If crop B brings in a profit of $210 per acre, and crop A brings in a profit of $110 per acre, how many acres of each crop should the farmer plant maximize her profit?

Question

A farmer is going to divide her 30-acre farm between two crops. The seed for crop A costs $25 per acre. Seed for crop B costs $50 per acre. The farmer can spend at most $1250 on the seed. 

If crop B brings in a profit of $210 per acre, and crop A brings in a profit of $110 per acre, how many acres of each crop should the farmer plant maximize her profit?

........?   acres of crop A

  ........? acres of crop B

Answer 1

There is a niche topic in algebra called "linear programming." Apparently farmers need to know this technique. The idea is to construct line equations, find intersections, and test each intersection (and smallest x an y axis intercepts points that are part of the connected polygon created by the lines) for the Return equation. There is a theorem somewhere that guarantees either an intersection or axis intercept part of polygon maximizes return.

Return equation

R=110x+210y ,,,,,,,,,,x is acre of A, y is acre of B

capital line equation,,,,,,,,, y=25-(x/2)

land line equation,,,,,,,,, y=30-x

intersection is x=10 acre

at intersection, R=5300

at x=0 y=25, R= 5250 ,,,,, it is impossible with capital to buy more than y=25

at x=30 y=0 R= 3300

Answer 2

Hello, Ngoc,

I tried to derive a single equation that calculated net profits (Profit minus seed expenses) but am having difficulty deciding what to do with it. I had hoped to find the first derivative, but the process was becoming tedious. In trying to understand the relationships, I set up a spreadsheet that would calculate the net profit once I've entered the number of acres for crop A (crop B is (30 - A) acres). We are limited by the available funds for seed ($1250), In trying a few numbers, I arrived at these values. (I was unable to fit the spreadsheet image here).

Crop A: 10 acres

Crop B: 20 acres

Seed cost = $1250

Net Profits (Sales - Costs) = $4050

This isn't the result of an optimization - I simply tried a combination that gave the highest profit for the $1250 cost of seeds. I gave up on the derivation, but wanted to include this, in case you see something helpful.

Bob