Answer
Change in surface area is $\dfrac{dA}{dt}=-180\dfrac{m^2}{min}$ and $\dfrac{dA}{dt}=-135\dfrac{m^3}{min}$
Work Step by Step
As we are given that Side of cube is $x=3m $ and $\dfrac{dx}{dt}=-5\dfrac{m}{min}$
Find surface area.
$A=6x^2$
or, $\dfrac{dA}{dt}=12x\dfrac{dx}{dt}= 12\times 3\times{-5}=-180\dfrac{m^2}{min}$
For rate of change in volume, we have :
$V=x^3$
Now, differentiating with respect to $t$, we have:
Thus, $\dfrac{dV}{dt}=3x^2\dfrac{dx}{dt}=3\times3^2\times(-5)=-135\dfrac{m^3}{min}$
Hence, the Change in surface area is $\dfrac{dA}{dt}=-180\dfrac{m^2}{min}$ and $\dfrac{dA}{dt}=-135\dfrac{m^3}{min}$
