NCERT Exemplar for Class 12 Maths Chapter 12

NCERT Exemplar for Class 12 Maths to help students advance their knowledge on the concept, formulas, theorems, principles,etc. discussed in the chapter. Solutions of the NCERT exemplar for class 12 Maths Chapter 12 Linear Programming meet all the requirements of students and familiarise them with the most asked question. Students will be able to practise more and cover all the sections in a precise manner which will help them to improve their performance in the upcoming 12th board examinations.

NCERT exemplar for Class 12 Maths Chapter 12 Linear Programming contains a total of 22 important questions of short and long type. Students can easily access them by scrolling below and practise them consistently to score maximum marks. Along with this, there are several NCERT exemplar for class 12 maths of all the chapters provided on this platform.

Access Solutions of the NCERT Exemplar Class 12 Maths Chapter 12 Linear Programming – Short Answer (S.A.)

1. Determine the maximum value of \( Z = 11x + 7y \) subject to the constraints: \[ 2x + y \geq 6, \quad x \geq 2, \quad x \geq 0, \quad y \geq 0. \] \section*

{Solution} We are required to maximize \( Z = 11x + 7y \) \hspace{2mm} \text{(i)}\\ subject to the constraints: \[ 2x + y \geq 6 \quad \text{(ii)} \] \[ x \geq 2 \quad \text{(iii)} \] \[ x \geq 0, \quad y \geq 0 \quad \text{(iv)} \] The feasible region determined by the system of constraints (ii) to (iv) is bounded, and it is represented by the region \(OABC\). We will now use the corner point method to determine the maximum value of \(Z\). \subsection*{Corner Points and Corresponding Values of \( Z \)} \begin{tabular}{|c|c|} \hline Corner Points & Corresponding Value of \( Z \) \\ \hline \( (0, 0) \) & \( 0 \) \\ \( (2, 0) \) & \( 22 \) \\ \( (2, 2) \) & \( 36 \) \\ \( (0, 6) \) & \( 42 \) \quad \text{(Maximum)} \\ \hline \end{tabular} \section*{Conclusion} Hence, the maximum value of \( Z \) is \( 42 \) at the point \( (0, 6) \).

2. Maximize \( Z = 3x + 4y \) subject to the constraints: \[ x + y \leq 4, \quad x \geq 0, \quad y \geq 0. \] \section*

{Solution} The feasible region determined by the constraints \( x + y \leq 4 \), \( x \geq 0 \), and \( y \geq 0 \) is as follows. The corner points of the feasible region are: \[ O(0, 0), \quad A(4, 0), \quad \text{and} \quad B(0, 4). \] The values of \( Z \) at these points are as follows: \subsection*{Corner Points and Corresponding Values of \( Z \)} \begin{tabular}{|c|c|} \hline Corner Point & \( Z = 3x + 4y \) \\ \hline \( O(0, 0) \) & \( 0 \) \\ \( A(4, 0) \) & \( 12 \) \\ \( B(0, 4) \) & \( 16 \quad \rightarrow \text{Maximum} \) \\ \hline \end{tabular} \section*{Conclusion} Therefore, the maximum value of \( Z \) is \( 16 \) at the point \( B(0, 4) \).

3. Maximize \( Z = 3x + 4y \) subject to the constraints: \[ x + y \leq 4, \quad x \geq 0, \quad y \geq 0. \] \section*

{Solution} The feasible region determined by the constraints \( x + y \leq 4 \), \( x \geq 0 \), and \( y \geq 0 \) is as follows. The corner points of the feasible region are: \[ O(0, 0), \quad A(4, 0), \quad \text{and} \quad B(0, 4). \] The values of \( Z \) at these points are as follows: \begin{center} \begin{tabular}{|c|c|} \hline \textbf{Corner Point} & \( Z = 3x + 4y \) \\ \hline \( O(0, 0) \) & \( 0 \) \\ \( A(4, 0) \) & \( 12 \) \\ \( B(0, 4) \) & \( 16 \) (Maximum) \\ \hline \end{tabular} \end{center} \section*{Conclusion} Therefore, the maximum value of \( Z \) is \( 16 \) at the point \( B(0, 4) \).

