Mathematics - Application of Derivative Question with Solution | TestHub

$\frac{1}{2}xy\sin A=\frac{1}{2}(\frac{1}{2}bc\sin A)$

$\Rightarrow xy=\frac{1}{2}bc$

$z_A^2=(PQ{)}^2=x^2+y^2-2xy\cos A=x^2+\frac{b^2{c}^2}{4x^2}-bc\cos A$

$\Rightarrow 2Z_A(\frac{d{Z}_A}{dx})=2x-\frac{b^2{c}^2}{2x^3}$

$\frac{d{Z}_A}{dx}=0\Rightarrow x=\sqrt{\frac{bc}{2}},and\frac{d^2{Z}_A}{d{x}^2}>0$

Hence $Z_A$ is minimum if $x=\sqrt{\frac{bc}{2}}$ and the minimum value of $Z_A$ is,

$\sqrt{\frac{bc}{2}+\frac{bc}{2}-bc\cos A}=\sqrt{2\Delta \tan \frac{A}{2}}$