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?
Answer & explanation
Correct answer: option 4
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 $72\, cm^2/sec$.