Question

Maximise $z=8x+9y$ subject to the constraints given below :
$2x+3y\le 6;3x-2y\le 6;y\le 1;x\ge 0;y\ge 0$

Answer 1

Given constraints are
$2x+3y\le 6$
$3x-2y\le 6$
$y\le 1$
$x,y\ge 0$
For graph of $2x+3y\le 6$
We draw the graph of $2x+3y=6$
$x$ $0$ $3$
$y$ $2$ $0$

Putting Origin in the equation, we get:
$2\times 0+3\times 0\le 6⟹(0,0)$ satisfy the constraints.
Hence, feasible region lie towards origin side of line.
For graph of $3x-2y\le 6$
We draw the graph of line $3x-2y=6$.
$x$ $0$ $2$
$y$ $-3$ $0$
Putting Origin in the equation, we get:
$3\times 0-2\times 0\le 6$ $⟹$ Origin $(0,0)$ satisfy $3x-2y=6$.
Hence, feasible region lie towards origin side of line.
For graph of $y\le 1$
We draw the graph of line $y=1$, which is parallel to $x-$axis and meet $y-$axis at $1$.
Putting Origin in the equation, we get:
$0\le 1⟹$ feasible region lie towards origin side of $y=1$.
Also, $x\ge 0,y\ge 0$ says feasible region is in Ist quadrant
Therefore, $OABCDO$ is the required feasible region, having corner point $O(0,0),A(0,1),B(\frac{3}{2},1),C(\frac{30}{13},\frac{6}{13}),D(2,0)$.
Here, feasible region is bounded. Now the value of objective function $Z=8x+9y$ is obtained as.
Corner Points $Z=8x+9y$
$O(0,0)$ $0$
$A(0,1)$ $9$
$B(\frac{3}{2},1)$ $21$
$C(\frac{30}{13},\frac{6}{13})$ $22.6$
$D(2,0)$ $16$
$Z$ is maximum when $x=\frac{30}{13}$ and $y=\frac{6}{13}$.