2023-M2-06

$\begin{array}{cl}
= & \pi \dint_0^h x^2dy \\
= & \pi \dint_0^h [r^2-(y-r)^2] dy \\
= & \pi \dint_0^h \left( r^2-y^2+2ry-r^2 \right) dy \\
= & \pi \dint_0^h \left(-y^2+2ry \right) dy \\
= & \pi\left[ \dfrac{-1}{3}y^3+ry^2 \right]_0^h \\
= & \pi \left(\dfrac{-1}{3}h^3+rh^2 \right) \\
= & \dfrac{\pi}{3}h^2(3r-h)
\end{array}$

Sub. $r=10$ into the result of (a), we have

$\begin{array}{rcl}
V & = & \pi(11)^2 h -\dfrac{\pi}{3}h^2(30-h) \\
V & = & 121\pi h -10\pi h^2+\dfrac{\pi}{3}h^3 \\
\dfrac{dV}{dt} & = & 121\pi \dfrac{dh}{dt}-20\pi h\dfrac{dh}{dt}+\pi h^2 \dfrac{dh}{dt} \\
1 & = & (121\pi-20\pi h+\pi h^2)\dfrac{dh}{dt} \text{, }\because \dfrac{dV}{dt}=1 \\
\dfrac{dh}{dt} & = & \dfrac{1}{121\pi-20\pi h+\pi h^2}
\end{array}$

Let $f(h)=121\pi-20\pi h+\pi h^2$. Then we have

$\begin{array}{rcl}
f'(h) & = & 0 \\
-20\pi +2\pi h & = & 0 \\
h & = & 10
\end{array}$

Since $f'(h)<0$ for $h<10$ and $f'(h)>0$ for $h>10$, then $f(h)$ attains its minimum value when $h=10$.

Note that $f(h)$ attains it minimum value implies that $\dfrac{1}{f(h)}$ attains it maximum value.

Hence, the greatest value of $\dfrac{dh}{dt}$

$\begin{array}{cl}
= & \dfrac{1}{121\pi-20\pi (10)+\pi(10)^2} \\
= & \dfrac{1}{21\pi}
\end{array}$