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(D) Given: $\frac{d(2r)}{dt} = 4 	ext{ cm/s} \implies \frac{dr}{dt} = 2 	ext{ cm/s}$.Volume of cylinder $V = \pi r^2 h$. Since volume is constant, $

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$-48$

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(D) Given: $\frac{d(2r)}{dt} = 4 	ext{ cm/s} \implies \frac{dr}{dt} = 2 	ext{ cm/s}$.Volume of cylinder $V = \pi r^2 h$. Since volume is constant, $\frac{dV}{dt} = 0$.$\frac{dV}{dt} = \pi \left( r^2 \frac{dh}{dt} + 2rh \frac{dr}{dt} ight) = 0$.$r^2 \frac{dh}{dt} + 2rh(2) = 0 \implies r \frac{dh}{dt} + 4h = 0 \implies \frac{dh}{dt} = -\frac{4h}{r}$.At $r = 4 	ext{ cm}$ and $h = 12 	ext{ cm}$, $\frac{dh}{dt} = -\frac{4(12)}{4} = -12 	ext{ cm/s}$.Lateral surface area $S = 2 \pi rh$.$\frac{dS}{dt} = 2 \pi \left( r \frac{dh}{dt} + h \frac{dr}{dt} ight)$.Substituting the values: $\frac{dS}{dt} = 2 \pi \left( 4(-12) + 12(2) ight) = 2 \pi (-48 + 24) = 2 \pi (-24) = -48 \pi 	ext{ cm}^2/	ext{s}$.

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