4. Minimize \( Z = 13x – 15y \) subject to the constraints: \[ x + y \leq 7, \quad 2x – 3y + 6 \geq 0, \quad x \geq 0, \quad y \geq 0. \] \section*

{Solution} We are required to minimize \( Z = 13x – 15y \) subject to the constraints: \[ x + y \leq 7, \quad 2x – 3y + 6 \geq 0, \quad x \geq 0, \quad y \geq 0. \] The shaded region shown as \( OABC \) is bounded, and the coordinates of its corner points are \( (0,0) \), \( (7,0) \), \( (3,4) \), and \( (0,2) \), respectively. The corresponding values of \( Z \) at these points are as follows: \begin{center} \begin{tabular}{|c|c|} \hline \textbf{Corner Point} & \( Z = 13x – 15y \) \\ \hline \( (0,0) \) & \( 0 \) \\ \( (7,0) \) & \( 91 \) \\ \( (3,4) \) & \( -21 \) \\ \( (0,2) \) & \( -30 \) \\ \hline \end{tabular} \end{center} \section*{Conclusion} Hence, the minimum value of \( Z \) is \( -30 \) at the point \( (0,2) \).

5. Determine the maximum value of \( Z = 3x + 4y \), if the feasible region (shaded) for a linear programming problem (LPP) is shown in the following figure. \section*

{Solution} As is clear from the graph, the corner points are \( O \), \( A \), \( E \), and \( D \) with coordinates \( (0,0) \), \( (52,0) \), \( (44,16) \), and \( (0,38) \), respectively. Also, the given region is bounded. Given: \( Z = 3x + 4y \). Using the constraints \( 2x + y = 104 \) and \( 2x + 4y = 152 \): \[ \Rightarrow -3y = -48 \] \[ \Rightarrow y = 16 \quad \text{and} \quad x = 44 \] The corresponding values of \( Z \) at these points are as follows: \begin{center} \begin{tabular}{|c|c|} \hline \textbf{Corner Point} & \( Z = 3x + 4y \) \\ \hline \( (0,0) \) & \( 0 \) \\ \( (52,0) \) & \( 156 \) \\ \( (44,16) \) & \( 196 \) \quad \(\leftarrow\) \text{Maximum} \\ \( (0,38) \) & \( 152 \) \\ \hline \end{tabular} \end{center} \section*{Conclusion} Hence, \( Z \) is maximum at \( (44,16) \) and its maximum value is \( 196 \).

6. Feasible region (shaded) for a LPP is shown in Fig. 12.8. Maximize Z = 5x + 7y.

Solution:

Given: Z = 5x + 7y and feasible region OABC.

The corner points of the feasible region are O(0, 0), A(7, 0), B(3, 4) and C(0, 2).

On evaluating the value of Z, we get

Corner points	Value of Z
O(0, 0)	Z = 5(0) + 7(0) = 0
A(7, 0)	Z = 5(7) + 7(0) = 35
B(3, 4)	Z = 5(3) + 7(4) = 43
C(0, 2)	Z = 5(0) + 7(2) = 14

From the above table it’s seen that the maximum value of Z is 43.

Therefore, the maximum value of the function Z is 43 at (3, 4).

7. Determine the maximum value of \( Z = 3x + 4y \), if the feasible region (shaded) for a linear programming problem (LPP) is shown in the following figure. \section*

{Solution} As is clear from the graph, the corner points are \( O \), \( A \), \( E \), and \( D \) with coordinates \( (0,0) \), \( (52,0) \), \( (44,16) \), and \( (0,38) \), respectively. Also, the given region is bounded. Given: \( Z = 3x + 4y \). Using the constraints \( 2x + y = 104 \) and \( 2x + 4y = 152 \): \[ \Rightarrow -3y = -48 \] \[ \Rightarrow y = 16 \quad \text{and} \quad x = 44 \] The corresponding values of \( Z \) at these points are as follows: \begin{center} \begin{tabular}{|c|c|} \hline \textbf{Corner Point} & \( Z = 3x + 4y \) \\ \hline \( (0,0) \) & \( 0 \) \\ \( (52,0) \) & \( 156 \) \\ \( (44,16) \) & \( 196 \) \quad \(\leftarrow\) \text{Maximum} \\ \( (0,38) \) & \( 152 \) \\ \hline \end{tabular} \end{center} \section*{Conclusion} Hence, \( Z \) is maximum at \( (44,16) \) and its maximum value is \( 196 \).

