Question

If the surface area of a cube is increasing at a rate of $3.6{\text{cm}}^2/\text{sec}$ retaining its shape, then the rate of change of its volume (in ${\text{cm}}^3/\text{sec}$) when the length of a side of the cube is $10\text{cm}$ is:

• A. $9$
• B. $10$
• C. $18$
• D. $20$

Answer 1

The correct option is A $9$

Let $a$ be side of cube
$A=$ Surface area and $V=$ Volume of cube
$A=6a^2$
$\Rightarrow \frac{dA}{dt}=6\left(2a\frac{da}{dt}\right)$
$\Rightarrow 3.6=12\times 10\frac{da}{dt}$
$\Rightarrow \frac{da}{dt}=\frac{3}{100}\cdots \left(1\right)$

$V=a^3$
$\Rightarrow \frac{dV}{dt}=3a^2\frac{da}{dt}$
Using equation $\left(1\right)$, we get
$\frac{dV}{dt}=3\times 100\times \frac{3}{100}$
$=9{\text{cm}}^3/\text{sec}$