Question

Maximize $f=3x+y$ subject to the constraints
$8x+5y\le 40$; $4x+3y\ge 12$; $x\ge 0$ and $y\ge 0$.

Answer 1

Solution:

Converting inequalities into equations:

$8x+5y=40$

$x$	$0$	$8$
$y$	$5$	$0$

$4x+3y=12$

$x$	$0$	$4$
$y$	$3$	$0$

At $(3,0):f\left(x\right)=3x+y=3\times 3+0=9$

At $(5,0):f\left(x\right)=3x+y=3\times 5+0=15$

At $(0,8):f\left(x\right)=3x+y=3\times 0+8=8$

At $(0,4):f\left(x\right)=3x+y=3\times 0+4=4$

Thus, at $(5,0),f$ is maximum.