The Shortest CreaseDate: 12/29/97 at 03:14:48 From: Kathy Subject: The shortest crease Q: A paper which is 6 units one side and 25 units another side. Put the shorter side to face you. Fold the lower right corner to the left side, making a crease. Our task is to find the minimum crease by folding the right corner to the left side. HINT: this question is related to derivative - you may use derivative to solve it. (You are required to write down the equations that you use and explain how you solve it.) This question is a project given in my calculus class. We are supposed to write equations to solve it. We have to write a bunch of explanations and equations to complete a project paper. So far, I have no idea how to solve it. Could you help me? Thank you very much for your help. Sincerely, Kathy Date: 12/29/97 at 12:22:00 From: Doctor Anthony Subject: Re: The shortest crease I have taken a general case with width equal to k, so you can see how the working can be adjusted for different widths of paper. If you take a length x along the width, and fold it, leaving (k-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 length of the fold and the area of the folded triangle in terms of k 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 k/cos(theta) We also have x.sin(theta) = k-x x(1+sin(theta)) = k k x = ------------- 1 + sin(theta) The length of the fold, by Pythagoras is given by FOLD^2 = x^2 + [k/cos(theta)]^2 = [k/(1+sin(theta)]^2 + [k/cos(theta)]^2 = k^2[(1+sin(theta))^(-2) + sec^2(theta)] Differentiate with respect to theta and we get k^2[(-2)(1+sin(theta)^(-3)(cos(theta) + 2.sec(theta)sec(theta)tan(theta)] k^2[-2.cos(theta)/(1+sin(theta))^3 + 2.sin(theta)/cos^3(theta)] and equating this to zero, we get cos^4(theta) = sin(theta)(1+sin(theta))^3 putting sin(theta) = 1/3 and then cos(theta) = sqrt(8)/3, this equation is satisfied. So the minimum fold occurs with sin(theta) = 1/3 x = k/(1+1/3) = k/(4/3) = (3/4)k FOLD^2 = (9/16)k^2 + (9/8)k^2 = (27/16)k^2 Fold = (3.Sqrt(3)/4)k -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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