Question

Solve graphically the linear inequalities
$2x+3y\le 7,x+2y\le 4$
$x\ge 0,y\ge 0$
If $z=6x+5y$ is the objective function, find its maximum value.

Answer 1

$z=6x+5y$

Convert the linear inequalities into equalities

$2x+3y=7$ ......... $\left(i\right)$

$x+2y=4$ ........... $\left(ii\right)$

From $\left(i\right)$

$x=0⟹y=2.3$ and $y=0$ when $x=3.5$

So, the points $(0,2.3)$ and $(3.5,0)$ lie on the line given in $\left(i\right)$

From $\left(ii\right)$, we get the points

$(0,2)$ and $(4,0)$

Let's plot these point and we get the graph shown above.

The shaded part shows the feasible region.

The lines intersect at $(2,1)$ and other corner points of the region are $(0,2),(3.5,0),(0,0)$.

To find the maximum value of $z$, we need to find the value of $z$ at the corner points

Corner points $z=6x+5y$

$(0,0)$ $0$

$(3.5,0)$ $21$

$(0,2)$ $10$

$(2,1)$ $17$

Thus, $z$ is maximum at $(3.5,0)$ and its maximum value is $21$.