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June 26th, 2017, 09:17 PM   #1
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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.
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June 26th, 2017, 09:46 PM   #2
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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.
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June 26th, 2017, 10:28 PM   #3
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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.
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June 27th, 2017, 07:39 AM   #4
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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. $\displaystyle \nabla f=\, <\partial f/\partial a, \partial f/\partial b>\,=\, <b/2, a/2>$. $\displaystyle \nabla g=\, <\partial g/\partial a, \partial g/\partial b>\,=\, <2a, 2b>$. 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 $\displaystyle <b/2, a/2>$= $\displaystyle \lambda <2a, 2b>$ 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.
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June 27th, 2017, 10:01 AM   #5
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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.
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