Chapter 12: Linear Programming

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 $B(0,4)$. 

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 $A(4,0)$.

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$C(\frac{20}{19},\frac{45}{19})$.

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 $B(\frac{3}{2},\frac{1}{2})$. 

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 $B(4,3)$. 

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 $x+2y=6$.

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 $C(60,30)$.

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 $B(20,40)$.

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.

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.