Mathematics - Application of Derivative Question with Solution | TestHub

Let $x=2+\cos \theta$ and $y=2+\sin \theta$ $x+y=4+\sin \theta +\cos \theta \Rightarrow$ maximum of $(x+y)$ is $4+\sqrt{2}$ $x-y=\cos \theta -\sin \theta \Rightarrow$ maximum of $x-y$ is $\sqrt{2}$ $f(\theta )=xy=4+2(\sin \theta +\cos \theta )+\sin \theta \cos \theta$ $f^{\prime }(\theta )=2(\cos \theta -\sin \theta )+\cos 2\theta =0\Rightarrow \tan \theta =1$ or $2+\sin \theta +\cos \theta =0$ rejected $f\left(\frac{\pi }{4}\right)=4+2\sqrt{2}+\frac{1}{2}$ $g(\theta )=\frac{x}{y}=\frac{2+\cos \theta }{2+\sin \theta }\Rightarrow g^{\prime }(\theta )=\frac{(2+\sin \theta )(-\sin \theta )-(2+\cos \theta )\times \cos \theta }{(2+\sin \theta {)}^2}$ $g^{\prime }(\theta )=0\Rightarrow 2\sin \theta +{\sin }^2\theta +2\cos \theta +{\cos }^2\theta =0\Rightarrow \sin \theta +\cos \theta =-\frac{1}{2}$ $\theta =\frac{\pi }{4}-{\cos }^{-1}\left(\frac{-1}{2\sqrt{2}}\right)$. Let $\alpha ={\cos }^{-1}\left(\frac{-1}{2\sqrt{2}}\right)$ $\cos \left(\frac{\pi }{4}-\alpha \right)=\frac{\sqrt{7}-1}{4}$ and $\sin \left(\frac{\pi }{4}-\alpha \right)=\frac{-\sqrt{7}-1}{4}$ Maximum value of $\frac{x}{y}$ is $\frac{7+\sqrt{7}}{7-\sqrt{7}}=\frac{4+\sqrt{7}}{3}$