A farm co-op wants to grow corn and soybeans. Each acre of corn requires 9 gallons of fertilizer and 0.75 hours to harvest. Each acre of soybeans requires 3 gallons of fertilizer and 1 hour to harvest. The co-op has available at most 40,500 gallons of fertilizer and 5,250 hours of labor for harvesting. If the profits per acre are $60 for corn and $40 for soybeans, how many acres of each should the co-op plant to maximize their profit? What is the maximum profit?

Question

Mathematics

Deacon

12 October, 22:21

A farm co-op wants to grow corn and soybeans. Each acre of corn requires 9 gallons of fertilizer and 0.75 hours to harvest. Each acre of soybeans requires 3 gallons of fertilizer and 1 hour to harvest. The co-op has available at most 40,500 gallons of fertilizer and 5,250 hours of labor for harvesting. If the profits per acre are $60 for corn and $40 for soybeans, how many acres of each should the co-op plant to maximize their profit? What is the maximum profit?

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Mallory Lam · Answer 1

3,900 acres of corn and 3,300 acres of soybeans

Use c = corn acres and s = soybean acres. Then the hours are split by the equation:

0.5c + s = 5,250

and the fertilizer is split by the equation:

7c + 4s = 40,500

If you solve the hours for "s" and substitute into the second, you have:

7c + 4 * (5,250 - 0.5c) = 40,500 ... clear the parentheses

7c + 21,000 - 2c = 40,500 ... subtract 21,000 from each side; simplify the left side

5c = 19,500 ... divide each side by 5

c = 3,900

and substitute that back into the first equation:

0.5 * 3,900 + s = 5,250 ... simplify the left side

1,950 + s = 5,250 ... subtract 1,950 from each side

s = 3,300