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(A) Let $r$ be the radius and $V$ be the volume of the spherical mothball. The volume $V = \frac{4}{3} \pi r^3$ and surface area $S = 4 \pi r^2$.Given

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$6 	ext{ months}$

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(A) Let $r$ be the radius and $V$ be the volume of the spherical mothball. The volume $V = \frac{4}{3} \pi r^3$ and surface area $S = 4 \pi r^2$.Given that the rate of change of volume is proportional to the surface area: $\frac{dV}{dt} = -k S$, where $k > 0$.Since $V = \frac{4}{3} \pi r^3$, $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Substituting this into the equation: $4 \pi r^2 \frac{dr}{dt} = -k (4 \pi r^2)$.This simplifies to $\frac{dr}{dt} = -k$.Integrating with respect to $t$: $r(t) = -kt + C$.At $t = 0$, $r = 3$, so $C = 3$. Thus, $r(t) = 3 - kt$.At $t = 4$, $r = 1$, so $1 = 3 - 4k$, which gives $4k = 2$, or $k = 0.5$.The mothball evaporates completely when $r = 0$: $0 = 3 - 0.5t$.Solving for $t$: $0.5t = 3$, so $t = 6 	ext{ months}$.

$A$ spherical mothball has an initial radius of $3 \text{ cm}$. Due to evaporation, the radius of the ball reduces to $1 \text{ cm}$ in $4 \text{ months}$. In how many months would the mothball evaporate completely if the volume is lost at a rate proportional to the surface area?

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