The rate of change of the total surface area of a cube with respect to its volume, when the edge of the cube is 8 cm, is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{2}\)
Let a be the edge of the cube. Its total surface area S and volume V are
\(S = 6a^{2}\), \(V = a^{3}\)
Both depend on a, so the rate of change of S with respect to V is found through a:
\( \displaystyle \frac{dS}{dV} = \frac{dS/da}{dV/da} \)
\( \displaystyle = \frac{12a}{3a^{2}} = \frac{4}{a} \)
When a = 8:
\( \displaystyle \frac{dS}{dV} = \frac{4}{8} = \frac{1}{2} \)