8. Refer to Exercise 7 above. Find the maximum value of Z.

Solution:

In the evaluating table for the value of Z, it’s clearly seen that the maximum value of Z is 47 at (3, 2)

9. The feasible region for a linear programming problem (LPP) is shown in the following figure. Evaluate \( Z = 4x + y \) at each of the corner points of this region. Find the minimum value of \( Z \), if it exists. \section*

{Solution} From the shaded region, it is clear that the feasible region is unbounded with the corner points \( A(4,0) \), \( B(2,1) \), and \( C(0,3) \). Also, we have \( Z = 4x + y \). \[ \text{[since, } x + 2y = 4 \text{ and } x + y = 3 \Rightarrow y = 1 \text{ and } x = 2] \] \begin{center} \begin{tabular}{|c|c|} \hline \textbf{Corner Point} & \( Z = 4x + y \) \\ \hline \( (4,0) \) & \( 16 \) \\ \( (2,1) \) & \( 9 \) \\ \( (0,3) \) & \( 3 \) \quad \(\leftarrow\) \text{Minimum} \\ \hline \end{tabular} \end{center} Now, we see that 3 is the smallest value of \( Z \) at the corner point \( (0,3) \). Note that the region is unbounded, therefore 3 may or may not be the minimum value of \( Z \). To decide this issue, we graph the inequality \( 4x + y < 3 \) and check whether the resulting open half-plane has no point in common with the feasible region; otherwise, \( Z \) has no minimum value. From the graph above, it is clear that there is no point in common with the feasible region, and hence \( Z \) has a minimum value of 3 at \( (0,3) \).

10. In the following figure, the feasible region (shaded) for a linear programming problem (LPP) is shown. Determine the maximum and minimum values of \( Z = x + 2y \). \section*

{Solution} From the shaded bounded region, it is clear that the coordinates of the corner points are \[ \left(\frac{3}{13}, \frac{24}{13}\right), \left(\frac{18}{7}, \frac{2}{7}\right), \left(\frac{7}{2}, \frac{3}{4}\right), \text{ and } \left(\frac{3}{2}, \frac{15}{4}\right). \] Also, we have to determine the maximum and minimum values of \( Z = x + 2y \). \begin{center} \begin{tabular}{|c|c|} \hline \textbf{Corner Point} & \( Z = x + 2y \) \\ \hline \( \left(\frac{3}{13}, \frac{24}{13}\right) \) & \( \frac{3}{13} + \frac{48}{13} = \frac{51}{13} = 3\frac{12}{13} \) \\ \( \left(\frac{18}{7}, \frac{2}{7}\right) \) & \( \frac{18}{7} + \frac{4}{7} = \frac{22}{7} = 3\frac{1}{7} \) \quad \(\leftarrow\) \text{Minimum} \\ \( \left(\frac{7}{2}, \frac{3}{4}\right) \) & \( \frac{7}{2} + \frac{6}{4} = \frac{20}{4} = 5 \) \\ \( \left(\frac{3}{2}, \frac{15}{4}\right) \) & \( \frac{3}{2} + \frac{30}{4} = \frac{36}{4} = 9 \) \quad \(\leftarrow\) \text{Maximum} \\ \hline \end{tabular} \end{center} Hence, the maximum and minimum values of \( Z \) are 9 and \( 3\frac{1}{7} \), respectively.

11. A manufacturer of electronic circuits has a stock of 200 resistors, 120 transistors, and 150 capacitors and is required to produce two types of circuits, A and B. Type A requires 20 resistors, 10 transistors, and 10 capacitors. Type B requires 10 resistors, 20 transistors, and 30 capacitors. If the profit on type A circuit is Rs 50 and that on type B circuit is Rs 60, formulate this problem as a Linear Programming Problem (LPP) so that the manufacturer can maximize his profit. \section*

{Solution} Let the manufacturer produce \( x \) units of type A circuits and \( y \) units of type B circuits. From the given information, we have the following constraint table: \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textbf{Component} & \textbf{Type A (x)} & \textbf{Type B (y)} & \textbf{Maximum Stock} \\ \hline Resistors & 20 & 10 & 200 \\ \hline Transistors & 10 & 20 & 120 \\ \hline Capacitors & 10 & 30 & 150 \\ \hline Profit & Rs 50 & Rs 60 & — \\ \hline \end{tabular} \end{center} Thus, the total profit \( Z = 50x + 60y \) (in Rs). Now, we have the following mathematical model for the given problem. \textbf{Objective Function:} \[ \text{Maximize } Z = 50x + 60y \quad \text{(i)} \] \textbf{Subject to the constraints:} \[ \begin{aligned} 20x + 10y & \leq 200 & \quad \text{[Resistor constraint]} \\ & \Rightarrow 2x + y \leq 20 & \quad \text{(ii)} \\ 10x + 20y & \leq 120 & \quad \text{[Transistor constraint]} \\ & \Rightarrow x + 2y \leq 12 & \quad \text{(iii)} \\ 10x + 30y & \leq 150 & \quad \text{[Capacitor constraint]} \\ & \Rightarrow x + 3y \leq 15 & \quad \text{(iv)} \\ x & \geq 0, \, y \geq 0 & \quad \text{[Non-negativity constraint]} \quad \text{(v)} \end{aligned} \] Therefore, the maximum profit \( Z = 50x + 60y \) is subject to the constraints: \[ \begin{aligned} 2x + y & \leq 20 \quad \text{(ii)} \\ x + 2y & \leq 12 \quad \text{(iii)} \\ x + 3y & \leq 15 \quad \text{(iv)} \\ x & \geq 0, \, y \geq 0 \quad \text{(v)} \end{aligned} \]

