SOLUTION: A farmer is going to divide her 50 acre farm between two crops. Seed for crop A costs $50 per acre. Seed for crop B costs $25 per acre. The farmer can spend at most $2,250 on seed.
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-> SOLUTION: A farmer is going to divide her 50 acre farm between two crops. Seed for crop A costs $50 per acre. Seed for crop B costs $25 per acre. The farmer can spend at most $2,250 on seed.
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Question 1176645: A farmer is going to divide her 50 acre farm between two crops. Seed for crop A costs $50 per acre. Seed for crop B costs $25 per acre. The farmer can spend at most $2,250 on seed. If crop B brings in a profit of $90 per acre, and crop A brings in a profit of $220 per acre, how many acres of each crop should the farmer plant to maximize her profit? Write the objective function then use the feasible region shown in the graph below to maximize it.
Let x = the number of acres of crop A
Let y = the number of acres of crop B
Objective Function: P =
Constraints:
x≥0, y≥0
x+y≤50
50x+25y2250
The Farmer should Plant acres of crop A and acres of crop B
On the same set of axes, graph all four of the constraint inequalities. However, I recommend that you graph them by using the opposite sense, that is if the inequality is "less than or equal", graph "greater than" instead. That way, instead of the feasible area being where all four solution sets overlap which is sometimes hard to see, what you will have is a feasibility area that is completely clear of shading and is easy to discern.
Once you have the feasibility area defined, check the value of the objective function at each of the vertices to determine which one maximizes the objective and therefore is the solution to the problem.
John
My calculator said it, I believe it, that settles it
From
I > Ø
