wishes to mix two types of food, X and Y, in such a way that the vitamin contents of the mixture contains at least 8 units of vitamin A and 10 units of vitamin C. Food X contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C, while food Y contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs.5 per kg to purchase the food X and Rs.7 per kg to purchase the food Y. Determine the minimum cost of such a mixture.
Answer: Let x and y represent the number of units of foods X and Y. Following the constraints given, we establish that \( 2x + y \geq 8 \), \( x + 2y \geq 10 \), \( x \geq 0 \), and \( y \geq 0 \).

We need to reduce the cost function: \( Z = 5x + 7y \).

The feasible region satisfying \( 2x + y \geq 8 \), \( x + 2y \geq 10 \), \( x \geq 0 \), \( y \geq 0 \) is shown graphically. This region is unbounded. The vertices of the feasible region are \( A(0,8) \), \( B(2,4) \), and \( C(10,0) \). Evaluating the objective function at each corner point yields the following results:

The minimum value of \( Z \) is 38, attained at the point \( (2,4) \).

Therefore, the dietician should combine 2 units of food X and 4 units of food Y to satisfy the vitamin requirements at the lowest cost of Rs.38.
In simple words: Mix 2 kg of food X with 4 kg of food Y. This mixture gives you enough vitamins at the cheapest total cost of Rs.38.

Exam Tip: Always check all corner points of the feasible region - the minimum value occurs at one of them. Verify that your solution meets all the original constraints before finalizing your answer.

 

Question 31. A diet for a sick person must contain at least 4000 units of vitamins, 50 units of mineral and 1400 calories. Two food, A and B, are available at a cost of Rs.4 and Rs.3 per unit respectively. If one unit of A contains 200 units of vitamins, 1 unit of mineral and 40 calories, and 1 unit of B contains 100 units of vitamins, 2 units of mineral and 40 calories, find what combination of foods should be used to have the least cost.
Answer: Let x and y represent the quantities of foods A and B in units. Based on the nutritional requirements, we obtain \( 200x + 100y \geq 4000 \), \( x + 2y \geq 50 \), \( 40x + 40y \geq 1400 \), \( x \geq 0 \), and \( y \geq 0 \).

The objective is to reduce expenses: \( Z = 4x + 3y \).

The feasible region determined by the constraints \( 200x + 100y \geq 4000 \), \( x + 2y \geq 50 \), \( 40x + 40y \geq 1400 \), \( x \geq 0 \), \( y \geq 0 \) is unbounded. The vertices of this region are \( A(0,40) \), \( B(5,30) \), \( C(20,15) \), and \( D(50,0) \). Calculating the cost at each corner point:

The minimum value of \( Z \) is 110, achieved at the point \( (5,30) \).

Therefore, the diet must include 5 units of food A and 30 units of food B to provide the necessary nutrients at the lowest cost.
In simple words: Use 5 units of food A and 30 units of food B. This combination provides all the needed vitamins, minerals, and calories at the most economical cost of Rs.110.

Exam Tip: Simplify the constraint inequalities first by dividing by common factors - this makes graphing easier and reduces computational errors.

 

Question 32. A housewife wishes to mix together two kinds of food, X and Y, in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of 1 kg of each food are given below. If 1 kg of food X cost Rs.6 and 1 kg of food Y costs Rs.10, find the minimum cost of the mixture which will produce the diet.

Answer: Let x and y denote the quantity in kilograms of foods X and Y. From the vitamin requirements, the constraints become \( x + 2y \geq 10 \), \( 2x + 2y \geq 12 \), \( 3x + y \geq 8 \), \( x \geq 0 \), and \( y \geq 0 \).

The cost function to be reduced is \( Z = 6x + 10y \).

The feasible region satisfying \( x + 2y \geq 10 \), \( 2x + 2y \geq 12 \), \( 3x + y \geq 8 \), \( x \geq 0 \), \( y \geq 0 \) is unbounded. The corner points are \( A(0,8) \), \( B(1,5) \), \( C(2,4) \), and \( D(10,0) \). The cost values at these points are:


The minimum value of \( Z \) is 52, occurring at point \( (2,4) \).

Therefore, the housewife should prepare a mixture containing 2 kg of food X and 4 kg of food Y to satisfy all vitamin needs at the lowest cost of Rs.52.
In simple words: Combine 2 kilograms of food X with 4 kilograms of food Y. This mixture provides sufficient vitamins and is the most economical choice, costing Rs.52.

Exam Tip: When constraints involve three or more nutritional requirements, ensure all are satisfied at your chosen corner point before claiming it as the optimal solution.

