Question

The surface area of a cube is increasing at the rate of $2c{m}^2/sec$. When its edge is 90 cm, the volume is increasing at the rate of.

• A. $1620c{m}^3/sec$
• B. $810c{m}^3/sec$
• C. $405c{m}^3/sec$
• D. $45c{m}^3/sec$

Answer 1

The correct option is D $45c{m}^3/sec$

Given $\frac{dA}{dt}=2c{m}^2/sec$ , $x=90cm$
Surface area of cube is $A=6x^2$
$\frac{dA}{dt}=12\times x\times \frac{dx}{dt}$
$2=12\times 90\times \frac{dx}{dt}$
$\Rightarrow \frac{1}{6\times 90}=\frac{dx}{dt}$
Now, $V=x^3$
$\frac{dV}{dx}=3x^2\frac{dx}{dt}$
$\Rightarrow \frac{dV}{dt}=3\times 90\times 90\times \frac{1}{6\times 90}=45c{m}^3/sec$