Linear Programming Question 8
Question: $ Z=7x+y, $ subject to $ 5x+y\ge 5,x+y\ge 3,x\ge 0,y\ge 0. $ The minimum value of Z occurs at
Options:
A) (3, 0)
B) $ ( \frac{1}{2},\frac{5}{2} ) $
C) (7, 0)
D) (0, 5)
Show Answer
Answer:
Correct Answer: D
Solution:
We have, maximize $ Z=7x+y. $ subject to: $ 5x+y\ge 5,x+y\ge 3,x,y\ge 0. $ Let $ {\ell_1}:5x+y=5 $
$ {\ell_2}:x+y=3 $
$ {\ell_3}:x=0 $ and $ {\ell_4}:y=0 $
Shaded portion is the feasible region,
Where $ A(3,0),B( \frac{1}{2},\frac{5}{2} ),C(0,5) $
For B: solving $ {\ell_1} $ and $ {\ell_2} $ , we get $ B( \frac{1}{2},\frac{5}{2} ) $
Now maximize $ Z=7x+y $
Z at B $ ( \frac{1}{2},\frac{5}{2} )=7( \frac{1}{2} )+\frac{5}{2}=6 $
Z at C (0, 5) =7(0) +5=5
Thus Z, is minimized at C (0, 5) and its minimum value is 5
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