Prove that the area of right-angled triangle of given hypotenuse is maximum when the triangle is isosceles.

Explanation

Let h be the hypotenuse and `theta` be the angle between the hypotenuse and the base.
Then, base `= h cos theta` and altitude `= h sin theta`
Let P be the perimeter
Then, `P = h + h cos theta + h sin theta`
`:. (dP)/(d theta) = - h sin theta + h cos theta = h (cos theta - sin theta)`
And, `(d^(2)P)/(d theta^(2)) = h (-sin theta - cos theta) = - h (sin theta + cos theta)`
Now, `(dP)/(d theta) = 0 rArr h(cos theta - sin theta) = 0`
`rArr cos theta = sin theta " i.e., " tan theta = 1 rArr theta = (pi//4)`
Also, `[(d^(2)P)/(d theta^(2))]_(theta = (pi//4)) = - h ["sin"(pi)/(4) + "cos"(pi)/(4)] = - h sqrt2 lt 0`
`:. P` is maximum when `theta = (pi//4)`
In this case, base `= h " cos"(pi)/(4) = (h)/(sqrt2)` and altitude `= h "sin"(pi)/(4) = (h)/(sqrt2)`
`:.` base = altitude, and hence the triangle is isosceles