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 June 26th, 2017, 09:17 PM #1 Newbie   Joined: Jun 2017 From: Loughborough Posts: 5 Thanks: 0 How do I work out the maximum area of a right triangle with a fixed length hypotenuse How do I work out the maximum area of a right triangle given a fixed length hypotenuse? I have watched several YT videos and ALL are based upon the ASSUMPTION the other two angles are 45 degrees - and whilst this may be true, this is exactly what I need to prove - so I'm looking for a formula that using sine or cosine to PROVE which angles provide the maximum area of a right triangle given a fixed hypotenuse. Any ideas - links?﻿ Last edited by skipjack; June 26th, 2017 at 10:25 PM. June 26th, 2017, 09:46 PM #2 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics Let the length of the hypotenuse be c and the lengths of the other two sides be a and b. The problem then becomes maximising $\displaystyle \frac{ab}{2}$ given that $\displaystyle a^2 + b^2 = c^2$, i.e. $\displaystyle a = \sqrt{c^2 - b^2}$. $\displaystyle \frac{\mathrm{d}}{\mathrm{d}b}\frac{b\sqrt{c^2 - b^2}}{2}=\frac{\sqrt{c^2 - b^2}+\frac{1}{2}b(-2b)(c^2-b^2)^{-1/2}}{2}=\frac{\sqrt{c^2 - b^2}-(2b^2)(c^2-b^2)^{-1/2}}{2}$ $\displaystyle \frac{\sqrt{c^2 - b^2}-(2b^2)(c^2-b^2)^{-1/2}}{2} = 0 \Rightarrow \sqrt{c^2 - b^2}=(2b^2)(c^2-b^2)^{-1/2} \Rightarrow c^2 - b^2 = \frac{b^4}{c^2 - b^2} \Rightarrow b = \frac{c}{\sqrt 2}$ $\displaystyle a = \sqrt{c^2 - \left(\frac{c}{\sqrt 2}\right)^2} =\frac{c}{\sqrt 2}$ Maximum area is $\displaystyle \frac{1}{2}\left(\frac{c}{\sqrt 2 }\right)^2=\frac{c^2}{4}$. Last edited by skipjack; June 26th, 2017 at 09:58 PM. June 26th, 2017, 10:28 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,629 Thanks: 2077 The triangle's area is half the area of a rectangle whose diagonals are each of length c, so area(triangle) = $\frac14c^2\sin(\theta)$, where $\theta$ is the angle of intersection of the rectangle's diagonals. This area has a maximum of $\frac14c^2$ for $\theta =90^\circ\!$, the rectangle then being a square. June 27th, 2017, 07:39 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 You can also do this using "Lagrange multipliers". You want to maximized $\displaystyle f(a, b)= (1/2)ab$ with the condition $\displaystyle g(a, b)= a^2+ b^2= c^2$, a constant. $\displaystyle \nabla f=\, <\partial f/\partial a, \partial f/\partial b>\,=\,$. $\displaystyle \nabla g=\, <\partial g/\partial a, \partial g/\partial b>\,=\, <2a, 2b>$. At a maximum of $f$, subject to the condition that $g$ is a constant, those two vectors must be parallel: $\displaystyle \nabla f= \lambda \nabla g$ for some constant $\displaystyle \lambda$ (the "Lagrange multiplier"). That is, we must have $\displaystyle$= $\displaystyle \lambda <2a, 2b>$ or $\displaystyle b/2= 2\lambda a$ and $\displaystyle a/2= 2\lambda b>$. That, together with the requirement that $\displaystyle a^2+ b^2= c^2$, gives us three equations to solve for a, b, and $\displaystyle \lambda$. Since a value for $\displaystyle \lambda$ is not necessary for a solution to this problem, a good way to start is to eliminate $\displaystyle \lambda$ by dividing one equation by the other: $\displaystyle \frac{b/2}{a/2}= \frac{2a}{2b}$ so $\displaystyle \frac{b}{a}= \frac{a}{b}$. $\displaystyle b^2= a^2$ so $\displaystyle b= \pm a$. Since these are lengths, they must be positive, so $\displaystyle b= a$ (so that the two legs have equal length and the two angles are 45 degrees). Then the condition becomes $\displaystyle a^2+ b^2= 2a^2= c^2$ and $\displaystyle a= b= \frac{c}{\sqrt{2}}= \frac{\sqrt{2}}{2}c$. The area when $\displaystyle a= b= \frac{c}{\sqrt{2}}= \frac{\sqrt{2}}{2}c$ is $\displaystyle c^2$. Last edited by skipjack; June 27th, 2017 at 09:49 AM. June 27th, 2017, 10:01 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,629 Thanks: 2077 That could be clearer. You have an unpaired ">" in the line commencing with "That is", and you seem to have omitted "/4" at the end. Rewriting $\displaystyle \frac{c}{√2}$ as $\displaystyle \frac{√2}{2}c$ seems unnecessary. Tags area, fixed, hypotenus, hypotenuse, length, maximum, triangle, work Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Shariq Faraz Geometry 2 May 31st, 2017 12:22 AM panky Trigonometry 4 April 24th, 2017 03:10 PM Volle Calculus 4 July 27th, 2015 03:49 PM mfosar Geometry 9 June 7th, 2014 09:55 PM d8950093 Algebra 4 September 17th, 2013 08:11 PM

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