QPaperGen

Let the length of a side of the cube be \(x\), its volume be \(V\), and its surface area be \(S\). We know that \(x\) is a function of time \(t\). The formulas for the volume and surface area of a cube are:

\[\] Volume: \(V = x^3\) Surface Area: \(S = 6x^2\)\[\]

It is given that the rate of change of the volume with respect to time is 8 cm\(^3\)/s, i.e.:\[ \frac{dV}{dt} = 8 \text{ cm}^3/\text{s} \]To relate the rate of change of volume to the rate of change of the side length, we use the chain rule on the volume formula:\[ \frac{dV}{dt} = \frac{d}{dt}(x^3) = \frac{d}{dx}(x^3) \cdot \frac{dx}{dt} \]\[ 8 = 3x^2 \cdot \frac{dx}{dt} \]From this, we can express the rate of change of the side length, \(\frac{dx}{dt}\):\[ \frac{dx}{dt} = \frac{8}{3x^2} \quad \text{(1)} \]Now, we need to find the rate of change of the surface area with respect to time, \(\frac{dS}{dt}\). We apply the chain rule to the surface area formula:\[ \frac{dS}{dt} = \frac{d}{dt}(6x^2) = \frac{d}{dx}(6x^2) \cdot \frac{dx}{dt} \]\[ \frac{dS}{dt} = 12x \cdot \frac{dx}{dt} \]Substitute the expression for \(\frac{dx}{dt}\) from equation (1) into this equation:\[ \frac{dS}{dt} = 12x \cdot \left(\frac{8}{3x^2}\right) \]\[ \frac{dS}{dt} = \frac{96x}{3x^2} = \frac{32}{x} \text{ cm}^2/\text{s} \]Finally, we calculate the rate of change of the surface area when the length of the edge of the cube is 12 cm.When \(x = 12\) cm:\[ \frac{dS}{dt} = \frac{32}{12} = \frac{8}{3} \text{ cm}^2/\text{s} \]Therefore, if the length of the edge of the cube is 12 cm, its surface area is increasing at the rate of \(\frac{8}{3}\) cm\(^2\)/s.