The volume of a cube is increasing at the rate of $8 c{m}^3/sec$. How fast is the surface area increasing when the length of an side is $12 cm$?

Let $x$ be the side S be the surface area $V$ be the volume of cube

$\text{ G.T, }x=12,\frac{dv}{dt}=8{\mathrm{cm}}^2/\sec$

volume $V=x^3$

$\frac{dv}{dt}=3x^2\frac{dx}{dt}$

$\Rightarrow 8=3\times {12}^2\frac{\mathrm{d}x}{\mathrm{dt}}\Rightarrow \frac{\mathrm{d}x}{\mathrm{dt}}=\frac{8}{3\times 144}$

surface area $S=6x^2$

$\frac{ds}{dt}=12x\times \frac{dx}{dt}=12\left(12\right)\times \frac{8}{3\times 144}$

$\frac{ds}{dt}=\frac{8}{3}{\mathrm{cm}}^2/\sec$