Question

Find the approximate change in the volume $V$ of a cube of side $x$ meters caused by increasing the side by $2\%$.

Answer 1

Given that the side of a cube is $x$ meters

So, the volume of a cube will be $V=x^3$

Now the side $x$ is increased by $2\%$, so the new length of side will be $x(1+\frac{2}{100})=\frac{51x}{50}$

The new volume will be $V_1={\left(\frac{51x}{50}\right)}^3={\left(\frac{51}{50}\right)}^3{x}^3$

Therefore, the change in volume is $V_1-V=({\left(\frac{51}{50}\right)}^3-1)x^3$

The percentage change in volume is :

$=\frac{V_1-V}{V}\times 100=\left(\frac{{\left(\frac{51}{50}\right)}^3-1}{1}\right)\times 100=\left(\frac{{51}^3-{50}^3}{{50}^3}\right)\times 100=\frac{7651}{125000}\times 100=6.12\%$.