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A right triangle is to be constructed using a piece of wire of length 2L. Find its dimensions if its area to be maximum.

I have no idea how to solve this, all i know is that i will use local maximum rules.

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WLOG $$a+a\cos t+a\sin t=2L\iff\cos t+\sin t=\dfrac{2L-a}a$$

Now $\sqrt2\ge\cos t+\sin t=\dfrac{2L-a}a\iff a\ge\dfrac{2L}{\sqrt2+1}$

$\implies a_\text{min}=\dfrac{2L}{\sqrt2+1}=(\sqrt2-1)2L$

Squaring we get $\sin t\cos t=\dfrac{(2L-a)^2-a^2}{2a^2}=\dfrac{4L^2-4La}{2a^2}$

Now we need to maximize $=\dfrac{a^2\cos t\sin t}2=L^2-La=L(L-a)$

as $L$ is constant, the area will be maximized if $L-a$ is maximum $\iff a$ is minimum

Hope you can take it from here

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  • $\begingroup$ @Arthur, Thanks for your observation. Please find the updated answer to derive $t=\dfrac\pi4$ for maximum area i.e., isosceles right triangle will have the maximum area $\endgroup$ – lab bhattacharjee Mar 1 '16 at 9:17
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Another version my answer separated for the sake of clarity

$$a+b+c=2L\iff b+c=2L-a$$ where $b^2+c^2=a^2$

Squaring & on rearrangement, $$\triangle=\dfrac{bc}2=L(L-a)$$ which will be maximum $\iff a$ is minimum

Now $2(b^2+c^2)-(b+c)^2=(b-c)^2\ge0$

$\implies2a^2\ge(b+c)^2\iff\sqrt2a\ge(b+c)$

the equality occurs if $b=c\implies a=\sqrt2 b$

Hope you can take it from here

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I use the well-known result that area is maximized when all sides are equal and all angles are $60^0.$ If a right angle is a constraint then it has to be an isosceles triangle with equal sides adjoining the right angle, the hypotenuse making $45^0$ with the small side of length $x$.

$$ 2 x + \sqrt 2 x = 2 L; \; x = L( 2 - \sqrt 2). $$

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Let $x$ be the side of one leg, and $f(x)$ be the area.

If one leg is $x$, and the other leg has length $y$, then $f(x) = \frac{xy}{2}$, as you probably know. By the Pythagorean theorem, $$x + y + \sqrt{x^2 + y^2} = 2L$$With some work you can use this to express $y$ as a function of $x$, in order to help you express $f(x)$ in a differentiable form.

Or use trigonometry as the other answer does. That's probably easier.

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I find the perimeter $p$ wrt a fixed area $A$. Given the area is $A=\frac 1 2 bh$ we have base and height that measure $b$ and $h=\frac{2A}{b}$, so the perimeter is: $$ p(b)=b+\frac{2A}{b}+\sqrt{b^2+\frac{4A^2}{b^2}} $$ it is minimal when the derivative wrt $b$ is $0$ that happens in $b=\sqrt2A$ that means $h=b$ and the triangle is a half square.

This implies that the half square is also the shape of the triangle with maximal area when we fix a perimeter $p$.


Starting from the formula $p(b)$ above you can also isolate the square root on the right side, square left and right side of expression and then get a quite decent formula for $A$: $$ A=\frac{pb(2b-p)}{4(b-p)} $$ then you can derive and get $b$ wrt $p$, it will result to be $b=\left(1-\frac 1 {\sqrt{2}}\right)p$, then you need to put it back in the formula above and you will get $A=\frac 1 2 \left(1-\frac 1 {\sqrt{2}}\right)^2p^2$ (again the triangle is the half square).

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We have to maximize $\phi(x,y):=xy$ under the constraints $$x\geq0, \quad y\geq0, \quad x+y+\sqrt{x^2+y^2}\leq L\ .$$ Let $(x,y)$ be a feasible point with $x<y$, and consider the point $(x',y')$ given by $$x'= y':={x+y\over2}\ .$$ Then $x'+y'=x+y$, and $$x'^2+y'^2=x^2+y^2-{1\over2}(y-x)^2< x^2+y^2\ ,$$ hence $(x',y')$ is feasible as well. On the other hand $$x'y'={1\over4}(x+y)^2=xy+{1\over4}(y-x)^2>xy\ .$$ Therefore the objective function takes its maximum at a point with $x=y$, and this necessitates $$x=y={L\over 2+\sqrt{2}}\ .$$ The maximal area of the triangle then comes to $${1\over2}xy={3-2\sqrt{2}\over 4}L^2\ .$$

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