The volume of a cube is increasing at the rate of $27 ~cm^3/s$. How fast is the surface area increasing when the length of the cube is 12 cm.

$9 ~cm^2/s$

V → Volume of cube 

x → side of cube

t → time

in this case $\frac{dv}{dt} = 27 \frac{cm^3}{s}$  (rate of charge of volume) 

V = x³   so  $\frac{d(x^3)}{dt} = 27$

$\Rightarrow 3 x^2 \frac{d x}{d t}=27$           ........(1)

Surface area of cube  S(x) = 6x²

rate of charge of surface area = $\frac{ds(x)}{dt} = 12x \frac{dx}{dt}$

from (1)

$\frac{dx}{dt} = \frac{27}{3x^2}$

So $\frac{d(3(x))}{d t}=\frac{12 x \times 27}{3 x^2}=\frac{4 \times 27}{x}$

for x = 12

$\frac{d(s(x))}{d t}=\frac{4 \times 27}{12}=9 ~cm^2 / s$