SOLUTION: Minimize z=3x+2y 2y+6x>=20 6y+2y>=20 subject to 2y+2x>=12 x>=0 y>=0

Question 1147366: Minimize z=3x+2y
2y+6x>=20
6y+2y>=20
subject to 2y+2x>=12
x>=0
y>=0
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

A corrected statement of the problem:

Minimize z=3x+2y
subject to
(1) 2y+6x>=20
(2) 6y+2x>=20 (NOT 6y+2y>=20)
(3) 2y+2x>=12
(4) x>=0
(5) y>=0

A graph; (1) is red; (2) is green; (3) is blue....

A bit of algebra shows the corners of the feasibility region are
(0,10)
(2,4)
(4,2)
(10,0)

Virtually all resources will tell you to evaluate the objective function at all corners of the feasibility region:
(0,10) --> 3x+2y = 0+20 = 20
(2,4) --> 3x+2y = 6+8 = 14
(4,2) --> 3x+2y = 12+4 = 16
(10,0) --> 3x+2y = 30+0 = 30

The minimum value of the objective function is 14, at (2,4).

In fact, it is NOT necessary to evaluate the objective function at every corner of the feasibility region. The corner where the objective function has its minimum value can be determined by comparing the slopes of the constraint lines to the slope of the objective function.

(1): slope = -3
(2): slope = -1/3
(3): slope = -1
objective function: -3/2

The slope of the objective function is between the slopes of constraint lines (1) and (3); the minimum value of the objective function will be at the intersection of those two constraint lines -- at (2,4).
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