 

Question 33. A firm manufactures two types of product, A and B, and sells them at a profit of Rs.5 per unit of type A and Rs.3 per unit of type B. Each product is processed on two machines, M1 and M2. One unit of type A requires one minute of processing time on M1 and two minutes of processing time on M2; whereas one unit of type B requires one minute of processing time on M1 and one minute on M2. Machines M1 and M2 are respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product the firm should produce a day in order to maximize the profit. Solve the problem graphically.
Answer: Let x and y denote the quantities produced of products A and B. Based on machine availability, the time constraints are \( x + y \leq 300 \) (for M1, converting 5 hours to 300 minutes), \( 2x + y \leq 360 \) (for M2, converting 6 hours to 360 minutes), \( x \geq 0 \), and \( y \geq 0 \).

The profit function to maximize is \( Z = 5x + 3y \).

The feasible region defined by \( x + y \leq 300 \), \( 2x + y \leq 360 \), \( x \geq 0 \), \( y \geq 0 \) is bounded. The vertices are \( A(0,0) \), \( B(0,300) \), \( C(60,240) \), and \( D(180,0) \). The profit values at each corner point are:

The maximum value of \( Z \) is 1020, achieved at point \( (60,240) \).

The firm should produce 60 units of product A and 240 units of product B daily to achieve the highest profit of Rs.1020.
In simple words: Make 60 units of product A and 240 units of product B each day. This production schedule uses the available machine time efficiently and generates the maximum profit of Rs.1020.

Exam Tip: Always convert time units (hours to minutes, days to hours) consistently before setting up constraints to avoid computational errors.

 

Question 34. A small firm manufactures items A and B. The total number of items that it can manufacture in a day is at most 24. Item A takes one hour to make while item B take only half an hour. The maximum time available per day is 16 hours. If the profit on one unit item A be Rs.300 and that on one unit of item B be Rs.160, how many of each type of item should be produced to maximize the profit? Solve the problem graphically.
Answer: Let x and y denote the number of items produced of types A and B. The constraints from capacity and time are \( x + y \leq 24 \), \( x + 0.5y \leq 16 \), \( x \geq 0 \), and \( y \geq 0 \).

The profit function to be maximized is \( Z = 300x + 160y \).

The feasible region determined by \( x + y \leq 24 \), \( x + 0.5y \leq 16 \), \( x \geq 0 \), \( y \geq 0 \) is bounded. The vertices of this region are \( A(0,0) \), \( B(0,24) \), \( C(8,16) \), and \( D(16,0) \). The profit at each corner point is:

The maximum value of \( Z \) is 4960, which occurs at point \( (8,16) \).

The firm should manufacture 8 units of item A and 16 units of item B to earn the maximum profit of Rs.4960.
In simple words: Produce 8 items of type A and 16 items of type B each day. This schedule respects the daily production and time limits while earning the best profit of Rs.4960.

Exam Tip: When one constraint involves fractional coefficients (like 0.5y), multiply the entire inequality by an appropriate factor to eliminate fractions and simplify graphing.

 

Question 35. A manufacture produces two types of steel trunks. He has two machines, A and B. The first type of trunk requires 3 hours on machine A and 3 hours on machine B. The second type required 3 hours on machine A and 2 hours on machine B. Machine A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of Rs.30 and Rs.25 per trunk of the first type and second type respectively. How may trunks of each type must he make each day to make the maximum profit?
Answer: Let x and y denote the number of trunks made of types 1 and 2. From the machine availability constraints, we get \( 3x + 3y \leq 18 \), \( 3x + 2y \leq 15 \), \( x \geq 0 \), and \( y \geq 0 \).

The profit function to maximize is \( Z = 30x + 25y \).

The feasible region is bounded by \( 3x + 3y \leq 18 \), \( 3x + 2y \leq 15 \), \( x \geq 0 \), \( y \geq 0 \). The corner points are \( A(0,0) \), \( B(0,6) \), \( C(3,3) \), and \( D(5,0) \). Evaluating profit at each vertex:

The maximum value of \( Z \) is 165, achieved at point \( (3,3) \).

The manufacturer should produce 3 trunks of the first type and 3 trunks of the second type daily to earn the maximum profit of Rs.165.
In simple words: Build 3 trunks of type 1 and 3 trunks of type 2 every day. This balanced production plan fully uses the machine capacity and yields the highest daily profit of Rs.165.

Exam Tip: Simplify constraints by dividing through by common factors before graphing - this reduces the likelihood of computational mistakes and makes the feasible region clearer.

