The volume of a cube is increasing at a rate of 9 cubic centimeter per second. Then rate of change of surface area when the length of an edge is 10 centimeter is:

3.6 cm²/s

The correct answer is Option (4) → 3.6 cm²/s

S → slop of cube

$\frac{dv}{dt}=9$, $v=s^3$

so $3s^2\frac{ds}{dt}=9$

$\frac{ds}{dt}=\frac{3}{s^2}$

Surface area $A = 6s^2$

$\frac{dA}{dt}=12s\frac{ds}{dt}$

$⇒\frac{dA}{dt}=\frac{125×3}{s^2}=\frac{36}{s}$

at $s=\frac{dA}{dt}=3.6cm^2/s$