Question

Consider the linear inequalities
$2x+3y\le 6,2x+y\le 4,x\ge 0,y\ge 0$
(a) Mark the feasible region.
(b) Maximise the function $z=4x+5y$ subject to the given constraints.

Answer 1

Convert the linear inequalities into equalities to find the corner points

$2x+3y=6$ ......... $\left(i\right)$

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

From $\left(i\right)$

$x=0⟹y=2$ and $y=0$ when $x=3$

So, the points $\left(i\right)$ are $(0,2)$ and $(3,0)$ lie on the line given in $\left(i\right)$

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

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

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

The shaded part shows the feasible region.

The lines intersect at $(\frac{3}{2},1)$ and other corner points of the region are $(0,2),(2,0),(0,0)$.

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

Corner points $z=4x+5y$

$(0,0)$ $0$

$(2,0)$ $8$

$(0,2)$ $10$

$(\frac{3}{2},1)$ $11$

Thus, $z$ is maximum at $(\frac{3}{2},1)$