Question

Maximize Z = - x + 2y, subject to the constraints are x $\ge$ 3, x + y $\ge$ 5, x + 2y $\ge$ 6 and x, y $\ge$ 0.

Answer 1

Our problem is to maximize Z = -x + 2y .........(i)

Subject to constraints are x $\ge$ 3 ..........(ii)

x + y $\ge$ 5 ..........(iii)

x + 2y $\ge$ 6 ......(iv)

x $\ge$ 0, y $\ge$ 0 .........(v)

Firstly, draw the graph of the line x + y = 5

$\begin{array}{ccc}x& 0& 5\\ y& 5& 0\end{array}$

Putting (0, 0) in the inequality x + y $\ge$ 5, we have

0 + 0 $\ge$ 5

$\Rightarrow$ 0 $\ge$ 5 (which is false)

So, the half plane is away from the origin.

Secondly, draw the graph of line x + 2y = 6

$\begin{array}{ccc}x& 0& 6\\ y& 3& 0\end{array}$

Putting (0, 0) in the inequality x + 2y $\ge$ 6, we have

$0+2\times 0\ge 6$

$\Rightarrow 0\ge 6$ (which is false)

So, the half plane is away from the origin,

Thirdly, draw the graph of the line - x + 2y = 1

$\begin{array}{ccc}x& 0& -1\\ y& \frac{1}{2}& 0\end{array}$

Putting (0, 0) in the inequality

- x + 2y > 1, we have

$-0+2\times 0>1$

$\Rightarrow 0>1$ (which is false)

So, the half plane is away from the origin.

Since, $x\ge 3,y\ge 0$

So, the feasible region lies in the first quadrant. The points of intersection of lines x = 3 and - x + 2y = 1 is C(3, 2) and lines x + 2y = 6 and x + y = 5 is E(4, 1).

It can be seen that the feasible region is unbounded.

The corner points of the feasible region are A(6, 0), B(4, 1) and C(3, 2). The values of Z at these points are as follows:

$\begin{array}{cc}Cornerpoint& Z=-x+2y\\ A(6,0)& -6\\ B(4,1)& -2\\ C(3,2)& 1\leftarrow Maximum\end{array}$

As the feasible region is unbounded therefore, Z = 1 may or may not be the maximum value. For this, we graph the inequality, - x + 2y > 1 and check whether the resulting half plane has points in common with the feasible region or not.

The resulting feasible region has points in common with the feasible region. Therefore, Z = 1 is not the maximum value.

Hence, Z has no maximum value.