Afarm 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?

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) = 40,500clear the parentheses

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

5c = 19,500divide each side by 5

c = 3,900

and substitute that back into the first equation:

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

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

s = 3,300

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.