Linear Programming Question 120
Question: The maximum value of $ z=2x+5y $ subject to the constraints $ 2x+5y\le 10,x+2y\ge 1,x-y\le 4,x\ge y\ge 0, $ Occurs at
Options:
A) Exactly one pint
B) Exactly two points
C) Infinitely many points
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
We find that the feasible region is on the same side of the line $ 2x+5y=10 $ as the origin, on the same side of the line $ x-y=4 $ as the origin and on the opposite side of the line $ x+2y=1 $ from the origin. Moreover, the lines meet the coordinate axes at (5, 0), (0, 2); (1, 0), (0, 1/2) and (4, 0). The lines $ x=4 $ and $ 2x+5y=10 $ intersect at $ ( \frac{30}{7},\frac{2}{7} ) $
The values of the objective function at the vertices of the pentagon are:
(i) $ Z=0+\frac{5}{2}=\frac{5}{2} $
(ii) $ Z=2+0=2 $
(iii) $ Z=8+0=8 $
(iv) $ Z=\frac{60}{7}+\frac{10}{7}=\frac{70}{7}=10 $
(v) $ Z=0+10=10 $
The maximum value 10 occurs at the points D (30/7, 2/7) and E (0, 2). Since D and E are adjacent vertices, the objective function has the same maximum value 10 at all the points on the line segment DE.
BETA
