Question

The surface area of a cube is increasing at the rate of

2 cm² /sec. When its edge is 90 cm, the volume is increasing at the rate of –

• A. 1620 cm³ /sec
• B. 810 cm³/sec
• C. 405 cm³/sec
• D. 45 cm³/sec

Answer 1

Answer: (D)

45 cm³/sec

Answer 2

Let at any time t, the length of each edge of the cube be x cm. Let S denote the surface area and V the volume of the cube. Then,

S = 6x² and V = x³

⇒ $\frac{dS}{dt}$ = 12 x $\frac{dx}{dt}$ and $\frac{dV}{dt}$ = 3x² $\frac{dx}{dt}$

⇒ 2 = 12 × 90 $\frac{dx}{dt}$ and $\frac{dV}{dt}$= 3 × 90² ×$\frac{dx}{dt}$

$\left\lbrack \because x = 90cmand\frac{dS}{dt} = 2 \right\rbrack$

⇒ 90 × $\frac{dx}{dt}$=$\frac{1}{6}$ and $\frac{dV}{dt}$ = 3 × 90 × $\left( 90 \times \frac{dx}{dt} \right)$

⇒  $\frac{dV}{dt}$ = 3 × 90 × $\frac{1}{6}$ cm³/sec = 45 cm³/sec.

Hence (4) is the correct answer.