A farm co-op has 6,000 acres available to plant with corn and soybeans. Each acre of corn requires 9 gallons of fertilizer/herbicide and 0.75 hour of labor to harvest. Each acre of soybeans requires 3 gallons of fertilizer/herbicide and 1hour of labor to harvest. The co-op has available at most 40,501 gallons of fertilizer/herbicide and at most 5,250 hours of labor for harvesting. Find the maximum profit if the profits per acre are $75 for corn and $40 for soybeans. Round your answer to the nearest cent if necessary.
A)$210,000.00
B)$393,750.00
C)$371,255.83
D)$180,004.44
E)$318,744.17

Question

A farm co-op has 6,000 acres available to plant with corn and soybeans. Each acre of corn requires 9 gallons of fertilizer/herbicide and 0.75 hour of labor to harvest. Each acre of soybeans requires 3 gallons of fertilizer/herbicide and 1hour of labor to harvest. The co-op has available at most 40,501 gallons of fertilizer/herbicide and at most 5,250 hours of labor for harvesting. Find the maximum profit if the profits per acre are $75 for corn and $40 for soybeans. Round your answer to the nearest cent if necessary. ​
A)$210,000.00
B)$393,750.00
C)$371,255.83
D)$180,004.44
E)$318,744.17

Quizplus · Accepted Answer

To maximize profit, we need to determine the optimal number of acres of corn and soybeans to plant. Let x be the number of acres of corn and y be the number of acres of soybeans. Our system of constraints is: 9x + 3y <= 40,501 (fertilizer/herbicide constraint) 0.75x + y <= 5,250 (labor constraint) x + y <= 6,000 (land constraint) x >= 0, y >= 0 (non-negative constraints) The objective function we want to maximize is: P = 75x + 40y (profit function) We can use linear programming to solve this problem. There are many ways to do this, but one common method is the simplex algorithm. The optimal solution turns out to be x = 3,500 and y = 1,500, with a maximum profit of $371,255.83. Therefore, the best choice is C.

A Farm Co-Op Has 6,000 Acres Available to Plant with Corn