Question

Minimize Z = x + 2y, subject to constraints are 2x + y $\ge$ 3, x + 2y $\ge$ 6 and x, y $\ge$ 0. Show that the minimum of Z occurs at more than two points.

Answer 1

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

Subject to the constraints are

2x + y $\ge$ 3 ......(ii)

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

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

Firstly, draw the graph of the line 2x + y = 3

$\begin{array}{ccc}x& 0& \frac{3}{2}\\ y& 3& 0\end{array}$

Putting (0, 0) in the inequality 2x + y $\ge$ 3, we have

$2\times 0+0\ge 3\Rightarrow 0\ge 3$ (which is false)

So, the half plane is away from the origin. Since, x, y $\ge$ 0

So, the feasible region lies in the first quadrant.

Secondly, draw the graph of the 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.

The intersection point of the lines x + 2y = 6 and 2x + y = 3 is B (0, 3).

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

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

It can be seen that the value of Z at points A and B is same. If we take any other point such as (2, 2) on line x + 2y = 6, then z = 6, Thus, the minimum value of Z occurs for more than 2 points. Therefore, the value of Z is minimum at every point on the line, x + 2y = 6.