Finding the maximum area of a triangle with a perimeter constrain

Question

Using graphical methods, determine the dimensions of a right triangle that has the largest possible area, given that the perimeter cannot be larger than $P$. The final answer should be in terms of $P$.

I got the equation max $$\frac{1}2 xy$$ s.t. $$ x + y + \sqrt{ x ^2 + y^2} \leq P$$

Answer 1

Here is a triangle with constant perimeter (12 units):

Observe how the function of area changes.

Answer 2

Hint: if $z$ is the hypotenuse, then $\frac12xy=\frac14(P^2-2Pz)$, so you need the right triangle with minimum hypotenuse. Now show $(x+y)^2\le 2z^2$ or otherwise conclude this happens when the triangle is isosceles.