The hypotenuse of a triangle is shortest when the other two legs are equal. Prove this using the Calculus of Variations.

Question

I started off my proof by of course stating that the different right triangles I would be comparing should have the same area A. I was able to show that what the question is asking is true visually and computationally using the Pythagorean Theorem, and even using the triangle inequality, but I don't really know how to set it up in such a way that I can use the Euler-Lagrange equation in order to prove that this is true.

Equations:

$$\frac{1}{2}xy=A$$

So

$$y(x) = 2A/x \Longrightarrow y'(x) = -\frac{2A}{x^2}$$

Answer 1

Let $A=\frac{xy}{2}$ be constant. Then we have $x^2 = (\frac{2A}{y})^2$.

The hypothenus is $h(y)=\sqrt{x^2 + y^2} = \sqrt{y^2 + \frac{4A^2}{y^2}}$, so $h'(y)=\frac{2y-\frac{8A^2}{y^3}}{2\sqrt{y^2 + \frac{4A^2}{y^2}}}$.

Then $h$ has got a minimum when $h'(y)=0$, so when $2y=\frac{8A^2}{y^3}$, so when $y^4 = 4\cdot \frac{x^2y^2}{4}$, so when $y=x$.

Answer 2

$$\min_{xy=2A}\sqrt{x^2+y^2} = \sqrt{\min_{xy=2A}x^2+y^2} $$ and $(x-y)^2\geq 0$ is equivalent to $x^2+y^2\geq 2xy$, with equality attained only at $x=y$. This implies

$$\min_{xy=2A}\sqrt{x^2+y^2} = \sqrt{4A}. $$