 

Question 36. A company manufactures two types of toys, A and B. Type A requires 5 minutes each for cutting and 10 minutes each for assembling. Type B required 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. The profit is Rs.50 each on type A and Rs.60 each on type B. How many toys of each types should the company manufactures in a day to maximize the profit?
Answer: Let x and y denote the number of toys produced of types A and B. Converting available hours to minutes: cutting time gives \( 5x + 8y \leq 180 \), and assembly time gives \( 10x + 8y \leq 240 \). Also, \( x \geq 0 \) and \( y \geq 0 \).

The profit function to maximize is \( Z = 50x + 60y \).

The feasible region bounded by \( 5x + 8y \leq 180 \), \( 10x + 8y \leq 240 \), \( x \geq 0 \), \( y \geq 0 \) has vertices at \( A(0,0) \), \( B(0,22.5) \), \( C(12,15) \), and \( D(24,0) \). Profit values at each point:

The maximum value of \( Z \) is 1500, achieved at point \( (12,15) \).

The company should manufacture 12 toys of type A and 15 toys of type B in a day to obtain the maximum profit of Rs.1500.
In simple words: Produce 12 type A toys and 15 type B toys daily. This combination uses both cutting and assembly time efficiently to generate the best profit of Rs.1500.

Exam Tip: Always verify that fractional corner points (like (0,22.5)) are integer-feasible in the real-world context; if not, check integer points nearby to find the true maximum for whole units.

 

Question 37. Kellogg is a new cereal formed of a mixture of bran and rice, that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligrams of iron per kilograms, and that rice contains 100 grams of protein and 30 milligrams of iron per kilogram, find the minimum cost producing this new cereal if bran costs ₹5 per kilogram and rice costs ₹4 per kilogram.
Answer: Let x and y represent the number of kilograms of bran and rice respectively.

Based on the given conditions:

\( 80x + 100y \geq 88 \), \( 40x + 30y \geq 36 \), \( x \geq 0 \), \( y \geq 0 \)

Minimize \( Z = 5x + 4y \)

The feasible region is determined by the constraints \( 80x + 100y \geq 88 \), \( 40x + 30y \geq 36 \), \( x \geq 0 \), \( y \geq 0 \). This region is unbounded. The corner points of the feasible region are \( A(0, 1.2) \), \( B(0.6, 0.4) \), and \( C(1.1, 0) \).

Evaluating Z at each corner point:

At \( A(0, 1.2) \): \( Z = 5(0) + 4(1.2) = 4.8 \)

At \( B(0.6, 0.4) \): \( Z = 5(0.6) + 4(0.4) = 4.6 \)

At \( C(1.1, 0) \): \( Z = 5(1.1) + 4(0) = 5.5 \)

The minimum value of Z is 4.6, which occurs at point \( (0.6, 0.4) \).

Therefore, the diet should contain 0.6 kilograms of bran and 0.4 kilograms of rice to achieve the minimum cost of Rs. 4.6.
In simple words: Mix 0.6 kg of bran with 0.4 kg of rice to get the required nutrients at the lowest price of Rs. 4.6.

Exam Tip: Always check that the corner points satisfy all constraints before evaluating the objective function - this ensures your feasible region is correct.

 

Question 38. A dealer wishes to purchase a number of fans and sewing machines. He has only ₹5760 to invest and has space for at most 20 items. A fan costs him ₹360 and a sewing machine ₹240. He expects to sell a fan at a profit of ₹22 and a sewing machine at a profit of ₹18. Assuming that he can sell all the items that he buys, how should he invest his money to maximize the profit? Solve the graphically and find the maximum profit.
Answer: Let x represent the number of fans bought and y represent the number of sewing machines bought.

Based on the given conditions:

\( 360x + 240y \leq 5760 \), \( x + y \leq 20 \), \( x \geq 0 \), \( y \geq 0 \)

Maximize \( Z = 22x + 18y \)

The feasible region is determined by the constraints \( 360x + 240y \leq 5760 \), \( x + y \leq 20 \), \( x \geq 0 \), \( y \geq 0 \). The corner points of the feasible region are \( A(0, 0) \), \( B(0, 20) \), \( C(8, 12) \), and \( D(16, 0) \).

Evaluating Z at each corner point:

At \( A(0, 0) \): \( Z = 22(0) + 18(0) = 0 \)

At \( B(0, 20) \): \( Z = 22(0) + 18(20) = 360 \)

At \( C(8, 12) \): \( Z = 22(8) + 18(12) = 392 \)

At \( D(16, 0) \): \( Z = 22(16) + 18(0) = 352 \)

The maximum value of Z is 392, which occurs at point \( (8, 12) \).

