Rotate right triangle with perimeter 1 about the hypotenuse

Question

We rotate every right triangle with perimeter 1 about its hypotenuses. Is it true that we can choose a solid from so obtained solids that has maximum volume? If yes, what's the volume? I guess I should use derivative in the end, but what to do at first?

Answer 1

In a right triangle with legs $x$ and $y$ and hypotenuse $z$, the altitude, perpendicular to the hypotenuse has length $$ r = \frac{xy}{z} = \frac{xy}{\sqrt{x^2 + y^2}}. $$ When you rotate the right triangle about the hypotenuse, the solid shape consists of two cones of radius $r$ and combined height $z$. So the total volume (using Archimedes' formula) is $$ V = \frac{\pi}{3}zr^2 = \frac{\pi}{3}z \Bigl( \frac{xy}{z} \Bigr)^2 = \frac{\pi x^2 y^2}{3z} = \frac{\pi x^2 y^2}{3\sqrt{x^2 + y^2}}. $$ Now the perimeter constraint $P = 1$ amounts to $$ x + y + z = 1 \qquad\text{or}\qquad x + y + \sqrt{x^2 + y^2} = 1. $$

This is a standard constrained optimization problem, which can be solved using the method of Lagrange multipliers. The objective function that you're trying to maximize is $V = V(x,y)$ and the constraint is $P(x, y) = 1$. Solving the system of equations $$ \left\{ \begin{align} \dfrac{dV}{dx} &= \lambda \dfrac{dP}{dx}, \\ \dfrac{dV}{dy} &= \lambda \dfrac{dP}{dy}, \\ P &= 1 \end{align} \right. $$ yields $$ x = y = \frac{1}{2+\sqrt{2}}, $$ which corresponds to the case of the isosceles right triangle (half of a square).