Maximum Volume of a Right Circular Cone with Given Slant Height?

[checkitagain]

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What is the maximum volume of a right circular cone with a slant height of \sqrt{3} \ units?Feel free to use Volume \ = \ \dfrac{1}{3}\pi r^2h.

 

[Ackbach]

Here's my solution.

Spoiler

The volume of the cone is, as given, $V=\pi r^{2}h/3$, and the slant height is given as $\sqrt{3}$. Since the cone is a right circular cone, we have from the Pythagorean Theorem that
$r^{2}+h^{2}=\left(\sqrt{3}\right)^{2}=3.$ Hence, $r^{2}=3-h^{2}$. Writing the volume in terms of the height alone yields
$V=\pi(3-h^{2})h/3.$
We now have a maximization problem in one variable, the height. The height can range from a minimum of $0$ to a maximum of $\sqrt{3}$. That is, $h\in\left[0,\sqrt{3}\right]$. We are then maximizing a continuous function over a compact set. We are guaranteed that a maximum exists. I choose to take the Calculus I approach:

We have that
$$\frac{dV}{dh}=\frac{\pi}{3}\,(3-3h^{2})=\pi(1-h^{2}).$$
This equals zero when $h=\pm 1$. Since $h=-1\not\in\left[0,\sqrt{3}\right]$, the only critical point is at $h=1$. Evaluating our original function $V$ at the critical point and the two endpoints yields:

$$V(0)=0,\quad V(1)=\frac{2\pi}{3},\quad V\left(\sqrt{3}\right)=0.$$

Hence, the maximum volume is $2\pi/3$, and occurs when $h=1$ and $r=\sqrt{2}$.