# The volume of a cube is increasing at the rate of $8 cm^{3}/S$. How fast is the surface area increasing when the length of its edge is 12 cm?

Let x be the strength of side, v be the volume & s be the surface area of cube
Then $v= x^{3}\: and\: s= 6x^{2}$,where x is a function of
line t
Now $\frac{dv}{dt}= 8\frac{cm^{3}}{s} \left [ given \right ]$
$\therefore 8= \frac{dv}{dt}= \frac{d}{dx}\left ( x^{3} \right )= 3x^{2}\frac{dx}{dt}\left ( chain \: rule \right )$
$8= 3x^{2}\frac{dx}{dt}$
$\frac{8}{3x^{2}}= \frac{dx}{dt}-(i)$
Now, $\frac{ds}{dt}= 12x\frac{8}{3x^{2}}\left [ using(i) \right ]\Rightarrow \frac{32}{x}$
Hence when $x= 12cm$
then, $\frac{ds}{dt}= \frac{32}{12}= \frac{8}{3}\: cm^{2}/S$

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