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

$$r = \frac{ab}{10}$$ The total surface area then becomes

$$S = \frac{\pi \cdot ab(a+b)}{10}$$ 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.


1 Answer 1


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 $$A_{\max}={\pi c^2\over\sqrt{2}}\ .$$


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