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Question Number 208167 by hardmath last updated on 06/Jun/24

$$\mathrm{y}=3{\cos }^2\alpha +2\cos \alpha$$$$\mathrm{find}:\max \left(\mathrm{y}\right)=?$$

Answered by A5T last updated on 07/Jun/24

$$y⩽3\times 1+2\times 1=5\Rightarrow max\left(y\right)=5(Equalityat\alpha =0]$$

Commented by A5T last updated on 07/Jun/24

$${[\sqrt{3}cos\alpha +\frac{1}{\sqrt{3}}]}^2=3{cos}^2\alpha +2cos\alpha +\frac{1}{3}$$$$\Rightarrow y=3{cos}^2\alpha +2cos\alpha ={[\sqrt{3}cos\alpha +\frac{1}{\sqrt{3}}]}^2-\frac{1}{3}$$$$\Rightarrow max\left(y\right)={[\sqrt{3}\times 1+\frac{1}{\sqrt{3}}]}^2-\frac{1}{3}=\frac{16-1}{3}=5$$$$\Rightarrow min\left(y\right)=\frac{-1}{3}[Equalityatcos\alpha =\frac{-1}{3}]$$

Commented by hardmath last updated on 07/Jun/24

$$\mathrm{thamk}\mathrm{you}\mathrm{dear}\mathrm{professor}$$$$\min \left(\mathrm{y}\right)=1?$$

Answered by mr W last updated on 07/Jun/24

$$y=3{\cos }^2\alpha +2\cos \alpha$$$$y^{\prime }=-6\cos \alpha \sin \alpha -2\sin \alpha$$$$=-2\sin \alpha (3\cos \alpha +1)\stackrel{!}{=}0$$$$\Rightarrow \sin \alpha =0\Rightarrow \cos \alpha =\pm 1$$$$\Rightarrow 3\cos \alpha +1=0\Rightarrow \cos \alpha =-\frac{1}{3}$$$$with\cos \alpha =1:y=3+2=5\to max$$$$with\cos \alpha =-1:y=3-2=1$$$$with\cos \alpha =-\frac{1}{3}:y=\frac{3}{9}-\frac{2}{3}=-\frac{1}{3}\to min$$

Commented by hardmath last updated on 07/Jun/24

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professors}$$

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