$[\underline{[\text { Area and Volume (Extremum Problems) }}]$

Find the maximum volume of a cone with a slant height equal to $a$.

#
Let $h$ be the height of the cone, $r$ be the radius of the base, and $V(h)$ be the volume of the cone. It is clear that $0 < h < a$. If $0 < h < \frac{a}{\sqrt{3}}$, then $V^{\prime}(h) > 0$. Therefore, on the interval $\left(0 ; \frac{a}{\sqrt{3}}\right)$, the function $V(h)$ is increasing. If $\frac{a}{\sqrt{3}} < h < a$, then $V^{\prime}(h) < 0$. Therefore, on the interval $\left(\frac{a}{\sqrt{3}} ; a\right)$, the function $V(h)$ is decreasing. Consequently, at the point $h = \frac{a}{\sqrt{3}}$, the function $V(h)$ has a maximum, and

$$
V_{\max } = V\left(\frac{a}{\sqrt{3}}\right) = \frac{1}{3} \pi \left(a 2 - \left(\frac{a}{\sqrt{3}}\right) 2\right) \frac{a}{\sqrt{3}} = \frac{1}{3} \pi \cdot \frac{2}{3} a 2 \cdot \frac{a}{\sqrt{3}} = \frac{2 \pi a^{2} \sqrt{3}}{27}.
$$

## Answer

$\frac{2 \pi a^{3} \sqrt{3}}{27}$.