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### Cost of Fencing a Yard

```
Date: 02/24/97 at 18:21:33
From: Anees Bensassi
Subject: Calculus, related rates

A yard with an area of at least 500 square meters must be fenced on
three sides (two widths, one length). Fencing the widths will cost \$15
per meter and fencing the length will cost \$25 per meter. What
dimensions should you use to enclose 500 square meters at a minimum
cost?

Thank you
```

```
Date: 03/02/97 at 23:50:06
From: Doctor Luis
Subject: Re: Calculus, related rates

Here's a diagram of the fence (drawn with a wall as the fourth side):

---------------------------
|                  |
|                  |
|     500 m^2      |   width w
|                  |
|__________________|

length l

Now, the cost of fencing the yard is:

c(l,w) = (25 dollars/meter)(l meters) + (15 dollars/meter)(2w meters)

or

c(w,l) = 30w + 25l

Since the area of the yard is 500 m^2 we know that:

lw = 500

Solving for l and substituting into c(w,l):

c(w) = 30w + 25*(500/w)
= 30w + 12500/w

This last expression gives the cost of fencing the yard as a function
of the width. Taking c'(w) and setting it equal to zero, we obtain:

c'(w) = 30 - 12500/w^2 = 0
30 = 12500/w^2
w^2 = 1250/3
w = 20.412 m (approximately)

Since wl = 500:

l = 500/w
l = 24.495 m

The cost of building the fence would be:

c(20.412) = 30(20.412) + 12500/(20.412)
= \$1224.74

Note that there are two critical points of c(w), namely
w_1 = sqrt(1250/3) and w_2 = -sqrt(1250/3). One can readily reject w_2
as a possible answer based on a physical argument (negative lengths
are meaningless). However, one may legitimately ask whether w_1 is a
minimum or a maximum critical point of c(w). This question can be
settled by examining the concavity of c''(w):

c''(w) = 12500/(2w^3)
= 6250/w^3

Since c''(w) > 0 for w > 0, c(w) is concave upward for w > 0, which
proves that the point (w_1,c(w_1)) is a minimum point of c(w).

-Doctor Luis,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus

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