Question

Minimize and miximize Z = x + 2y subject to constraints are x + 2y $\ge$ 100, 2x - y $\le$ 0, 2x + y $\le$ 200 and x, y $\ge$ 0.

Answer 1

Our problem is to minimize and miximize

Z = x + 2y ...(i)

Subject to constraints are x + 2y $\ge$ 100 .....(ii)

2x - y $\le$ 0 ...(iii)

2x + y $\le$ 200 .....(iv)

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

Firstly, draw the graph of the line x + 2y = 100

$\begin{array}{ccc}x& 0& 100\\ y& 50& 0\end{array}$

Putting (0, 0) in the inequality x + 2y $\ge$ 100, we have

$0+2\times 0\ge 100$

$\Rightarrow 0\ge 100$ (which is false)

So, the half plane is away from the origin.

Secondly, draw the graph of line 2x - y = 0

$\begin{array}{ccc}x& 0& 10\\ y& 0& 20\end{array}$

Putting (5, 0) in the inequality 2x - y $\le$ 0, we have $2\times 5-0\le 0\Rightarrow 10\le 0$ (which is false)

So, the half plane is towards Y-axis.

Thirdly, draw the graph of line 2x + y = 200

$\begin{array}{ccc}x& 0& 100\\ y& 200& 0\end{array}$

Putting (0, 0) in the inequality 2x + y $\le$ 200,

we have $2\times 0+0\le 200\Rightarrow 0\le 200$ (which is true)

So, the half plane is towards the origin, Since, x, y $geq$ 0

So, the feasible region lies in the first quadrant.

On solving equations 2x - y = 0 and x + 2y = 100, we get B(20, 40) and on solving equations 2x - y = 0 and x + 2y = 200, we get C(50, 100)

$\therefore$ Feasible region is ABCDA,

The corner points of the feasible region are A(0, 50), B(20, 40), C(50, 100) and D(0, 200). The values of Z at these points are as follows:

$\begin{array}{cc}Cornerpoint& Z=x+2y\\ A(0,50)& 100\to Minimum\\ B(20,40)& 100\to Minimum\\ C(50,100)& 250\\ D(0,200)& 400\to Minimum\end{array}$

The maximum value of Z is 400 at D(0, 200) and the minimum value of Z is 100 at all the points on the line segment joining A(0, 50) and B(20, 40).