12. A firm has to transport 1200 packages using large vans that can carry 200 packages each and small vans that can take 80 packages each. The cost for engaging each large van is Rs 400 and each small van is Rs 200. Not more than Rs 3000 is to be spent on the job, and the number of large vans cannot exceed the number of small vans. Formulate this problem as a Linear Programming Problem (LPP) given that the objective is to minimize cost. \section*

{Solution} Let the firm use \( x \) large vans and \( y \) small vans. Based on the given information, we have the following constraint table: \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textbf{Resource} & \textbf{Large Vans (x)} & \textbf{Small Vans (y)} & \textbf{Maximum/Minimum} \\ \hline Packages & 200 & 80 & 1200 \\ \hline Cost & 400 & 200 & 3000 \\ \hline \end{tabular} \end{center} Thus, the objective function for minimizing the cost is: \[ Z = 400x + 200y \] \textbf{Subject to the constraints:} \[ \begin{aligned} 200x + 80y & \geq 1200 & \quad \text{[Package constraint]} \\ & \Rightarrow 5x + 2y \geq 30 & \quad \text{(i)} \\ 400x + 200y & \leq 3000 & \quad \text{[Cost constraint]} \\ & \Rightarrow 2x + y \leq 15 & \quad \text{(ii)} \\ x & \leq y & \quad \text{[Van constraint]} \quad \text{(iii)} \\ x & \geq 0, \, y \geq 0 & \quad \text{[Non-negativity constraint]} \quad \text{(iv)} \end{aligned} \] \textbf{Therefore, the required LPP to minimize cost is:} \[ \text{Minimize } Z = 400x + 200y \] \textbf{Subject to:} \[ \begin{aligned} 5x + 2y & \geq 30 & \quad \text{(i)} \\ 2x + y & \leq 15 & \quad \text{(ii)} \\ x & \leq y & \quad \text{(iii)} \\ x & \geq 0, \, y \geq 0 & \quad \text{(iv)} \end{aligned} \]

13. A company manufactures two types of screws, A and B. All the screws have to pass through a threading machine and a slotting machine. A box of type A screws requires 2 minutes on the threading machine and 3 minutes on the slotting machine. A box of type B screws requires 8 minutes on the threading machine and 2 minutes on the slotting machine. In a week, each machine is available for 60 hours. On selling these screws, the company gets a profit of Rs 100 per box on type A screws and Rs 170 per box on type B screws. Formulate this problem as a Linear Programming Problem (LPP) given that the objective is to maximize profit. \section*

{Solution} Let the company manufacture \( x \) boxes of type A screws and \( y \) boxes of type B screws. From the given information, we have the following constraint table: \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textbf{Resource} & \textbf{Type A (x)} & \textbf{Type B (y)} & \textbf{Maximum Time Available (in minutes)} \\ \hline Time required on threading machine & 2 & 8 & 60 $\times$ 60 \\ \hline Time required on slotting machine & 3 & 2 & 60 $\times$ 60 \\ \hline Profit (in Rs) & 100 & 170 & \\ \hline \end{tabular} \end{center} Thus, the objective function for maximizing profit is: \[ Z = 100x + 170y \] \textbf{Subject to the constraints:} \[ \begin{aligned} 2x + 8y & \leq 60 \times 60 & \quad \text{[Time constraint for threading machine]} \\ & \Rightarrow x + 4y \leq 1800 & \quad \text{(i)} \\ 3x + 2y & \leq 60 \times 60 & \quad \text{[Time constraint for slotting machine]} \\ & \Rightarrow 3x + 2y \leq 3600 & \quad \text{(ii)} \\ x & \geq 0, \, y \geq 0 & \quad \text{[Non-negativity constraints]} \quad \text{(iii)} \end{aligned} \] \textbf{Therefore, the required LPP is:} \[ \text{Maximize } Z = 100x + 170y \] \textbf{Subject to:} \[ \begin{aligned} x + 4y & \leq 1800 & \quad \text{(i)} \\ 3x + 2y & \leq 3600 & \quad \text{(ii)} \\ x & \geq 0, \, y \geq 0 & \quad \text{(iii)} \end{aligned} \]

