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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 

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)


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
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Associated Topics:
College Calculus

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