If an edge of a variable cube is increasing at the rate of 0.8 cm/sec, at what rate its surface area is increasing when its edge is 10 cm?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 96 cm2/sec
For a cube of edge x, the surface area is \( S=6x^2 \). Differentiating with respect to time t:
\( \displaystyle \frac{dS}{dt}=12x\,\frac{dx}{dt} \)
With x = 10 cm and \( \frac{dx}{dt}=0.8 \) cm/sec:
\( \displaystyle \frac{dS}{dt}=12\times 10\times 0.8=96\ \text{cm}^2/\text{sec} \)
Why the other options are incorrect
Option (1) 72 cm2/sec — Incorrect. At x = 10 the rate is 12 × 10 × 0.8 = 96, not 72.
Option (3) 48 cm2/sec — Incorrect. 48 uses \( S=3x^2 \) (half the faces) or \( 6x\frac{dx}{dt} \) without the factor 2 from differentiating \( x^2 \).
Option (4) 90 cm2/sec — Incorrect. At x = 10 the rate is 96, not 90.