NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1

Exercise 12.1

1: Maximise $Z=3x+4y$

Subject to the constraints: $x+y\ge 4$, $x\ge 0$, $y\ge 0$  

Ans: The constraints: $x+y\ge 4$, $x\ge 0$, and $y\ge 0$, determine the feasible region as shown below: 

The values of $Z$at the corner points $O(0,0),A(4,0),B(0,4)$ of the feasible region are: 

Hence, the maximum value of $Z$ is  $16$ at the point $B(0,4)$. 

2: Minimise $Z=3x+4y$.

Subject to constraints $x+2y\le 8$, $3x+3y\le 12$,$x\ge 0$ and $y\ge 0$.

Ans: The constraints $x+2y\le 8$, $3x+3y\le 12$,$x\ge 0$ and $y\ge 0$ determine the feasible region as shown below: 

The values of $Z$at the corner points $O(0,0),A(4,0),B(2,3)$ and $C(0,4)$ of the feasible region are: 

Hence, the minimum value of $Z$ is $-12$ at the point $A(4,0)$. 

3: Maximise $Z=5x+3y$

Subject to the constraints: $3x+5y\le 15$, $5x+2y\le 10$, $x\ge 0$, $y\ge 0$  

Ans: The constraints: $3x+5y\le 15$, $5x+2y\le 10$,$x\ge 0$, and $y\ge 0$ determine the feasible region as shown below:

The values of $Z$at the corner points $O(0,0),A(2,0),B(0,3)$ and $C(\frac{20}{19},\frac{45}{19})$ of the feasible region are: 

Hence, the maximum value of $Z$ is $\frac{235}{19}$ at the point$C(\frac{20}{19},\frac{45}{19})$. 

4: Minimise $Z=3x+5y$

Subject to the constraints: $x+3y\ge 3$, $x+y\ge 2$, $x\ge 0$, $y\ge 0$  

Ans: The constraints $x+3y\ge 3$, $x+y\ge 2$,$x\ge 0$, and $y\ge 0$ determine the feasible region as shown below: 

Therefore, the feasible region is unbounded.

The values of $Z$at the corner points $A(3,0),B(\frac{3}{2},\frac{1}{2})$ and $C(0,2)$ of the feasible region are: 

Since, the feasible region is unbounded, $7$ may or may not be the minimum value of  $Z$. 

So, we draw the graph of the inequality, $3x+5y<7$, and check if the resulting half-plane has common points with the feasible region.

As the feasible region has no common point with $3x+5y<7$, the minimum value of $Z$ is $7$ at the point $B(\frac{3}{2},\frac{1}{2})$. 

5: Maximise $Z=3x+2y$

Subject to the constraints: $x+2y\le 10$, $3x+5y\le 15$, $x\ge 0$, $y\ge 0$  

Ans: The constraints, $x+2y\le 10$, $3x+5y\le 15$,$x\ge 0$, and $y\ge 0$ determine the feasible region as shown below: 

The values of $Z$at the corner points $A(5,0),B(4,3)$ and $C(0,5)$ of the feasible region are: 

Hence, the maximum value of $Z$ is $18$ at the point $B(4,3)$. 

6: Minimise $Z=x+2y$

Subject to the constraints: $2x+y\ge 8$, $x+2y\ge 6$, $x\ge 0$, $y\ge 0$  

Ans: The constraints $2x+y\ge 8$, $x+2y\ge 6$,$x\ge 0$, and $y\ge 0$ determine the feasible region as shown below: 

The values of $Z$at the corner points $A(6,0)$ and $B(0,3)$ of the feasible region are: 

As the value of $Z$ at points $A(6,0)$ and $B(0,3)$ is same, we need to take any other point, for example, $(2,2)$ on line $x+2y=6$

Now, $Z=6$ .

Thus, the minimum value of $Z$ occurs at more than $2$ points. Hence, we can say that the value of $Z$ is minimum at every point on the line $x+2y=6$.

7: Minimise and Maximise $Z=5x+10y$

Subject to the constraints: $x+2y\le 120$, $x+y\ge 60$, $x-2y\ge 60$,$x\ge 0$, $y\ge 0$  

Ans: The constraints $x+2y\le 120$, $x+y\ge 60$,$x-2y\ge 60$, $x\ge 0$, and $y\ge 0$ determine the feasible region as shown below:  

The values of $Z$at the corner points $A(60,0),B(120,0),C(60,30)$ and $D(40,20)$ of the feasible region are:

Hence, the minimum value of $Z$ is $300$ at the point $A(60,0)$and the maximum value of $Z$ is $600$ at all the points lying on the line segment joining the points $B(120,0)$ and $C(60,30)$. 

8: Minimise and Maximise $Z=x+2y$

Subject to the constraints: $x+2y\ge 100$, $2x-y\le 0$, $2x+y\le 200$,$x\ge 0$, $y\ge 0$  

Ans: The constraints $x+2y\ge 100$, $2x-y\le 0$,$2x+y\le 200$, $x\ge 0$, and $y\ge 0$ determine the feasible region as shown below: 

The values of $Z$at the corner points $A(0,50),B(20,40),C(50,100)$ and $D(0,200)$ of the feasible region are: 

Hence, the maximum value of $Z$ is $400$ at the point $D(0,200)$and the minimum value of $Z$is $100$ at all the points lying on the line segment joining the points $A(0,50)$ and $B(20,40)$.

9: Maximise $Z=-x+2y$

Subject to the constraints: $x\ge 3$, $x+y\ge 5$, $x+2y\ge 6$ , $y\ge 0$  

Ans: The constraints $x\ge 3$, $x+y\ge 5$,$x+2y\ge 6$, and $y\ge 0$ determine the feasible region as shown below: 

It is visible that the feasible region is unbounded. 

The values of $Z$ at corner points $A(6,0),B(4,1)$ and $C(3,2)$are given as:

Since, the feasible region is unbounded, $Z=1$ may or may not be the maximum value. 

So, we will draw the graph of the inequality, $-x+2y>1$, and check if the resulting half-plane has common points with the feasible region.

As the resulting feasible region has common points with the feasible region. Therefore, $Z=1$ is not the maximum value. Also, $Z$ has no maximum value. 

10: Maximise $Z=x+y$

Subject to the constraints: $x-y\le -1$, $-x+y\le 0$, $x\ge 0$ , and $y\ge 0$  

Ans: The constraints, $x-y\le -1$, $-x+y\le 0$,$x\ge 0$, and $y\ge 0$ determine the feasible region as shown below: 

It can be seen that there is no feasible region. Thus, $Z$ does not have any maximum value.

Conclusion

Exercise 12.1 of Chapter 12 in Class 12 Maths is essential for mastering linear programming. This exercise focuses on formulating linear programming problems, using the graphical method to solve them, and identifying feasible and optimal solutions. Understanding objective functions, constraints, and the feasible region is crucial. Vedantu's NCERT Solutions offer clear, step-by-step guidance, helping students grasp these concepts effectively. Practicing these solutions will enhance problem-solving skills and exam readiness. Concentrate on the graphical method and evaluating objective functions at corner points for optimal solutions.

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