question archive Consider the following linear programming problem: Maximize Z = 4X1 + 10X2 Subject to: 3X1 + 4X2 =< 480 4X1 + 2X2 =< 360 all variables => 0 The feasible corner points are (48,84),(0,120),(0,0),(90,0)
Consider the following linear programming problem: Maximize Z = 4X1 + 10X2 Subject to: 3X1 + 4X2 =< 480 4X1 + 2X2 =< 360 all variables => 0 The feasible corner points are (48,84),(0,120),(0,0),(90,0)
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Consider the following linear programming problem:
Maximize Z = 4X1 + 10X2
Subject to: 3X1 + 4X2 =< 480
4X1 + 2X2 =< 360
all variables => 0
The feasible corner points are (48,84),(0,120),(0,0),(90,0). What is the maximum possible value for the objective function?
a. 1032
b. 1200
c. 360
d. 1600
e. none of the above.
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Answer: b .
Substitute the vertices to the objective function
?Z=4X1?+10X2??
(48,84)
Z = 4(48) + 10(84) = 192 + 840 = 1,032
(0,120)
Z = 4(0) + 10(120) = 1,200
(0, 0)
Z = 4(0) + 10(0) = 0
(90, 0)
Z = 4(90) + 10(0) = 360
Thus, the maximum possible value for the objective function is 1,200
The answer is B.
