Question

Maximize Z = 3x + 2y, subject to constraints are $x+2y\le 10,3x+y\le 15andx,y\ge 0$.

Answer 1

Our problem is to maximize,

Z = 3x + 2y .......(i)

Subject to constraints x + 2y $\le$ 10 ....(ii)

3x + y $\le$ 15 ....(iii)

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

Firstly, draw the graph of the line x + 2y = 10

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

Putting (0, 0) in the inequality x + 2y $\le$ 10, we have

$0+2\times 0\le$ 10 $\Rightarrow$ 0 $\le$ 10 (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.

Secondly, draw the graph of the line $3x+y=15$

$\begin{array}{ccc}x& 0& 5\\ y& 15& 0\end{array}$

Putting (0, 0) in the inequality 3x + y $\le$ 15, we have $3\times 0+0\le 15\Rightarrow 0\le 15\left(whichistrue\right)$

So, the half plane is towards the origin.

On solving equations $x+2y=10and3x+y=15$,

we get x = 4 and y = 3

$\therefore$ Intersection point B is (4, 3).

$\therefore$ Feasible region is OABCO. The corner points of the feasible region are O(0, 0), A(5, 0), B(4, 3) and C(0, 5). The values of Z at these points are as follows:

$\begin{array}{cc}Cornerpoint& Z=3x+2y\\ O(0,0)& 0\\ A(5,0)& 15\\ B(4,3)& 18\to Maximum\\ C(0,5)& 10\end{array}$

Therefore, the maximum value of Z is 18 at the point B(4, 3).