Question

Minimize Z = -3x + 4y, subject to constraints are $x+2y\le 8,3x+2y\le 12,x\ge 0andy\ge 0$.

Answer 1

Our problem is to minimize

Z = -3x + 4y .........(i)

Subject to constraints x + 2y $\le$ 8 .......(ii)

3x + 2y $\le$ 12 ............(iii)

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

Firstly, draw the graph of the line, x + 2y = 8

$\begin{array}{ccc}x& 0& 8\\ y& 4& 0\end{array}$

Putting (0, 0) in the inequality x + 2y $\le$ 8, we have

$0+0\le 8\Rightarrow 0\le 8$ (which is true)

So, the half plane is towards the origin.

Since, x, y $\le$ 0

So, the feasible region lies in the first quadrant.

Secondly, draw the graph of the line, 3x + 2y = 12

$\begin{array}{ccc}x& 0& 4\\ y& 6& 0\end{array}$

Putting (0, 0) in the inequality 3x + 2y $\le$ 12, we have $3\times 0+2\times 0\le 12$

$\Rightarrow 0\le 12$ (which is true)

So, the half plane is towards the origin.

$\therefore$ Feasible region is OABCO.

On solving equations x + 2y = 8 and 3x + 2y = 12, we get x = 2 and y = 3

$\therefore$ Intersection point B is (2, 3)

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

$\begin{array}{cc}Cornerpoint& Z=-3x+4y\\ O(0,0)& 0\\ A(4,0)& -12\to Mininum\\ B(2,3)& 6\\ C(0,4)& 16\end{array}$

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