Optimization Problem

[nezenic]

Hello,

I am having some trouble getting started with this certain problem and was hoping for a push in the right direction.

Here is the original statement:

A cattle ranch currently allows 20 steers per acre of grazing land, on average its steers weigh 2000 pounds. Estimates by the Agriculture Department indicate that the average market weight per steer will be reduced by 50 pounds for each additional steer added per acre of grazing land. How many steers per acre should be allowed in order for the ranch to get the largest possible TOTAL market weight for its cattle?

This is the exact phrasing, along with the uppercased "total". Currently I am just confused as to coming up with an equation that relates these values together, as well as an equation that might as a constraint.

So...
20 steers/acre
2000 lbs/acre or 2000 lbs per 20 steers
Minus 50 lbs per additional steer

I have come up with a relationship, but I am not sure if it is correct or incorrect...

2000 - 50t = 20x
where t = number of steers
and x = number of acres

So I am having a hard time visualizing this equation being right or wrong, and I am not sure how I would come up with a restraint to find one of the unknowns so that I could make a valid substitution.

Any ideas and help is appreciated!

 

[soroban]

Hello, nezenic!

A cattle ranch currently allows 20 steers per acre of grazing land, on average
its steers weigh 2000 pounds. .Estimates by the Agriculture Department
indicate that the average market weight per steer will be reduced by 50 pounds
for each additional steer added per acre of grazing land.
How many steers per acre should be allowed in order for the ranch
to get the largest possible TOTAL market weight for its cattle?

Presently, the ranch has 20 steer/acre; the steer weigh 2000 pounds each.
. . There is a total of: \(\displaystyle \,20\,\times 2000 \:=\:400,000\) pounds of beef.

Suppose we add \(\displaystyle x\) more steer per acre, \(\displaystyle 20\,+\,x\) steer/acre.
. . Their average weight will drop to \(\displaystyle 2000\,-\,50x\) pounds each.

The total weight will be: \(\displaystyle \:W \:=\20\,+\,x)(2000\,-\,50x)\) pounds.

. . And that is the function we want to maximize.

 

[nezenic]

Wow! That makes so much sense. I couldn't figure out how I was going about it wrong...but thanks so much!

(20x2000=40,000)