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I've attempted this problem many times but I have no idea how to set up the problem at all.

Suppose you work on Rich Mann's ranch. Rich 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. Rich wants to use all 1200 yards of fencing. The area between the fence and the lake and the are of the square corral will be sodded with grass that costs 30 cents per square yard. How would build the fences to minimize the price of the sodded grass?

b. If price was not a concern, how would you advise Rich to construct the fences so that the two areas of the two corrals is a maximum?

Questioner:Karina

Country:Illinois, United States

Category:Calculus

Private:No

Subject:Optimizatoin

Question:I've attempted this problem many times but I have no idea how to set up the problem at all.

Suppose you work on Rich Mann's ranch. Rich 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. Rich 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 build the fences to minimize the price of the sodded grass?

b. If price was not a concern, how would you advise Rich to construct the fences so that the two areas of the two corrals is a maximum?

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For starters, I suggest you look at:

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

for a collection of these problems I have answered for others , and, when you get to this topic, another collection:

http://en.allexperts.com/q/Calculus-2063/2009/11/Related-Rates-87.htm

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You will build a circular fence with radius R. That will require 2 pi R yards of fencing.

[Very important: Note boundary values: R1: 2 pi R >= 100, and R3: 2 pi R <= 1200 ]

The remainder fencing is 1200 - 2 pi R

The total grass area will be:

Outer circle - inner circle + square corral.

Outer circle: pi R^2

Inner circle: pi (pond radius)^2 = pi (100/2pi)^2 = pi(50/pi)^2 = 2500/pi

Square corral uses 1200 - 2 pi R fencing for four sides. so

one side = 300 - pi R/2, and area = (300 - pi R/2)^2

To minimize the cost, you just minimize the area. (Who cares what sod goes for?)

To maximize the area, you, er.. maximize the area. (duh...)

Total area A = pi R^2 - 2500/pi + (300 - pi R/2)^2

Basically, that's it. You have A = f(R). So you will now:

Simplify a bit.

Find dA/dR.

Set it = 0 and solve for R. Call that R2

Set R1 = the first boundary value.(above)

Set R3 = the other boundary value.(above)

Find A(R1), A(R2), A(R3).

One is your max, one is your min.

Calculus

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All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.**Education/Credentials**

(See above.)