# A right triangle with hypotenuse 10 and two sides of variable length are rotated about its hypotenuse.What is the maximum possible area of solid?

The complete question is "A right triangle with hypotenuse 10 and two other sides of variable length is rotated about its longest side.What is the maximum possible area of such a solid?

The solid formed is two cones.

My approach:

I took the length of the two sides as 'a' and 'b'. Then the radius of the cone is the perpendicular from hypotenuse to the opposite side given by

$$r = \frac{(a)(b)}{10}$$ The total surface area then becomes

$$S = \frac{(pi*(ab)(a+b)}{(10)}$$ I now need to find the values of a and b for which S is maximum and this feels quite long and cumbersome. Could anyone think of a more optimized solution. I feel it should happen when a=b but cant prove that in anyway.

Any help is greatly appreciated. Thanks in advance.

Denote the legs of the triangle by $a$ and $b$, its height by $h$. The two cones have a common rim of length $2\pi h$. The total area $A$ of the two cones is therefore given by $$A={1\over2}\cdot 2\pi h\cdot(a+b)={\pi\over c}ab(a+b)\ .$$ The point $(a,b)$ lies on the circle of radius $c$, and of course in the first quadrant. It is obvious that $ab$ as well as $a+b$ take their maximum on the midpoint $\bigl({c\over\sqrt{2}},{c\over\sqrt{2}}\bigr)$ of this arc. The maximal possible area is therefore given by $$A_{\max}={\pi c^2\over\sqrt{2}}\ .$$