|


Cost of Fencing a YardDate: 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/
|
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]


Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/