The surface area of a cube increases at the rate of 72 cm^2 / sec. Find the rate of change of its volume, when the edge of the cube measures 3 cm.

Let edge of cube be 'x cm' $$\begin{aligned} & S=6 x^{2}, \frac{d S}{d t}=72 \mathrm{~cm}^{2} / \mathrm{sec} \Rightarrow 12 x \frac{d x}{d t}=72 \Rightarrow \frac{d x}{d t}=\frac{6}{x} \\ & \text { Volume, } \left.V=x^{3} \Rightarrow \frac{d V}{d t}=3 x^{2} \frac{d x}{d t}=3 \times x^{2} \times \frac{6}{x}=18 x, \frac{d V}{d t}\right]_{x=3}=54 \mathrm{~cm}^{3} / \mathrm{sec} \end{aligned} $$ ∴ Volume is increasing at the rate of $\displaystyle \mathbf{5 4} \mathbf{~ c m}^{\mathbf{3}} / \mathbf{s e c}$