Question

Minimize and maximize Z = 5x + 10y subject to constraints are x + 2y $\le$ 120, x + y $\ge$ 60, x - 2y $\ge$ 0 and x, y $\ge$ 0.

Answer 1

Our problem is to minimize and maximize

Z = 5x + 10y ........(i)

Subject to constraints x + 2y $\le$ 120 ......(ii)

x + y $\ge$ 60 .........(iii)

x - 2y $\ge$ 0 .......(iv)

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

Firstly, draw the graph of the lines x + 2y = 120

$\begin{array}{ccc}x& 0& 120\\ y& 60& 0\end{array}$

Putting (0, 0) in the inequality x + 2y $\le$ 120, we have 0+0 $\le$ 120 (which is true)

So, the half plane is towards the origin. Secondly, draw the graph of the line x + y = 60

$\begin{array}{ccc}x& 0& 60\\ y& 60& 0\end{array}$

Putting (0, 0) in the inequality x + y $\ge$ 60, we have 0 + 0 $\ge$ 60 (which is false)

So, the half plane is away from the origin.

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

$\begin{array}{ccc}x& 0& 10\\ y& 0& 5\end{array}$

Putting (5, 0) in the inequality x - 2y $\ge$ 0, we have

$5-2\times 0\ge 0\Rightarrow 5\ge 0$ (which is true)

So, the half plane is towards the X - axis, Since, x, y $\ge$ 0

So, the half feasible region lies in the first quadrant.

$\therefore$ Feasible region is ABCDA.

On solving equations x - 2y = 0 and x + y = 60, we get D(40, 20)

and on solving equations x - 2y = 0 and x + 2y = 120, we get C(60, 30).

The corner points of the feasible region are A(60, 0), B(120, 0), C(60, 30) and D(40, 20). The values of Z at these points are as follows:

$\begin{array}{cc}Cornerpoint& Z=5x+10y\\ A(60,0)& 300\to Minimum\\ B(120,0)& 600\to Minimum\\ C(60,30)& 600\to Minimum\\ D(40,20)& 400\end{array}$

The minimum value of Z is 300 at (60, 0) and the maximum value of Z is 600 at all the points on the line segment joining the points (120, 0) and (60, 30).