Calculation of Max. and Min. value of $\displaystyle f(\phi) = \frac{1+\cos \phi}{2+\sin \phi}.$

$\bf{My\; Solution::}$ Let $\displaystyle y = \frac{1+\cos \phi}{2+\sin \phi}\Rightarrow 2y+y\sin \phi = 1+\cos \phi$

So $y\cdot \sin \phi - 1\cdot \cos \phi=1-2y\;,$ Now Using cauchy-Schwatrz Inequality, We get

$\displaystyle \left((y)^2+(-1)^2\right)\cdot \left(\sin^2 \phi+\cos^2 \phi\right)\geq \left(2y\sin \phi-\cos \phi\right)^2\Rightarrow \left(y^2+1\right)\geq \left(1-2y\right)^2$

So $\displaystyle y^2+1\geq 1+4y^2-4y\Rightarrow 3y^2-4y\leq 0\Rightarrow 3y\left(y-\frac{4}{3}\right)\leq 0 $

So We get $\displaystyle 0\leq y \leq \frac{4}{3}\Rightarrow y\in \left[0,\frac{4}{3}\right]$

My Question is can we solve it Using Geometrically or using Derivative Test,

If Yes Then plz explain me, Thanks

  • 2
    $\begingroup$ Geometrically, consider the slope of the straight line in the plane $\mathbb{R}^2$ connecting a point on unit cirlce and the point $(-2,-1)$ $\endgroup$ – Petite Etincelle Nov 15 '14 at 18:54

$$\displaystyle f(\phi) = \frac{1+\cos \phi}{2+\sin \phi}\implies f'(\phi)=-\frac{2\sin\phi +\cos \phi +1}{(2+\sin \phi)^2}.$$


$$f'(\phi)=0\Leftrightarrow 2\sin\phi +\cos \phi +1=0.$$ Since $\sin^2\phi+\cos^2\phi=1,$ we get that $\sin\phi=-4/5,\cos\phi=3/5$ or $\sin \phi=0,\cos \phi=-1.$


$$\sin \phi=0,\cos \phi=-1\implies \phi=\pi+2n\pi=(2n+1)\pi, n\in\mathbb{Z}.$$ and

$$\sin \phi=-\frac45,\cos \phi=\frac 35 \implies \phi=-\arcsin(4/5)+2n\pi, n\in\mathbb{Z}.$$

Using the second derivative test we conclude that the function achieves its minimum at $\phi=(2n+1)\pi, n\in\mathbb{Z}$ and its maximum at $\phi=-\arcsin(4/5)+2n\pi, n\in\mathbb{Z}.$ The minimum value is $f(\pi)=0$ and the maximum value is $f(-\arcsin(4/5))=4/3.$


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