Expert: Paul Klarreich - 11/29/2008

Question
A commercial cattle ranch currently allows 20 steers per acre of grazing land. On average the steers weigh 1000 pounds at market. Estimates by the Department of Agriculture indicate that for every steer per acre added to graze, the average weight will fall by 50 pounds per steer. How many steers per acre should the ranch allow in order for the ranch to get the largest possible total market weight for its cattle?

Answer
Questioner:   Max
Category:  Calculus
Private:  No
 
Subject:  Min/Max problem
Question:  A commercial cattle ranch currently allows 20 steers per acre of grazing land. On average the steers weigh 1000 pounds at market. Estimates by the Department of Agriculture indicate that for every steer per acre added to graze, the average weight will fall by 50 pounds per steer. How many steers per acre should the ranch allow in order for the ranch to get the largest possible total market weight for its cattle?
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Hi, Max,

If this is your first time doing a Maximum-minimum problem, here is the scheme.

1. Identify the variables in the problem -- the things that change.  Give them names.

2. Find the one that is to be 'optimized'.  Write it as a function of the other variables.

3. If it is a function of more than one, use the other conditions (constraints) to eliminate all but one.

4. Differentiate, set that = 0, solve for your 'stationary' point.

5. Consider whether it is a maximum, minimum, or neither.  Check the logical endpoints, too.

6. Answer whatever other questions are asked.

AND, Please check the archives for other Maximum-minimum examples.  There are a TON of them INCLUDING THIS ONE OR SEVERAL LIKE IT.  Click BROWSE PAST ANSWERS and look for the subject line Maximum-minimum.

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Try:  n = number of additional steers per acre.
THEN:  w = 1000 - 50n

That should do it.