Find the maximum possible area of a certain right triangle I want to find the maximum possible area of a right triangle with hypotenuse $=10$.
My approach so far: let $x,y$ be the lengths of the two sides adjacent to the right angle; then $$100=x^2+y^2$$
Area $=\frac{xy}{2}$, so by the substitution method I got equality. But my result is wrong (probably I made a mistake in the equality), could someone show what to do?

Let the non-hypotenuse sides of the triangle be $(a,b)$.  Then the hypotenuse is $\sqrt{a^2+b^2}$ and the area is $A = \frac{ab}{2}$.  So
$$a^2+b^2 = 100\\
A^2 = \frac14 a^2b^2 = \frac14 a^2 (100-a^2) \\
$$
Maximizing $A^2$ maximizes $A$, and
$$
\frac{d(A^2)}{da} =\frac14( 200a -4a^3)
$$
which is zero at $a = \sqrt{50}$ making $b=\sqrt{50}$ as well and the area $A = \frac12\sqrt{50}^2 = 25$
