Answer to Question #237086 in Operations Research for lavanya

Question

. A tailor has the following material available: 16 sq. yd. cotton, 11 sq. yd. silk and 15 sq. yd. wool. A suit requires 2 sq. yd. cotton, 1 sq. yd. silk and 1 sq. yd. wool. A gown requires 1 sq. yd. cotton, 2 sq. td. Silk and 3 sq. yd. wool. If a suit sells for Rs. 300/- and a gown for Rs. 500/-, how many of each garment should the tailor mak to obtain the maximum amount of profit?

Answer 1

Let $x=$ the number of suits, $y=$ the number of gowns.

Maximize $Z=300x+500y$

subject to

$2x+y\leq16$

$x+2y\leq11$

$x+3y\leq 15$

$x\geq0, y\geq0$

$OA: x=0, 0\leq y\leq5, Z=500y$

$AB: y=5-\dfrac{1}{3}x, 0\leq x\leq3, Z=\dfrac{400}{3}x+2500$

$BC: y=\dfrac{11}{2}-\dfrac{1}{2}x, 3\leq x\leq7, Z=50x+2750$

$CD: y=16-2x, 7\leq x\leq8, Z=-700x+8000$

$OD: y=0, 0\leq x\leq8, Z=300x$

$\def\arraystretch{1.5} \begin{array}{c:c:c} Extreme & Lines\ through & Objective \ function \\ point & exreme\ point & value\\ \hline O (0,0) & x=0, & Z=500(0) \\ & y=0 &=0\\ \hdashline A (0,5) & x=0, & Z=500(5)\\ & y=5-\dfrac{1}{3}x & =2500\\ \hdashline B (3,4) & y=5-x/3, & Z=50(3)+2750\\ & y=11/2-x/2 & =2900\\ \hdashline C (7,2) & y=11/2-x/2, & Z=-700(7)+8000\\ & y=8-2x & =3100\\ \hdashline D (8,0) & y=8-2x & Z=300(8)\\ & y=0 & =2400\\ \hdashline \end{array}$

The maximum value of the objective function $Z=3100$  occurs at the extreme point $(7, 2).$

Hence, the optimal solution to the given LP problem is :

 $x=7,y=2$  and $max Z=3100.$