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Question 1058105: Linear Programming - The Graphical Method
A farmer is going to divide her 60 acre farm between two crops. Seed for crop A costs $20 per acre. Seed for crop B costs $10 per acre. The farmer can spend at most $700 on seed. If crop B brings in a profit of $60 per acre, and crop A brings in a profit of $180 per acre, how many acres of each crop should the farmer plant to maximize her profit?
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website! .
Linear Programming - The Graphical Method
A farmer is going to divide her 60 acre farm between two crops.
Seed for crop A costs $20 per acre. Seed for crop B costs $10 per acre.
The farmer can spend at most $700 on seed. If crop B brings in a profit of $60 per acre, and crop A brings in a profit of $180 per acre,
how many acres of each crop should the farmer plant to maximize her profit?
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Your restriction equations are
x + y = 60 acres (1) (for areas; x = the area for crop A; y = the area for crop B).
20x + 10y = 700 (2) for the seed cost.
Rewrite:
x + y = 60, (1')
2x + y = 70. (2')
The additional obvious restrictions are x >= 0 and y >= 0.
The plot is
Plots y = 60-x and y = 70-2x
Your area is a quadrilateral in QI restricted by the axes x- an y- and by the red and the green straight lines.
Your objective function is the profit z = 180x + 60y.
According to the conception/ideology of the linear programming method, you need to evaluate your
objective function 180x + 60y in three points:
y-intercept (x,y) = ( 0, 60);
x-intercept (x,y) = (35, 0);
and the point (x,y), which is the solution of the system (1'), (2') (the intersection point of the red and the green straight lines).
Then choose that point of the three, where the objective function is maximal.
Having this plan/instruction, you can complete the assignment on your own.
Good luck!
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