Question

Maximize $z=3x+5y$, Subject to $x+4y\le 24$, $3x+y\le 21,x+y\le 9,x\ge 0$. Also find maximum value of $z$.

Answer 1

Shaded portion OABCD is the feasible region,

Where $O(0,0),A(7,0),D(0,6)$

For B:

$3x+y=\mathrm{21....}\left(i\right)$

$x+y=\mathrm{9......}\left(ii\right)$

Subtract (ii) from (i) we get

$3x+y-x-y=21-9$

$2x=12$

$x=6$

$y=9-6=3$

$\therefore B(6,3)$

For C:

$x+y=\mathrm{9....}\left(iii\right)$

$x+4y=\mathrm{24......}\left(iv\right)$

Subtract (ii) from (i) we get

$x+y-x-4y=9-24$

$-3y=-15$

$y=5$

$x=9-5=4$

$\therefore C(4,5)$

$Z=3x+5y$

$ZatO(0,0)=3\left(0\right)+5\left(0\right)=0$

$ZatA(7,0)=3\left(7\right)+5\left(0\right)=21$

$ZatB(6,3)=3\left(6\right)+5\left(3\right)=33$

$ZatC(4,5)=3\left(4\right)+5\left(5\right)=37$

$ZatD(0,6)=3\left(0\right)+5\left(6\right)=30$

Thus, Z is maximized at C(4,5) and its maximum value is 37.