NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.1 - Linear Programming

Question 2: Solve the following Linear Programming Problems graphically: Minimise $z=-3x+4y$ Subject to . $x+2y\le 8,3x+2y\le 12,x\ge 0,y\ge 0.$ Show that the minimum of Z occurs at more than two points

Answer:

The region determined by constraints, $x+2y\le 8,3x+2y\le 12,x\ge 0,y\ge 0.$ is as follows,

The corner points of feasible region are $A(2,3),B(4,0),C(0,0),D(0,4)$

The value of these points at these corner points are :

The minimum value of Z is -12 at $B(4,0)$

Question 3: Solve the following Linear Programming Problems graphically: Maximise $Z=5x+3y$ Subject to $3x+5y\le 15$ , $5x+2y\le 10$ , $x\ge 0,y\ge 0$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, $3x+5y\le 15$ , $5x+2y\le 10$ , $x\ge 0,y\ge 0$ is as follows :

The corner points of feasible region are $A(0,3),B(0,0),C(2,0),D(\frac{20}{19},\frac{45}{19})$

The value of these points at these corner points are :

The maximum value of Z is $\frac{235}{19}$ at $D(\frac{20}{19},\frac{45}{19})$

Question 4: Solve the following Linear Programming Problems graphically: Minimise $Z=3x+5y$ Such that $x+3y\ge 3,x+y\ge 2,x,y\ge 0.$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints $x+3y\ge 3,x+y\ge 2,x,y\ge 0.$ is as follows,

The feasible region is unbounded as shown.

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

The value of these points at these corner points are :

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

For this, we draw $3x+5y<7$ and check whether resulting half plane has a point in common with the feasible region or not.

We can see a feasible region has no common point with. $Z=3x+5y$

Hence, Z has a minimum value of 7 at $B(\frac{3}{2},\frac{1}{2})$

Question 5: Solve the following Linear Programming Problems graphically: Maximise $Z=3x+2y$ Subject to $x+2y\le 10,3x+y\le 15,x,y\ge 0$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, $x+2y\le 10,3x+y\le 15,x,y\ge 0$ is as follows,

The corner points of feasible region are $A(5,0),B(4,3),C(0,5)$

The value of these points at these corner points are :

The maximum value of Z is 18 at $B(4,3)$

Question 6: Solve the following Linear Programming Problems graphically: Minimise $Z=x+2y$ Subject to $2x+y\ge 3,x+2y\ge 6,x,y\ge 0.$

Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints $2x+y\ge 3,x+2y\ge 6,x,y\ge 0.$ is as follows,

The corner points of the feasible region are $A(6,0),B(0,3)$

The value of these points at these corner points are :

Value of Z is the same at both points. $A(6,0),B(0,3)$

If we take any other point like $(2,2)$ on line $Z=x+2y$ , then Z=6.

Thus the minimum value of Z occurs at more than 2 points .

Therefore, the value of Z is minimum at every point on the line $Z=x+2y$ .

Question 7: Solve the following Linear Programming Problems graphically: Minimise and Maximise $z=5x+10y$ Subject to $x+2y\le 120,x+y\ge 60,x-2y\ge 0,x,y\ge 0$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, $x+2y\le 120,x+y\ge 60,x-2y\ge 0,x,y\ge 0$ is as follows,

The corner points of feasible region are $A(40,20),B(60,30),C(60,0),D(120,0)$

The value of these points at these corner points are :

The minimum value of Z is 300 at $C(60,0)$ and maximum value is 600 at all points joing line segment $B(60,30)$ and $D(120,0)$

Question 8: Solve the following Linear Programming Problems graphically: Minimise and Maximise $z=x+2y$ Subject to $x+2y\ge 100,2x-y\le 0,2x+y\le 200,x,y,\ge 0$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints $x+2y\ge 100,2x-y\le 0,2x+y\le 200,x,y,\ge 0$ is as follows,

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

The value of these points at these corner points are :

The minimum value of Z is 100 at all points on the line segment joining points $A(0,50)$ and $B(20,40)$ .

The maximum value of Z is 400 at $D(0,200)$ .

Question 9: Solve the following Linear Programming Problems graphically: Maximise $Z=-x+2y$ Subject to the constraints: $x\ge 3,x+y\ge 5,x+2y\ge 6,y\ge 0.$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints $x\ge 3,x+y\ge 5,x+2y\ge 6,y\ge 0.$ is as follows,

The corner points of the feasible region are $A(6,0),B(4,1),C(3,2)$

The value of these points at these corner points are :

The feasible region is unbounded, therefore 1 may or may not be the maximum value of Z.

For this, we draw $-x+2y>1$ and check whether resulting half plane has a point in common with a feasible region or not.

We can see the resulting feasible region has a common point with a feasible region.

Hence , Z =1 is not maximum value , Z has no maximum value.

Question 10: Solve the following Linear Programming Problems graphically: Maximise $Z=x+y,$ Subject to $x-y\le -1,-x+y\le 0,x,y,\ge 0.$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints $x-y\le -1,-x+y\le 0,x,y,\ge 0.$ is as follows,

There is no feasible region and thus, Z has no maximum value.