What is the minimum value of 3x + 5y in the feasible region? (0, 7), (3, 7), (6, 3), (6, 0)

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asked Jul 24, 2019 6.5k views

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William Chan · Answer 1 · 2019-07-24T17:22:52+0000

Answer:

The minimum value is

It occurs at
P(6,0)

Explanation:

The given objective function is
P(x,y)=3x+5y

The vertices of the feasible region are
(0,7),(3,7),(6,3),(6,0) .

We plug in the vertices in to feasible region and evaluate to obtain,

P(0,7)=3(0)+5(7)

P(0,7)=0+35

P(0,7)=35

P(3,7)=3(3)+5(7)

P(3,7)=9+35

P(3,7)=44

P(6,3)=3(6)+5(3)

P(6,3)=18+15

P(6,3)=33

P(6,0)=3(6)+5(0)

P(6,0)=18+0

P(6,0)=18

Hartmut Pfarr · Answer 2 · 2019-07-28T10:20:06+0000

Answer:

The minimum value is

It occurs at
P(6,0)

Explanation:

The given objective function is
P(x,y)=3x+5y

The vertices of the feasible region are
(0,7),(3,7),(6,3),(6,0) .

We plug in the vertices in to feasible region and evaluate to obtain,

P(0,7)=3(0)+5(7)

P(0,7)=0+35

P(0,7)=35

P(3,7)=3(3)+5(7)

P(3,7)=9+35

P(3,7)=44

P(6,3)=3(6)+5(3)

P(6,3)=18+15

P(6,3)=33

P(6,0)=3(6)+5(0)

P(6,0)=18+0

P(6,0)=18

What is the minimum value of 3x + 5y in the feasible region? (0, 7), (3, 7), (6, 3), (6, 0)

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