Optimization: Minimum AreaDate: 11/07/97 at 03:04:44 From: kiyoshi yamashita Subject: Optimization A piece of paper (rect. with width a and unlimited length) is folded so one corner just reaches the righthand side. How do you fold to make the area of the folded paper a minimum. I've tried using calculus and derivatives and even just finding a function for the area of the triangle. I can't, however, find a relation between the sides. I've tried the law of sines and the Pythagorean theorem. I ended up with the fact that the two legs are always equal but I know that's not true. Here's some of my work: Given: right triangle with Legs x,y and hyp. z; angleYZ=theta; and area A. A = .5xy z^2 = x^2+y^2 cos(theta) = x/y. sin(theta)/x = sin(90)/z ---> x = z*sin(theta) z^2 = z^2*sin^2(theta) + y^2 z^2(cos^2(theta)) = y^2 z^2*(x^2/z^2) = y^2 x^2 = y^2 x = y therefore: A = .5x^2. I've also worked to the fact that A = .5*x*z*cos(theta) My final attempt included the fact that tan(theta) = x/y. Thank you so much. Kiyoshi Date: 11/07/97 at 08:53:08 From: Doctor Anthony Subject: Re: Optimization If you take a length x along the width, and fold it, leaving (a-x) unfolded, and let the corner meet the righthand edge somewhere along its length such that the side x makes an angle (theta) with the length, then we can calculate the area of the folded triangle in terms of a and theta. The other edge of the folded portion (at right angles to x) will make an angle theta with the width, and it is easy to see that this length is given by a/cos(theta) We also have x.sin(theta) = a-x x(1+sin(theta)) = a a x = ----------- 1 + sin(theta) The area of the folded triangle = (1/2)x.a/cos(theta) a^2 A = ------------------------ 2cos(theta)[1+sin(theta)] a^2 = ------------------------- 2cos(theta) + sin(2.theta) dA/d(theta) = a^2[2cos(theta)+sin(2.theta)] x 0 - (-2sin(theta) + 2cos(2.theta)] ------------------------------------------------------------------- [2cos(theta) + sin(2.theta)]^2 and putting this equal to 0 we have 2sin(theta) - 2cos(2.theta) = 0 sin(theta - (1- 2sin^2(theta) = 0 2sin^2(theta) + sin(theta) - 1 = 0 (2sin(theta) -1)(sin(theta) + 1) = 0 and the one practical solution in terms of this problem is: 2sin(theta) = 1 sin(theta) = 1/2 so theta = 30 degrees and then x = a/(1+sin(theta)) x = a/(1+1/2) = a /(3/2) = 2a/3 So to get the minimum area you must fold a length 2/3 of 'a' over to meet the righthand long edge of the paper. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 11/07/97 at 12:59:42 From: Kiyoshi Yamashita Subject: Re: Optimization Thank you, thank you, thank you! |
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