Question

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 \(12 cm\)?

Answer 1

The correct answer is \(\frac{8}{3} cm^2 /s.\)
Let \(x\) be the length of a side, \(V\) be the volume, and \(s\) be the surface area of the cube. Then, \(V = x^3\) and \(S = 6x^2\) where \(x\) is a function of time \(t\).
It is given that \(\frac{dv}{dt}=8cm^3/s\)
Then, by using the chain rule, we have:
\(∴ 8=\frac{dv}{dt}=\frac{d}{dt}(x^3)=\frac{dx}{dt}(x^3).\frac{dx}{dt}=3x^2.\frac{dx}{dt}\)
\(\frac{dx}{dt}=\frac{8}{3x^2}....(1)\)
Now, \(\frac{ds}{dt}=\frac{d}{dt}(6x^2)=\frac{d}{dt}(6x^2).\frac{dx}{dt}\) ...[By chain rule]
\(=12x.\frac{dx}{dt}=12x.\frac{dx}{dt}=12x.(\frac{8}{3x^2})=\frac{32}{x}\)
\(\frac{ds}{dt}=\frac{32}{12}cm^2/s=\frac{8}{3}cm^2/s.\)
Thus, when \(x = 12 cm,\) Hence, if the length of the edge of the cube is \(12 cm\), then the surface area is increasing at the rate of \(\frac{8}{3} cm^2 /s.\)