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Consider the following linear program.

Min 3A + 4B
s.t.
1A + 3B ≥ 9
1A + 1B ≥ 7
A, B ≥ 0

User LampShade
by
7.2k points

1 Answer

6 votes

Answer:

Minimum z =22 at A =6, B=1

See the graph below variable A is on the horizontal axis and variable B on the vertical axis

Explanation:

We can solve using the graphical method

Min z = 3A + 4B

s.t.

1A + 3B ≥ 9

1A + 1B ≥ 7

A, B ≥ 0

Draw two lines using the constraint equations

1A + 3B = 9

1A + 1B = 7

The feasible region for the inequality constraints is the region that satisfies all 4 constraints below

1A + 3B ≥ 9

1A + 1B ≥ 7

A, B ≥ 0

This is the heavily shaded region in the graph. The minimum will be at one of the corner points

There are three corner points on the graph

(0,7), (6,1) and (9,0)

Plug in these coordinate values with A being the first and B being second in the parentheses and see which set of values makes the objective function the minimum

Point (7,0) ==> z = 3(0) + 4(7) = 28

Point (6,1) ==> z = 3(6) + 4(1) = 22

Point (9,0) ==> z = 3(9)+4(0) = 27

Consider the following linear program. Min 3A + 4B s.t. 1A + 3B ≥ 9 1A + 1B ≥ 7 A-example-1
User Rittergig
by
6.4k points