Theorem 1   If a right triangle has perimeter $P$ , then its area is bounded as $$ A \le \frac{3 - 2 \sqrt{2}}{4} P^2 $$ with equality when one has an isosceles right triangle.

Proof. Suppose a triangle has legs of length $x$ and $y$ . Then its hypotenuse has length $\sqrt{x^2 + y^2}$ , so the perimeter is given as $$ P = x + y + \sqrt{x^2 + y^2} $$ The area, of course, is $$ A = \frac{1}{2} x y $$

We want to maximize $A$ subject to the constraint that $P$ be constant. This means that the gradient of $A$ will be proportional to the gradient of $P$ . That is to say, for some constant $\lambda$ , we will have \begin{eqnarray*} \frac{\partial A}{\partial x} &=& \lambda \frac{\partial P}{\partial x} \\ \frac{\partial A}{\partial y} &=& \lambda \frac{\partial P}{\partial y} \end{eqnarray*}Together with the constraint, these form a system of three equations for the three quantities $x$ , $y$ , and $\lambda$ . Writing them out explicitly, \begin{eqnarray*} \frac{1}{2} y &=& \lambda \left( 1 + \frac{x}{\sqrt{x^2 + y^2}} \right) \\ \frac{1}{2} x &=& \lambda \left( 1 + \frac{y}{\sqrt{x^2 + y^2}} \right) \\ P &=& x + y + \sqrt{x^2 + y^2} \end{eqnarray*}Not that we cannot have $\lambda = 0$ because that would mean that all sides of our triangle would have zero length. Hence, we may eliminate $\lambda$ between the first two equations to obtain $$ x \left( 1 + \frac{x}{\sqrt{x^2 + y^2}} \right) = y \left( 1 + \frac{y}{\sqrt{x^2 + y^2}} \right), $$ which may be manipulated to yield $$ (x - y) \left( 1 + \frac{x+y}{\sqrt{x^2 + y^2}} \right) = 0. $$ We have two case to consider -- either the first factor or the second factor may equal zero. If the second factor equals zero, $$ 1 + \frac{x+y}{\sqrt{x^2 + y^2}} = 0, $$ move the ``1'' to the other side of the equation and cross-multiply to obtain $$ x + y = - \sqrt{x^2 + y^2}. $$ Since we want $x \ge 0$ and $y \ge 0$ but the right-hand side is non-positive, the only option would be to have a trianagle of zero area.

The other possibility was to have the second factor equal zero, which would give $$ x - y = 0. $$ In this case, $x$ equals $y$ . Imposing this condition on the constraint, we see that $$ P = (2 + \sqrt{2}) x, $$ so we have the solution \begin{eqnarray*} x &=& \frac{P}{2 + \sqrt{2}} = \frac{2 - \sqrt{2}}{2} P \\ y &=& \frac{P}{2 + \sqrt{2}} = \frac{2 - \sqrt{2}}{2} P . \end{eqnarray*}