Answer
change in surface area $\frac{dA}{dt}=-180\frac{m^2}{min}$ and $\frac{dA}{dt}=-135\frac{m^3}{min}$
Work Step by Step
The given side of cube is $x=3m $ and $\frac{dx}{dt}=-5\frac{m}{min}$
for surface area:
$A=6x^2$
on diffrentiating above:
$\frac{dA}{dt}=12x\frac{dx}{dt}$
$\frac{dA}{dt}=12\times 3\times{-5}=-180\frac{m^2}{min}$
for rate of change in volume:
$V=x^3$
differentiating with respect to t:
$\frac{dV}{dt}=3x^2\frac{dx}{dt}=3\times3^2\times(-5)=-135\frac{m^3}{min}$
so change in surface area is $\frac{dA}{dt}=-180\frac{m^2}{min}$ and $\frac{dA}{dt}=-135\frac{m^3}{min}$
