Question

Mrs. Griffin wants to plant soybeans and corn on 100 acres of land. Soybeans require 6 hours of labor per acre, and corn requires 8 hours of labor per acre. If Mrs. Griffin has 660 hours available, how many acres of each crop should she plant?

Please include how to solve the problem.

Answer 1

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Answer:

• soybeans: 70 acres
• corn: 30 acres

Explanation:

Let x represent acres of soybeans. Then 100-x is the number of acres of corn. All of Mrs. Griffin's labor hours will be used when ...

6x +8(100-x) = 660

-2x +800 = 660 . . . . . simplify

x -400 = -330 . . . . . . . divide by -2

x = 70 . . . . . . . . . . . . . add 400

Mrs. Griffin should plant 70 acres of soybeans and 30 acres of corn.

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Additional comment

The problem is not well-specified, as there is no indication of the reason why Mrs Griffin should plant any number of acres of either crop. So, we assume the intent is to use all of the acres and labor hours available.

This makes the problem equivalent to a system of two equations in two variables. We effectively substituted (100-x) for the acres of corn, which might otherwise be its own variable.

Often, you're interested in maximizing or minimizing some objective function given these constraints. If we wanted to minimize labor, for example, we would plant all soybeans, since there is less labor involved. That would only require 600 hours of labor, instead of 660.