Answer
the rate of change in volume of cube is = $54\frac{in^3}{s}$
Work Step by Step
We are given that the rate of surface area change is $\frac{dA}{dt}=72\frac{in^2}{s}$ and length of cube is $x=3 in$
$A=6x^2$
on differentiate above with respect to t:
$\frac{dA}{dt}=12x\frac{dx}{dt}=12\times3\frac{dx}{dt}$
so $\frac{dx}{dt}=2\frac{m}{s}$
volume of cube $V=x^3$
rate of change in volume of cube is:
$\frac{dV}{dt}=3x^2\frac{dx}{dt}=3\times3^2\times2=54\frac{in^3}{s}$
thus the rate of change in volume of cube is = $54\frac{in^3}{s}$
