Answer
The surface area is increasing at a rate of $~~\frac{4}{3}~cm^2/min$
Work Step by Step
$V = x^3$
$\frac{dV}{dt} = 3x^2\frac{dx}{dt} = 10~cm^3/min$
$\frac{dx}{dt} = \frac{10~cm^3/min}{3x^2}$
We can consider the surface area:
$A = 6x^2$
$\frac{dA}{dt} = 12x\frac{dx}{dt}$
$\frac{dA}{dt} = 12x(\frac{10~cm^3/min}{3x^2})$
$\frac{dA}{dt} = 4(\frac{10~cm^3/min}{x})$
$\frac{dA}{dt} = \frac{40~cm^3/min}{30~cm}$
$\frac{dA}{dt} = \frac{4}{3}~cm^2/min$
The surface area is increasing at a rate of $~~\frac{4}{3}~cm^2/min$
