We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

TEXT

# Putting It Together

A farmer has 100 acres of available land where he wants to plant a mixture of corn, wheat and spinach. It costs him $400 to produce an acre of corn,$160 to produce an acre of wheat, and $280 to produce an acre of spinach. He has a maximum of$20,000 to spend. He makes a profit of $120 per acre on corn,$40 per acre on wheat and \$60 per acre on spinach. How many acres of each crop should he plant? We first start by making a table for the various crops:
 Profits and Constraints for Crops Corn Wheat Spinach Total Number of Acres x1 x2 x3 ≤ 100 Cost (per acre) 400 160 280 ≤ 20000 Profit (per acre) 120 40 60
We can start making linear equations based on the table above:

x1 + x2 + x3 ≤ 100

Since we may not use all the acres, we can introduce a slack variable (say s1), which stands for the unused portion. Then we can create an actual equation instead of an inequality.

x1 + x2 + x3 + s1 = 100

We can also create an inequality for the cost per acre:

400x1 + 160x2 + 280x3 ≤ 20000

Dividing everything by 40 gives us:

10x1 + 4x2 + 7x3 ≤ 500

Since we may not use all the money, we can introduce a slack variable (say s2), which stands for the unused portion and then we can create an actual equation instead of an inequality.

10x1 + 4x2 + 7x3 + s2 = 500

We have a third function we can look at: The profit which we want to optimize:

z = 120x1 + 40x2 + 60x3

This gives us the following: Maximize: z = 120x1 + 40x2 + 60x3 → 0 = z − 120x1 − 40x2 − 60x3 Subject to: x1 + x2 + x3 + s1 = 100
 Maximise: z = 120x1 + 40x2 + 60x3 → 0 = z − 120x1 − 40x2 − 60x3 Subject to: x1 + x2 + x3 + s1 = 100 With x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, s1 ≥ 0 and s2 ≥ 0
This sets up our initial tableau
 x1 x2 x3 s1 s2 Z 1 1 1 1 0 0 100 10 4 7 0 1 0 500 -120 -40 -60 0 0 1 0
Our basic solutions have , but this will have farmer planting zero crops so this solution can be eliminated. We look at the bottom row and find the most negative number, which corresponds to x1. This becomes our pivot column. We have to decide which of the two numbers in this column to use. By doing some more calculations we actually find that the solution which maximizes profits is: x1 = 50, s1 = 50, and z = 6000 The other variables are zero. It may seem odd that leaving some ground unplanted will maximize profits, but this is often the case. (To see a full solution to this problem, email your instructor.)