The surface area of a sphere increases at the rate of 1 sq. ft per second, then the rate of increase of volume when the radius is 3 ft is

$$\begin{gathered} \mathrm{V}=3 \frac{\mathrm{ds}}{\mathrm{dt}}=1 \\ \mathrm{S}=4{\mathrm{\pi r}}^2 \Rightarrow \frac{\mathrm{ds}}{\mathrm{dt}}=8\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}} \\ \Rightarrow 1=24\mathrm{\pi }\frac{\mathrm{dr}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{24\mathrm{\pi }} \\ \mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^3 \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=4{\mathrm{\pi r}}^2\frac{\mathrm{dr}}{\mathrm{dt}} \\ =4\mathrm{\pi }\times 9\times \frac{1}{24\mathrm{\pi }} =\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \end{gathered}$$