SOLUTION: Solve using graphical method. Maximize: 7/6 x + 13/10 y Subject to: x/30 + y/40 ≤ 1 x/28 + y/35 ≤ 1 x/30 + y/25 ≤ 1

Question 1204861: Solve using graphical method.
Maximize: 7/6 x + 13/10 y
Subject to:
x/30 + y/40 ≤ 1
x/28 + y/35 ≤ 1
x/30 + y/25 ≤ 1
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

The boundary line of is
Compare that to the intercept template

The x and y intercepts for are located at (a,0) and (0,b) respectively.
Note that 'a' and 'b' must be nonzero.

So something like has its x and y intercepts at (30,0) and (0,40).
Draw a straight line through those two points.
This boundary line is solid because of the "or equal to" in the inequality sign.
Shade below the boundary due to the "less than" part.
Or you can use a test point like (0,0) to try out.

Follow this idea for the other inequalities as well. Then pay careful attention where the three regions overlap.
That overlapped region has corner vertex points that will be plugged into the objective function
I'll let the student take over from here.

Answer by ikleyn(52792)   (Show Source): You can put this solution on YOUR website!
.

At this site, there is a lesson
    - Solving minimax problems by the Linear Programming method
where many similar problems are solved with explanations.

Use this lesson to learn the method.

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