If crop B brings in a profit of $240 per acre, and crop A brings in a profit of $140 per acre, how many acres of each crop should the farmer plant to maximize her profit?

Question

A farmer is going to divide her 60 acre farm between two crops. Seed for crop A costs $25 per acre. Seed for crop B costs $50 per acre. The farmer can spend at most $2500 on seed.

If crop B brings in a profit of $240 per acre, and crop A brings in a profit of $140 per acre, how many acres of each crop should the farmer plant to maximize her profit?

Answer 1

Profit = R =140A +240B

Cost= C= =25A +50B=2500

divide by 25

A+2B =100

A +B = 60

B =40 acres

A = 20 acres

max Profit would seem to = 140A +240B = 140(20) + 240(40)

= 2800+9600

=$12,400

but devoting all acres to B would maximize Profits

= 240(60) = $14,400

the Costs are "irrelevant" since Profits already includes them, Profit =Revenue minus Costs

there may be a mistake in the problem, it might have been intended to read

Revenue =$120 per acre for A and $240 per acre for B

if so,

then P=R-C = 140A+240B - 25A -50B

= 115A +190B

= 155(20)+190(40)

= 3100+7600

=$10,700

but still B is more profitable per acre

and devoting all 60 to B has profits

= 190(60) = $11,400

No matter how you view the problem

Profits are seemingly maximized by

devoting all 60 acres to crop B

but there's the budget constraint of $2500

60x50 = costs of $3000, over the $2500 budget

she should devote 40 acres to crop B and 20 to crop A

with profit

= 20(140) + 40(240)

=2800+9600

= $11,400

if profits = $140 and $240 per A acres and B acres

or

if revenue = $140 and $240

then

max profits = 20(115)+40(190)

= 2300 + 7600

=$9,900