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(A) Let $x$ be the length of an edge,$V$ be the volume,and $S$ be the surface area of the cube.We have $V = x^{3}$ and $S = 6x^{2}$.Given that the rat

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$\frac{8}{3} \, cm^{2}/s$

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(A) Let $x$ be the length of an edge,$V$ be the volume,and $S$ be the surface area of the cube.We have $V = x^{3}$ and $S = 6x^{2}$.Given that the rate of change of volume is $\frac{dV}{dt} = 8 \, cm^{3}/s$.Using the chain rule,$\frac{dV}{dt} = \frac{d}{dt}(x^{3}) = 3x^{2} \cdot \frac{dx}{dt}$.Substituting the given values: $8 = 3x^{2} \cdot \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{8}{3x^{2}}$.Now,the rate of change of surface area is $\frac{dS}{dt} = \frac{d}{dt}(6x^{2}) = 12x \cdot \frac{dx}{dt}$.Substituting $\frac{dx}{dt}$ from the previous step: $\frac{dS}{dt} = 12x \cdot \left(\frac{8}{3x^{2}}\right) = \frac{32}{x}$.When the edge length $x = 12 \, cm$,the rate of change of surface area is $\frac{dS}{dt} = \frac{32}{12} = \frac{8}{3} \, cm^{2}/s$.

The volume of a cube is increasing at the rate of $8 \, cm^{3}/s$. How fast is the surface area increasing when the length of an edge is $12 \, cm$?

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