Prove that the perimeter of a right-angled triangle of given hypotenuse is maximum when the triangle is isosceles

Explanation

Let ABC be the right triangle with given hypotenuse h
Let, base BC = x and altitude `BA = a`
Then, `h^(2) = x^(2) + a^(2) rArr a = sqrt(h^(2) -x^(2))`
`:. P = a + x + h rArr P = sqrt((h^(2) -x^(2))) + x + h`
So, `(dP)/(dx) = (-x)/(sqrt(h^(2) -x^(2))) + 1 and (d^(2)P)/(dx^(2)) = (-h^(2))/((h^(2) -x^(2))^(3//2))`
Now, `(dP)/(dx) = 0 rArr x = (h)/(sqrt2)`
And for `x = (h)/(sqrt2), (d^(2)P)/(dx^(2)) = (-2^(3//2))/(h) lt 0`
`:.` P is maximum when `x = (h)/(sqrt2) and a = (h)/(sqrt2)` i.e., when x = a