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/ |
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