Mathematics - Application of Derivative Question with Solution | TestHub

${\mathrm{Q}}^3=\mathrm{P}-3\mathrm{Q}$

$\Rightarrow \frac{\mathrm{P}}{{\mathrm{Q}}^2}=\mathrm{Q}+\frac{3}{\mathrm{Q}}$

$f(Q)=Q+\frac{3}{Q}$

$f^{\prime }(Q)=1-\frac{3}{Q^2}\Rightarrow Q=\pm \sqrt{3}$

$f(Q)f$ will be minimum at $Q=\sqrt{3}$

So minimum value of $\mathrm{f}(\mathrm{Q})$ is $2\sqrt{3}$

i.e. minimum of $\left[\frac{\mathrm{P}}{{\mathrm{Q}}^2}\right]=[2\sqrt{3}]=3$