The volume of a cube is increasing at the rate of $6 \text{ cm}^3/\text{s}$. How fast is the surface area of cube increasing, when the length of an edge is $8 \text{ cm}$?

$3 \text{ cm}^2/\text{s}$

The correct answer is Option (3) → $3 \text{ cm}^2/\text{s}$ ##

Given, $\frac{dV}{dt} = 6 \text{ cm}^3/\text{s}$

$l = 8 \text{ cm}$

$V = \text{volume of cube} = l^3$

Find, $\frac{dS}{dt}$

$⇒\frac{dV}{dt} = 3l^2 \frac{dl}{dt}$

$⇒6 = 3l^2 \frac{dl}{dt}$

$⇒\frac{2}{l^2} = \frac{dl}{dt} \dots (i)$

Surface area of cube, $S = 6l^2$

$⇒\frac{dS}{dt} = 12l \frac{dl}{dt} \dots (ii)$

Now put eq. (i) and eq. (ii)

$⇒ \frac{dS}{dt} = 12l \times \frac{2}{l^2}$

$⇒\frac{dS}{dt} = \frac{24}{8}$

$⇒\frac{dS}{dt} = 3 \text{ cm}^2/\text{s}$