Optimisation: volume of a cone

Question

Question: A right triangle with hypotenuse of length a is rotated about one of its legs to generate a right circular cone. Find the greatest possible volume of such a cone.

My thoughts:

Given the formula for the volume of a cone and the Pythagorean Theorem, I can eliminate either the base or the height variable from the formula for the volume of a cone, giving me volume in terms of hypotenuse and height or volume in terms of hypotenuse and base.

I think that the only way to determine the greatest possible volume of such a cone is to assume that the length of the hypotenuse (a) is constant, as I will otherwise be unable to find the derivative (dV/dh or dV/db) and then determine the maximum.

Is this the case? Should I treat the length of hypotenuse as a constant?

Answer 1

YES.
You're told you have 'a right triangle with hypotenuse of length $a$'.
That means the length is constant. The value is not specified now and it's hidden behind the symbol $a$, but it is some constant.

Answer 2

Indicating with $x$ the angle of the hypotenuse with the base, we have

• radius $R=L\cos x$
• height $H=L\sin x$

then

• volume $V=\frac13\pi R^2H=\frac13 \pi L^3\cos^2x \sin x=\frac13 \pi L^3\sin x(1-\sin^2 x)$

and therefore

$$f(x)=\sin x(1-\sin^2 x)\implies f'(x)=\cos x-3\sin^2x\cos x=0\\\implies \cos x=0 \, \lor\,\sin x=\frac{\sqrt 3}3$$

that is

• max volume $V_{max}=\frac13 \pi L^3\frac{\sqrt 3}3\frac23=\frac{2\sqrt 3}{27}\pi L^3$

Answer 3

An Image will help you, let $x$ and $y$ the length of the legs, then the searched volume is $$V=\frac{1}{3}\pi x^2y$$ where $$x^2+y^2=a^2$$ plugging this in our formula, we get $$V=\frac{1}{3}\pi(-y^2+a^2)y$$ or $$V=\frac{1}{3}\pi(-y^3+a^2y)$$ let $$h(y)=\frac{1}{3}\pi(-y^3+a^2y)$$ then $$h'(y)=\frac{1}{3}\pi(-3y^2+a^2)$$ Can your proceed? Solving the equation $$h'(y)=0$$ then we get $$y=\frac{a}{\sqrt{3}}$$

Answer 4

Consider the triangle with legs $x$ and $y$. The hypotenuse is $a$. So,

$$a^2=x^2+y^2.\tag 1$$ The volume of the cone in question is $$\frac {x^2y}{12}\pi .$$

Substituting $x$ from $(1)$ we get $$\frac{a^2y-y^3}{12}\pi$$ for the volume as a function of $y$.

The derivative of the volume with respect to $y$ is

$$\frac{a^2-3y^2}{12}\pi.$$

Setting this equal to $0$ and solving for $y$ we get

$$y=\frac{a}{\sqrt 3}, \ \ x=a \sqrt{1-\frac1{3}}$$ as the lengths of the legs belonging to the triangle providing the largest volume.