Maximum Quadrilateral AreaDate: 05/15/2001 at 12:44:46 From: Martin Maister Subject: Quadrilateral area Dear Dr Math, Here is the question: Given a quadrilateral with sides of lengths a, b, c and d, prove that its area is maximized when opposite angles are supplementary. I have sat on this question for weeks. I tried calculus, using area as a*b*sin(angle between) and taking the derivative. I got stuck with too many variables. I have often seen this fact, but not how to prove it. My last resort? Dr Math! Thank you, and much appreciated. Martin Date: 05/15/2001 at 16:39:38 From: Doctor Rick Subject: Re: Quadrilateral area Hi, Martin. It can be proved using calculus. As you suggest, you start with the area formula: K = (1/2)a*b*sin(alpha) + (1/2)c*d*sin(beta) where alpha is the angle between sides a and b, and beta is the angle between sides c and d. The problem of too many variables is solved by using the Law of Cosines to relate alpha and beta. The diagonal x joining the other two vertices satisfies x^2 = a^2 + b^2 - 2*a*b*cos(alpha) from the one triangle, and x^2 = c^2 + d^2 - 2*c*d*cos(beta) from the other. Equating the two right sides, we get cos(beta) = (c^2 + d^2 - a^2 - b^2 + 2*a*b*cos(alpha))/(2*c*d) Take the derivative of cos(beta) with respect to alpha. Then go back to the area formula, and express sin(beta) as sin(beta) = sqrt(1 - cos^2(beta)) Take the derivative of K with respect to alpha, using the formula for the derivative of cos(beta) with respect to alpha where you need it. You'll end up with dK/d(alpha) = a*b/(2*sin(beta))*sin(alpha+beta) The derivative is zero when alpha+beta = 180 degrees. With a bit more work, I'm sure you can show that this condition gives the maximum of K. Another approach, not using calculus but using plenty of trigonometry, can be found in our Archives here: Bretschneider's Theorem and Cyclic Quadrilaterals http://mathforum.org/dr.math/problems/blumberg.11.30.00.html - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 05/15/2001 at 19:10:09 From: Martin Tutorials Subject: Re: Quadrilateral area Thank you very much for getting back to me! I do not take your wonderful service for granted! I hope you all have a wonderful summer! Best wishes, Martin |
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