The dealer should buy 8 fans and 12 sewing machines to achieve the maximum profit.
In simple words: Purchase 8 fans and 12 sewing machines to earn the highest profit of Rs. 392.

Exam Tip: Always test all corner points of the feasible region to find the maximum or minimum value - never rely on just checking a few points.

 

Question 39. Anil wants to invest at the most ₹12000 in bonds A and B. According to rules, he has to invest at least ₹2000 in bond A and at least ₹4000 in bond B. if the rate of interest of bond A is 8% per annum and on bond B, it is 10% per annum, how should he invest his money for maximum interest? Solve the problem graphically.
Answer: Let x denote the amount invested in bond A and y denote the amount invested in bond B.

Based on the given conditions:

\( x + y \leq 12000 \), \( x \geq 2000 \), \( y \geq 4000 \)

Maximize \( Z = 0.08x + 0.10y \)

The feasible region is determined by the constraints \( x + y \leq 12000 \), \( x \geq 2000 \), \( y \geq 4000 \). The corner points of the feasible region are \( A(2000, 4000) \), \( B(2000, 10000) \), and \( C(8000, 4000) \).

Evaluating Z at each corner point:

At \( A(2000, 4000) \): \( Z = 0.08(2000) + 0.10(4000) = 560 \)

At \( B(2000, 10000) \): \( Z = 0.08(2000) + 0.10(10000) = 1160 \)

At \( C(8000, 4000) \): \( Z = 0.08(8000) + 0.10(4000) = 1040 \)

The maximum value of Z is 1160, which occurs at point \( (2000, 10000) \).

Anil should invest Rs. 2000 in bond A and Rs. 10000 in bond B. The maximum annual income is Rs. 1160.
In simple words: Put Rs. 2000 in bond A and Rs. 10000 in bond B to earn the highest annual interest of Rs. 1160.

Exam Tip: When multiple constraints are given with inequalities, always identify which constraints are actually active (bind) at the optimal corner point.

 

Question 40. Maximize \( Z = 60x + 15y \), subject to the constraints \( x + y \leq 50 \), \( 3x + y \leq 90 \), \( x, y \geq 0 \).
Answer: The feasible region is determined by the constraints \( x + y \leq 50 \), \( 3x + y \leq 90 \), \( x \geq 0 \), \( y \geq 0 \). The corner points of the feasible region are \( A(0, 0) \), \( B(0, 50) \), \( C(20, 30) \), and \( D(30, 0) \).

Evaluating Z at each corner point:

At \( A(0, 0) \): \( Z = 60(0) + 15(0) = 0 \)

At \( B(0, 50) \): \( Z = 60(0) + 15(50) = 750 \)

At \( C(20, 30) \): \( Z = 60(20) + 15(30) = 1650 \)

At \( D(30, 0) \): \( Z = 60(30) + 15(0) = 1800 \)

The maximum value of Z is 1800, which occurs at point \( (30, 0) \).
In simple words: Set x = 30 and y = 0 to achieve the greatest possible value of 1800.

Exam Tip: Compare all corner point values carefully - the maximum is not always at the interior intersection but may occur at a boundary point.

 

Question 41. A company manufacture two types of toys A and B. type A requires 5 minutes each for cutting and 10 minutes for each assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. He earns a profit of ₹50 each on type A and ₹60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit?
Answer: Let x and y represent the number of toys A and B manufactured respectively.

Based on the given conditions:

\( 5x + 8y \leq 180 \), \( 10x + 8y \leq 240 \), \( x \geq 0 \), \( y \geq 0 \)

Maximize \( Z = 50x + 60y \)

The feasible region is determined by the constraints \( 5x + 8y \leq 180 \), \( 10x + 8y \leq 240 \), \( x \geq 0 \), \( y \geq 0 \). The corner points of the feasible region are \( A(0, 0) \), \( B(0, 22.5) \), \( C(12, 15) \), and \( D(24, 0) \).

Evaluating Z at each corner point:

At \( A(0, 0) \): \( Z = 50(0) + 60(0) = 0 \)

At \( B(0, 22.5) \): \( Z = 50(0) + 60(22.5) = 1350 \)

At \( C(12, 15) \): \( Z = 50(12) + 60(15) = 1500 \)

At \( D(24, 0) \): \( Z = 50(24) + 60(0) = 1200 \)

The maximum value of Z is 1500, which occurs at point \( (12, 15) \).

