Question

Maximize Z = 3x + 4y, subject to the constraints are x + y $\le$ 4, x $\ge$ 0 and y $\ge$ 0.

Answer 1

Our problem is to maximize Z = 3x + 4y ..............(i)

Subject to constraints x + y $\le$ 4 ..............(ii)

x $\ge$ 0 and y $\ge$ 0 ..............(iii)

Firstly, draw the graph of the line x + y = 4

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

Putting (0, 0) in the inequality x + y $\le$ 4, we have

0 + 0 $\le$ 4 $\Rightarrow$ 0 $\le$ 4 (which is true)

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

So, the feasible region lies in the first quadrant.

$\therefore$ Feasible region is OABO.

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

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

Therefore, the maximum value of Z is 16 at the point B(0, 4).

Note While plotting the graph, please be careful about the inequalities in which direction we have to plot this.