A spherical chocolate ball has a layer of ice-cream of uniform thickness around it.

Question

A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm , the ice-cream melts at the rate of $81 \mathrm{~cm}^3 / \mathrm{min}$ and the thickness of the ice-cream layer decreases at the rate of $\frac{1}{4 \pi} \mathrm{~cm} / \mathrm{min}$. The surface area (in $\mathrm{cm}^2$ ) of the chocolate ball (without the ice-cream layer) is :

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& \mathrm{v}=\frac{4}{3} \pi \mathrm{r}^3 \
& \frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}} \
& 81=4 \pi \mathrm{r}^2 \times \frac{1}{4 \pi} \
& \mathrm{r}^2=81 \
& \mathrm{r}=9
\end{aligned}\
\&\text { surface area of chocolate }=4 \pi(r-1)^2=256 \pi
\end{aligned}$$

Solutions · 1