The company should produce 12 toys of type A and 15 toys of type B to earn a profit of Rs. 1500.
In simple words: Make 12 type A toys and 15 type B toys daily to get the highest profit of Rs. 1500.

Exam Tip: Always convert time constraints to the same unit (here, all times were converted to minutes) before setting up the inequalities.

 

Question 42. One kind of cake requires 200 g of flour and 25 g of fat, another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat, assuming that there is no shortage of the other ingredients used in making the cakes. Make it an LPP and solve it graphically.
Answer: Let x represent the number of cakes of the first kind and y represent the number of cakes of the second kind.

Based on the given conditions:

\( 200x + 100y \leq 5000 \), \( 25x + 50y \leq 1000 \), \( x \geq 0 \), \( y \geq 0 \)

Maximize \( Z = x + y \)

The feasible region is determined by the constraints \( 200x + 100y \leq 5000 \), \( 25x + 50y \leq 1000 \), \( x \geq 0 \), \( y \geq 0 \). The corner points of the feasible region are \( A(0, 0) \), \( B(0, 20) \), \( C(20, 10) \), and \( D(25, 0) \).

Evaluating Z at each corner point:

At \( A(0, 0) \): \( Z = 0 + 0 = 0 \)

At \( B(0, 20) \): \( Z = 0 + 20 = 20 \)

At \( C(20, 10) \): \( Z = 20 + 10 = 30 \)

At \( D(25, 0) \): \( Z = 25 + 0 = 25 \)

The maximum value of Z is 30, which occurs at point \( (20, 10) \).

The company should bake 20 cakes of the first kind and 10 cakes of the second kind to maximize the total number of cakes.
In simple words: Prepare 20 of the first type and 10 of the second type to make the most cakes - a total of 30.

Exam Tip: When converting units (kilograms to grams), ensure consistency across all constraint equations to avoid calculation errors.

 

Question 43. A manufacturing company makes two types of teaching aids A and B of mathematics for class XII. Each type of A requires 9 labor hours of fabricating and 1 labor hour for finishing. Each type of B requires 12 labors hour for fabricating and 3 labor hour for finishing. For fabricating and finishing, the maximum labor hours available per week are 180 and 30 respectively. The company makes a profit of ₹80 on each piece of type A and ₹120 on each piece of type B. how many pieces of type A and type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Make it as an LLP and solve graphically. What is the maximum profit per week?
Answer: Let x represent the quantity of teaching aids of type A and y represent the quantity of teaching aids of type B.

Based on the given conditions:

\( 9x + 12y \leq 180 \), \( x + 3y \leq 30 \), \( x \geq 0 \), \( y \geq 0 \)

Maximize \( Z = 80x + 120y \)

The feasible region is determined by the constraints \( 9x + 12y \leq 180 \), \( x + 3y \leq 30 \), \( x \geq 0 \), \( y \geq 0 \). The corner points of the feasible region are \( A(0, 0) \), \( B(0, 10) \), \( C(12, 6) \), and \( D(20, 0) \).

Evaluating Z at each corner point:

At \( A(0, 0) \): \( Z = 80(0) + 120(0) = 0 \)

At \( B(0, 10) \): \( Z = 80(0) + 120(10) = 1200 \)

At \( C(12, 6) \): \( Z = 80(12) + 120(6) = 1680 \)

At \( D(20, 0) \): \( Z = 80(20) + 120(0) = 1600 \)

The maximum value of Z is 1680, which occurs at point \( (12, 6) \).

The company should manufacture 12 units of type A and 6 units of type B per week. The maximum profit per week is Rs. 1680.
In simple words: Make 12 of type A and 6 of type B weekly to achieve the best profit of Rs. 1680.

Exam Tip: Always double-check that the optimal corner point satisfies both labour hour constraints - this confirms your feasible region is correctly drawn.

 

Question 1. A company manufactures two types of teaching aids. The objective function is Z = 80x + 120y, where x represents the first type and y represents the second type. Determine the maximum value of Z and the optimal production quantities.
Answer: To find the maximum value, we evaluate the objective function at each corner point of the feasible region.

The highest value among all corner points is 1680, which takes place at the point (12, 6). Therefore, the company should produce 12 units of the 1st type teaching aid and 6 units of the 2nd type teaching aid to achieve the greatest profit of Rs. 1680.
In simple words: Test the profit formula at each corner of the allowed region. The point that gives the biggest number is your answer - make 12 of the first type and 6 of the second type for a profit of Rs. 1680.

Exam Tip: Always evaluate the objective function at all corner points of the feasible region - the maximum or minimum always occurs at a corner, never inside the region.