Expert: Paul Klarreich - 6/5/2006

Question
hi, please help!
find the dimensions of a rectangular region with the maximum area that can be enclosed by 60 feet of fencing.

please show me how u did it so that i can understand, thanks!

Answer
Hi, Hannah,

You wrote:
Subject:  differential calculus
Question:  hi, please help!
find the dimensions of a rectangular region with the maximum area that can be enclosed by 60 feet of fencing.

please show me how u did it so that i can understand, thanks!
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Aha!  The first of the 'Farmer Brown' problems -- Farmer Brown has (so much) fencing available and wants to maximize the area for his chickens using some clever arrangement of the fencing.

In this case the arrangement isn't very clever; it's just a rectangle.  

This is just a max-min problem and we do them like this:

A. Identify the variables in the problem and give them names.

B. Identify the quantity to be maximized (or minimized).

C. Express that quantity in terms of one variable.  If necessary, use a fact in the problem to eliminate all but one variable.  This is where you use your life experience, your vast knowledge of geometry, physics, and everything else you learned in school.

D. Differentiate, set the derivative equal to zero, and solve to find some of your 'critical' points.

E. Check the set of all critical points -- the one(s) you found in D, and the endpoints of the physical situation -- to find the actual max or min.

Part A:

Let  w = the width of the rectangle
Let  h = the height of the rectangle

Part B:

The area A is to be maximized.

C.  A = hw

Use the fact that the perimeter is 60, the amount of fencing.

2h + 2w = 60  -->  h = 30 - w

A = (30 - w)w = 30w - w^2

D.  dA/dw = 30 - 2w

Set  30 - 2w = 0,   w = 15.

E. The critical points are  w = 0, w = 30, and w = 15, which we just found.

Clearly  w = 0 or w = 30 give zero area (a minimum) and  w = 15 gives h = 15 and

an area of A = 15*15 = 225.