Linear Programming Question 4
Question: Maximize $z=4 x+6 y$, subject to $3 x+2 y \leq 12, x+y \geq 4, x \geq 0, y \geq 0$
Options:
A) 16 at (4,0)
B) 24 at (0,4)
C) 24 at (6,0)
D) 36 at (0,6)
Show Answer
Answer:
Correct Answer: D
Solution:
we have, minimized $ Z=4x+6y $ Subject to $ 3x+2y\le 12,x+y\ge 4,x,y\ge 0 $ Let $ {\ell_1}:3x+2y=12 $
$ {\ell_2}:x+y=4 $
$ {\ell_3}:x=0 $ and $ {\ell_4}:y=0 $ Shaded portion ABC is the feasible region, where A (4, 0), C (0, 4), B (0, 6). Now maximize $ Z=4x+6y $
$ ZatA(4,0)=4(4)+6(0)=16 $
$ ZatB(0,6)=4(0)+6(6)=36 $
$ ZatC(0,4)=4(0)+6(4)=24 $ Thus, Z is maximized at B (0, 6) and its maximum value is 36.
BETA
