If an edge of a variable cube is increasing at the rate of 0.5 cm/sec, at what rate is its surface area increasing when its edge is 12 cm?

$72\,cm^2/s$

The correct answer is Option (4) → $72\,cm^2/s$

Let a cm be the side of a cube and S its surface area at any time t seconds, then

$S = 6a^2$   …(i)

Differentiating (i) w.r.t. t, we get $\frac{dS}{dt}= 6.2a\frac{da}{dt}= 12a \frac{da}{dt}$   …(ii)

But $\frac{da}{dt}= 0.5\, cm/sec=\frac{1}{2}\,cm/sec$ (given)

From (ii), $\frac{dS}{dt}=12a ×\frac{1}{2} = 6a$.

When $a = 12 cm, \frac{dS}{dt}= 6 x 12 = 72$.

Hence, the area is increasing at the rate of $72\, cm^2/sec$.