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(C) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$.Given: $\frac{dV}{dt} = 40 	ext{ cm}^3/	ext{min}$ and

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$10$

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(C) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$.Given: $\frac{dV}{dt} = 40 	ext{ cm}^3/	ext{min}$ and $r = 8 	ext{ cm}$.The volume of a sphere is $V = \frac{4}{3}\pi r^3$.Differentiating with respect to $t$: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.Substituting the given values: $40 = 4\pi (8)^2 \frac{dr}{dt} \Rightarrow 40 = 256\pi \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{40}{256\pi} = \frac{5}{32\pi} 	ext{ cm/min}$.The surface area of a sphere is $S = 4\pi r^2$.Differentiating with respect to $t$: $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$.Substituting $r = 8$ and $\frac{dr}{dt} = \frac{5}{32\pi}$:$\frac{dS}{dt} = 8\pi 	imes 8 	imes \frac{5}{32\pi} = 64\pi 	imes \frac{5}{32\pi} = 2 	imes 5 = 10 	ext{ cm}^2/	ext{min}$.

The volume of a spherical balloon is increasing at a rate of $40 \text{ cm}^3/\text{min}$. Find the rate of change of its surface area when its radius is $8 \text{ cm}$ (in $\text{cm}^2/\text{min}$).

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