14. A company manufactures two types of sweaters: type A and type B. It costs Rs 360 to make a type A sweater and Rs 120 to make a type B sweater. The company can make at most 300 sweaters and spend at most Rs 72000 a day. The number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100. The company makes a profit of Rs 200 for each sweater of type A and Rs 120 for every sweater of type B.

Formulate this problem as a LPP to maximize the profit to the company.

Solution:

Let’s assume x and y to be the number of sweaters of type A and type B respectively.

From the question, the following constraints are:

360x + 120y ≤ 72000 ⇒ 3x + y ≤ 600 … (i)

x + y ≤ 300 … (ii)

x + 100 ≥ y ⇒ y ≤ x + 100 … (iii)

Profit: Z = 200x + 120y

Therefore, the required LPP to maximize the profit is

Maximize Z = 200x + 120y subject to constrains

3x + y ≤ 600, x + y ≤ 300, y ≤ x + 100, x ≥ 0, y ≥ 0.

15. A man rides his motorcycle at the speed of 50 km/hour. He has to spend Rs 2 per km on petrol. If he rides it at a faster speed of 80 km/hour, the petrol cost increases to Rs 3 per km. He has at most Rs 120 to spend on petrol and one hour’s time. He wishes to find the maximum distance that he can travel.

Express this problem as a linear programming problem.

Solution:

Let’s assume the man covers x km on his motorcycle at the speed of 50km/hr and covers y km at the speed of 50 km/hr and covers y km at the speed of 80 km/hr.

So, cost of petrol = 2x + 3y

The man has to spend Rs 120 atmost on petrol

⇒ 2x + 3y ≤ 120 …. (i)

Now, the man has only 1 hr time

So, x/50 + y/80 ≤ 1 ⇒ 8x + 5y ≤ 400 … (ii)

And, x ≥ 0, y ≥ 0

To have maximum distance Z = x + y.

Therefore, the required LPP to travel maximum distance is maximize Z = x + y, subject to the constraints

2x + 3y ≤ 120, 8x + 5y ≤ 400, x ≥ 0, y ≥ 0.

Long Answer (L.A.)

16. Refer to Question 11. How many circuits of type A and type B should be produced by the manufacturer to maximize his profit? Determine the maximum profit. \section*

{Solution} Referring to the solution of Question 11, we have the objective function and constraints as follows: Maximize \( Z = 50x + 60y \), subject to the constraints: \[ \begin{aligned} 2x + y & \leq 20, & \quad \text{(i)} \\ x + 2y & \leq 12, & \quad \text{(ii)} \\ x + 3y & \leq 15, & \quad \text{(iii)} \\ x & \geq 0, \, y \geq 0 & \quad \text{(iv)} \end{aligned} \] From the shaded region, it is clear that the feasible region determined by the system of constraints is \( OABCD \) and is bounded. The coordinates of the corner points are \( (0,0) \), \( (10,0) \), \( \left(\frac{28}{3}, \frac{4}{3}\right) \), \( (6,3) \), and \( (0,5) \) respectively. To find the corresponding value of \( Z \) at these corner points: \begin{center} \begin{tabular}{|c|c|} \hline \textbf{Corner Points} & \textbf{Corresponding value of \( Z = 50x + 60y \)} \\ \hline \( (0,0) \) & \( 0 \) \\ \hline \( (10,0) \) & \( 500 \) \\ \hline \( \left(\frac{28}{3}, \frac{4}{3}\right) \) & \( \frac{1400}{3} + \frac{240}{3} = \frac{1640}{3} \approx 546.67 \) \\ \hline \( (6,3) \) & \( 50(6) + 60(3) = 300 + 180 = 480 \) \\ \hline \( (0,5) \) & \( 60(5) = 300 \) \\ \hline \end{tabular} \end{center} Since the manufacturer is required to produce whole circuits of type A and type B (i.e., \( x \) and \( y \) must be integers), the maximum profit is achieved at the point \( (6,3) \) where \( Z = 480 \). \textbf{Conclusion:} The maximum profit is \( \textbf{Rs. 480} \) when the manufacturer produces \( \textbf{6 circuits of type A} \) and \( \textbf{3 circuits of type B} \).

