Answer
Rate of change in volume of cube is $54\dfrac{in^3}{s}$
Work Step by Step
Since, we are given that the rate of surface area change is $\dfrac{dA}{dt}=72\frac{in^2}{s}$ and length of cube is $x=3$ inch
$A=6x^2$
on differentiate above with respect to t:
$\dfrac{dA}{dt}=12x\dfrac{dx}{dt}=12\times3\dfrac{dx}{dt}$
This implies$\dfrac{dx}{dt}=2\frac{m}{s}$
Now, Volume of cube is $V=x^3$
Thus, the rate of change in volume of cube will be:
$\dfrac{dV}{dt}=3x^2(\dfrac{dx}{dt})=3\times3^2\times2=54\frac{in^3}{s}$
Hence,the Rate of change in volume of cube is $54\dfrac{in^3}{s}$
