Maximizing AreaDate: 05/02/97 at 19:06:07 From: Adrian Guerrero Subject: Calculus - Optimization A single wire, 80 cm long, is cut into two pieces. One piece is bent into the shape of a square and the other is bent into the shape of an equilateral triangle. Into what lengths should the wire be cut to maximize the total combined area enclosed by the the two pieces? Thanks. Date: 05/03/97 at 12:37:27 From: Doctor Anthony Subject: Re: Calculus - Optimization Let x be the length used to form the square, then 80-x will be the length used for the equilateral triangle. The area of the square will be: (x/4)^2 = x^2/16 The area of the triangle will be: (1/2)((80-x)/3)^2 sqrt(3)/2 = (sqrt(3)/36)(80-x)^2 Total area of these figures: A = x^2/16 + (sqrt(3)/36)(80-x)^2 Differentiating: dA/dx = 2x/16 + (2sqrt(3)/36)(80-x)(-1) = 0 (for max) = x/8 - (sqrt(3)/18)(80-x) = 0 x/8 = (sqrt(3)/18)(80-x) x/8 + (sqrt(3)/18)x = 80.sqrt(3)/18 x[1/8 + sqrt(3)/18] = 80.sqrt(3)/18 x = (80.sqrt(3)/18)/[1/8+sqrt(3)/18] x = 34.797 So we cut a length of 34.8 cm from the 80 cm length of wire to form the square, which leaves 45.2 cm for forming the triangle. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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