17. Refer to Question 12. What will be the minimum cost? \section*

{Solution} Referring to the solution of Question 12, we have the objective function and constraints as follows: Minimize \( Z = 400x + 200y \), subject to the constraints: \[ \begin{aligned} 5x + 2y & \geq 30, & \quad \text{(i)} \\ 2x + y & \leq 15, & \quad \text{(ii)} \\ x – y & \leq 0, & \quad \text{(iii)} \\ x & \geq 0, \, y \geq 0 & \quad \text{(iv)} \end{aligned} \] On solving the equations \( x – y = 0 \) and \( 5x + 2y = 30 \), we get: \[ \begin{aligned} y &= \frac{30}{7}, \\ x &= \frac{30}{7} \end{aligned} \] On solving the equations \( x – y = 0 \) and \( 2x + y = 15 \), we get: \[ \begin{aligned} x &= 5, \\ y &= 5 \end{aligned} \] From the shaded feasible region, it is clear that the coordinates of the corner points are \( (0,15) \), \( (5,5) \), and \( \left(\frac{30}{7}, \frac{30}{7}\right) \). To find the corresponding value of \( Z \) at these corner points: \begin{center} \begin{tabular}{|c|c|} \hline \textbf{Corner Points} & \textbf{Corresponding value of \( Z = 400x + 200y \)} \\ \hline \( (0,15) \) & \( 200(15) = 3000 \) \\ \hline \( (5,5) \) & \( 400(5) + 200(5) = 2000 + 1000 = 3000 \) \\ \hline \( \left(\frac{30}{7}, \frac{30}{7}\right) \) & \( 400 \times \frac{30}{7} + 200 \times \frac{30}{7} = \frac{18000}{7} \approx 2571.43 \) \\ \hline \end{tabular} \end{center} \textbf{Conclusion:} The minimum cost is \( \textbf{Rs. 2571.43} \) at the point \( \left(\frac{30}{7}, \frac{30}{7}\right) \).

18. Refer to Exercise 13. Solve the linear programming problem and determine the maximum profit to the manufacturer. \section*

{Solution} Referring to the problem in Exercise 13, we have to maximize the profit \( Z = 50x + 60y \), subject to the following constraints: \[ \begin{aligned} 2x + y &\leq 20, \\ x + 2y &\leq 12, \\ x + 3y &\leq 15, \\ x &\geq 0, \, y \geq 0. \end{aligned} \] \subsection*{Corner Points} The feasible region determined by the constraints has the following corner points: \[ \begin{aligned} (0, 0) \\ (10, 0) \\ \left(\frac{28}{3}, \frac{4}{3}\right) \\ (6, 3) \\ (0, 5) \end{aligned} \] \subsection*{Evaluation of Objective Function at Corner Points} We now evaluate \( Z = 50x + 60y \) at each of these corner points: \[ \begin{array}{|c|c|} \hline \text{Corner Points} & \text{Corresponding value of } Z = 50x + 60y \\ \hline (0, 0) & 50(0) + 60(0) = 0 \\ \hline (10, 0) & 50(10) + 60(0) = 500 \\ \hline \left(\frac{28}{3}, \frac{4}{3}\right) & 50\left(\frac{28}{3}\right) + 60\left(\frac{4}{3}\right) = \frac{1400}{3} + \frac{240}{3} = \frac{1640}{3} \approx 546.67 \\ \hline (6, 3) & 50(6) + 60(3) = 300 + 180 = 480 \\ \hline (0, 5) & 50(0) + 60(5) = 300 \\ \hline \end{array} \] \subsection*{Maximum Profit} The maximum value of \( Z \) is approximately \( 546.67 \), which occurs at the point \( \left(\frac{28}{3}, \frac{4}{3}\right) \). If only integer solutions are acceptable, the maximum profit is \( 500 \) at the point \( (10, 0) \). \textbf{Final Answer:} The maximum profit is \textbf{Rs 546.67} (for fractional values) or \textbf{Rs 500} (for integer values).

