Expert: Paul Klarreich - 1/15/2012

Question
Suppose you work on a ranch. The rancher tells you to build a circular fence around a 100 yard diameter circular pond and to use the remainder, if any, of your 1200 yards of fencing to build a square corral.

a) The rancher wants to use all 1200 yards of fencing.  The area between the fence and the lake and the area of the square corral will be sodded with grass that costs 30 cents per square yard. How would you build the fenes to minimize the price of the sodded grass?
b) If price was not a concern, how would you advise the rancher to construct the fences so that the two areas of the two corrals is a maximum?

**I really just don't know how to set up the equations/get started on this one**

Answer
Questioner:Mary
Country:Massachusetts, United States
Category:Calculus
Private:No
Subject:calculus AB--optimization

Question:

Suppose you work on a ranch. The rancher tells you to build a circular fence around a 100 yard diameter circular pond and to use the remainder, if any, of your 1200 yards of fencing to build a square corral.

a) The rancher wants to use all 1200 yards of fencing.  The area between the fence and the lake and the area of the square corral will be sodded with grass that costs 30 cents per square yard. How would you build the fenes to minimize the price of the sodded grass?
b) If price was not a concern, how would you advise the rancher to construct the fences so that the two areas of the two corrals is a maximum?

**I really just don't know how to set up the equations/get started on this one**
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You usually start with a diagram, AND by reading, as I suggest to questioners:

http://en.allexperts.com/q/Calculus-2063/2009/11/Maximum-minimum-problem-41.htm

So, assuming you did that,

r = radius of circle (at least 50)

A = total area of corrals

(we'll do (b) first)

Grass ring area = pi r^2 - pi(50)^2 = pi(r^2 - 50^2)

The circle will use 2 pi r  yards, leaving 1200 - 2 pi r  yards for the square.

Each side of that will be  (1200 - 2 pi r)/4 = 300 - pi r/2

The area of the square is  (300 - pi r/2)^2

Finally, the total:

A = pi(r^2 - 50^2) + (300 - pi r/2)^2
 
Now do our thing:

dA/dr = 2 pi r + 2(300 - pi r/2)(- pi/2)

dA/dr = 2 pi r - pi(300 - pi r/2)

Set that = 0:

2 pi r - pi(300 - pi r/2) = 0

2 r - (300 - pi r/2) = 0

2r - 300 + pi r/2 = 0

2r + pi r/2 = = 300

(2 + pi/2) r = 300

r = 300/(2  + pi/2)

Now check your endpoints for min/max and you are on your way.

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For (a), I think you just minimize the area.  Do the same stuff, but look four your min.