Find the dimensions of a triangle if its area to be maximum 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.
 A: 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
A: 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
A: 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 $x$.
$$ 2 x + \sqrt 2 x = 2 L; \; x = L( 2 - \sqrt 2). $$
A: 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.
A: 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).
A: 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
$${1\over2}xy={3-2\sqrt{2}\over 4}L^2\ .$$