19. Refer to question 14. How many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit? \section*

{Solution} Referring to question 14, we have the objective function: \[ \text{Maximise } Z = 200x + 120y \] subject to the constraints: \[ \begin{aligned} x + y &\leq 300, \\ 3x + y &\leq 600, \\ x – y &\geq -100, \\ x &\geq 0, \, y \geq 0. \end{aligned} \] \subsection*{Finding Corner Points} To find the corner points, we solve the following pairs of equations: 1. Solving \(x + y = 300\) and \(3x + y = 600\): \[ \begin{aligned} x &= 150, \\ y &= 150. \end{aligned} \] 2. Solving \(x – y = -100\) and \(x + y = 300\): \[ \begin{aligned} x &= 100, \\ y &= 200. \end{aligned} \] The coordinates of the corner points, based on the feasible region, are: \((0,0)\), \((200,0)\), \((150,150)\), \((100,200)\), and \((0,100)\). \subsection*{Evaluation of Objective Function at Corner Points} We now evaluate \( Z = 200x + 120y \) at each of these corner points: \[ \begin{array}{|c|c|} \hline \text{Corner Points} & \text{Corresponding value of } Z = 200x + 120y \\ \hline (0, 0) & 200(0) + 120(0) = 0 \\ \hline (200, 0) & 200(200) + 120(0) = 40000 \\ \hline (150, 150) & 200(150) + 120(150) = 30000 + 18000 = 48000 \, (\text{Maximum}) \\ \hline (100, 200) & 200(100) + 120(200) = 20000 + 24000 = 44000 \\ \hline (0, 100) & 200(0) + 120(100) = 12000 \\ \hline \end{array} \] \subsection*{Conclusion} The maximum profit is \( \textbf{Rs 48,000} \), which occurs when the company produces \textbf{150 sweaters of each type}.

20. Refer to question 15. Determine the maximum distance that the man can travel. \section*

{Solution} Referring to solution 15, we have the objective function: \[ \text{Maximise } Z = x + y \] subject to the constraints: \[ \begin{aligned} 2x + 3y &\leq 120, \\ 8x + 5y &\leq 400, \\ x &\geq 0, \, y \geq 0. \end{aligned} \] \subsection*{Finding Corner Points} To find the corner points, we solve the following system of equations: 1. Solving \(8x + 5y = 400\) and \(2x + 3y = 120\): \[ \begin{aligned} x &= \frac{300}{7}, \\ y &= \frac{80}{7}. \end{aligned} \] From the shaded feasible region, it is clear that the coordinates of the corner points are: \((0, 0)\), \((50, 0)\), \(\left(\frac{300}{7}, \frac{80}{7}\right)\), and \((0, 40)\). \subsection*{Evaluation of Objective Function at Corner Points} We now evaluate \( Z = x + y \) at each of these corner points: \[ \begin{array}{|c|c|} \hline \text{Corner Points} & \text{Corresponding value of } Z = x + y \\ \hline (0, 0) & 0 + 0 = 0 \\ \hline (50, 0) & 50 + 0 = 50 \\ \hline \left(\frac{300}{7}, \frac{80}{7}\right) & \frac{300}{7} + \frac{80}{7} = \frac{380}{7} \approx 54 \frac{2}{7} \text{ km (Maximum)} \\ \hline (0, 40) & 0 + 40 = 40 \\ \hline \end{array} \] \subsection*{Conclusion} The maximum distance that the man can travel is \(\textbf{54} \frac{2}{7} \textbf{ km}\).

21. Solve the following Linear Programming problems graphically: 1. Maximize \( Z = 3x + 4y \) subject to the constraints: \[ x + y \leq 4, \quad x \geq 0, \quad y \geq 0 \] 2. Minimize \( Z = -3x + 4y \) subject to: \[ x + 2y \leq 8, \quad 3x + 2y \leq 12, \quad x \geq 0, \quad y \geq 0 \] \section*

{Solution} \subsection*{1. Maximize \( Z = 3x + 4y \)} We first determine the corner points of the feasible region by solving the system of constraints: \[ \begin{aligned} x + y &\leq 4, \\ x &\geq 0, \\ y &\geq 0. \end{aligned} \] The corner points are: \[ (0,4), \quad (4,0), \quad (0,0) \] Now, we calculate the value of \( Z = 3x + 4y \) at each corner point: \[ \begin{array}{|c|c|} \hline \text{Corner Points} & \text{Value of } Z = 3x + 4y \\ \hline (0,4) & 3(0) + 4(4) = 16 \\ \hline (4,0) & 3(4) + 4(0) = 12 \\ \hline (0,0) & 3(0) + 4(0) = 0 \\ \hline \end{array} \] Hence, \( Z \) is maximum at \( (0,4) \) and the maximum value is \( \mathbf{16} \). \subsection*{2. Minimize \( Z = -3x + 4y \)} We first determine the corner points of the feasible region by solving the system of constraints: \[ \begin{aligned} x + 2y &\leq 8, \\ 3x + 2y &\leq 12, \\ x &\geq 0, \\ y &\geq 0. \end{aligned} \] The corner points are: \[ (0,4), \quad (2,3), \quad (4,0) \] Now, we calculate the value of \( Z = -3x + 4y \) at each corner point: \[ \begin{array}{|c|c|} \hline \text{Corner Points} & \text{Value of } Z = -3x + 4y \\ \hline (0,4) & -3(0) + 4(4) = 16 \\ \hline (2,3) & -3(2) + 4(3) = 6 \\ \hline (4,0) & -3(4) + 4(0) = -12 \\ \hline \end{array} \] Hence, \( Z \) is minimum at \( (4,0) \) and the minimum value is \( \mathbf{-12} \).

