Rotation around hypotenuse of a right triangle with a, b, c sides where $$a^2+b^2=c^2$$ and $$a+b+c=1$$ creates a solid figure. Find the length of each side of this triangle when the figure has the highest volume.
I've come to the part when $$V=\frac{\pi}{3}\cdot\frac{a^2b^2}{c}$$ and don't know what to do next. I'm not even sure if it's a proper way to solve this problem. Any help or hints will be appreciated.
If the legs of our right triangle are $a,b>0$, the constraint on the perimeter gives $$ \sqrt{a^2+b^2} = 1-(a+b),\qquad 0 = 1 +2ab -2a-2b$$ hence $$(1-a)(1-b)= 1-a-b+ab = \frac{2-2a-2b+2ab}{2}=\frac{1}{2}$$ and given $a\in\left(0,\frac{1}{2}\right)$ we have $b=\frac{1-2a}{2-2a}$. The volume of the "double cone" generated by the rotation of the right triangle around its hypotenuse is given by $$ \frac{\pi}{3}\left(\frac{ab}{\sqrt{a^2+b^2}}\right)^2\sqrt{a^2+b^2}=\frac{\pi a^2 b^2}{3\sqrt{a^2+b^2}}=\frac{\pi a^2(1-2a)^2}{6(1-a)(1-2a+2a^2)}=V(a) $$ and by solving $V'(a)=0$ one has that the absolute maximum of $V(a)$ over $\left(0,\frac{1}{2}\right)$ occurs at $a=1-\frac{1}{\sqrt{2}}$, i.e. that the maximum volume $V_{max}=\frac{\pi}{12}(\sqrt{2}-1)^3$ is attained by the only right and isosceles triangle with unit perimeter.