A farmer can plant a combination of two different crops on 20 acres of land. Seed costs $120 per acre for crop A and $200 per acre for crop B. Government restrictions limit acreage of crop A to 15 acres, but do not limit crop B. Crop A will take 15 hours of labor per acre at a cost of $5.60 per hur, and crpo B will require 10 hours of labor per acre at $5.00per hour. IF the expected income from crop A is $600 per acre and from crop B is $520 per acre, how many acres of crop A should be planted in order to maximize profit?
Please help me set this up. I am no good with word problems!
Thanks
acres of A = a
acres of B = b
Profit from A = 600a -120a -15(5.60)a
=396a
Profit from b = 520b - 200 b -10(5.00)b
=270b
Well, we want to plant as much A as possible so 15 acres of a and only 5 of b
No problem! Let's break down the information given in the problem to set up the equation.
Let x be the number of acres of crop A to be planted.
The cost of planting crop A on x acres is: $120/acre * x acres = $120x
The cost of planting crop B on (20 - x) acres is: $200/acre * (20 - x) acres = $200(20 - x)
The labor cost for crop A on x acres is: 15 hours/acre * $5.60/hour * x acres = $84x
The labor cost for crop B on (20 - x) acres is: 10 hours/acre * $5.00/hour * (20 - x) acres = $50(20 - x)
The income from crop A on x acres is: $600/acre * x acres = $600x
The income from crop B on (20 - x) acres is: $520/acre * (20 - x) acres = $520(20 - x)
To maximize profit, the total profit is calculated as total income - total cost.
Profit = (Income from crop A + Income from crop B) - (Cost of crop A + Cost of crop B + Labor cost for crop A + Labor cost for crop B)
Profit = ($600x + $520(20 - x)) - ($120x + $200(20 - x) + $84x + $50(20 - x))
Now, simplify the equation:
Profit = $600x + $520(20 - x) - $120x - $200(20 - x) - $84x - $50(20 - x)
Profit = $600x + $520*20 - $520x - $120x - $200*20 + $200x - $84x - $50*20 + $50x
Profit = $600x + $10,400 - $520x - $120x - $4,000 + $200x - $84x - $1,000 + $50x
Profit = $546x + $5,400
To maximize profit, we need to find the value of x that maximizes the Profit equation.
Finally, we can find the maximum profit by finding the value of x that gives the maximum value to the equation $546x + $5,400.