22. A manufacturer produces two models of bikes – model X and model Y. Model X takes 6 man-hours to make per unit, while model Y takes 10 man-hours per unit. There is a total of 450 man-hours available per week. Handling and marketing costs are Rs 2000 and Rs 1000 per unit for models X and Y, respectively. The total funds available for these purposes are Rs 80000 per week. Profits per unit for models X and Y are Rs 1000 and Rs 500, respectively. How many bikes of each model should the manufacturer produce, so as to yield a maximum profit? Find the maximum profit. \section*

{Solution} Let the manufacturer produce \(x\) units of model X and \(y\) units of model Y. \begin{itemize} \item Model X takes 6 man-hours to make per unit and model Y takes 10 man-hours to make per unit. There is a total of 450 man-hours available per week. \[ 6x + 10y \leq 450 \implies 3x + 5y \leq 225 \quad \text{(i)} \] \item The handling and marketing costs are Rs 2000 and Rs 1000 per unit for models X and Y, respectively. The total funds available for these purposes are Rs 80000 per week. \[ 2000x + 1000y \leq 80000 \implies 2x + y \leq 80 \quad \text{(ii)} \] \item Also, \(x \geq 0\), \(y \geq 0\). \end{itemize} Hence, the profits per unit for models X and Y are Rs 1000 and Rs 500, respectively. The required Linear Programming Problem (LPP) is: \[ \text{Maximize } Z = 1000x + 500y \] Subject to: \[ \begin{aligned} 3x + 5y &\leq 225 \quad \text{(i)} \\ 2x + y &\leq 80 \quad \text{(ii)} \\ x &\geq 0, \\ y &\geq 0 \end{aligned} \] From the shaded feasible region, it is clear that the coordinates of the corner points are \((0,0)\), \((40,0)\), \((25,30)\), and \((0,45)\). On solving \(3x + 5y = 225\) and \(2x + y = 80\), we get: \[ x = 25, \quad y = 30 \] The value of \(Z = 1000x + 500y\) at each corner point is calculated as follows: \[ \begin{array}{|c|c|} \hline \text{Corner Points} & \text{Value of } Z = 1000x + 500y \\ \hline (0,0) & 1000(0) + 500(0) = 0 \\ \hline (40,0) & 1000(40) + 500(0) = 40000 \\ \hline (25,30) & 1000(25) + 500(30) = 25000 + 15000 = 40000 \\ \hline (0,45) & 1000(0) + 500(45) = 22500 \\ \hline \end{array} \] So, the manufacturer should produce 25 bikes of model X and 30 bikes of model Y to get a maximum profit of Rs 40000, since it is required that each model of bikes should be produced.

NCERT Exemplar For Class 12 Maths

The NCERT exemplars are an effective study material for scoring higher marks in the examination paper. Students must practise these additional questions for their own benefits, as these are curated by the best subject-matter experts to boost both knowledge and confidence. Students can easily access the ncert exemplar for class 12 maths by visiting our website SimplyAcad and solve all the questions listed to secure maximum marks.

Here are some other NCERT exemplar for class 12 maths:

NCERT exemplar for class 12 maths Chapter 1	NCERT exemplar for class 12 maths Chapter 7
NCERT exemplar for class 12 maths Chapter 2	NCERT exemplar for class 12 maths Chapter 8
NCERT exemplar for class 12 maths Chapter 3	NCERT exemplar for class 12 maths Chapter 9
NCERT exemplar for class 12 maths Chapter 5	NCERT exemplar for class 12 maths Chapter 10
NCERT exemplar for class 12 maths Chapter 6	NCERT exemplar for class 12 maths Chapter 11