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Question Number 208167 by hardmath last updated on 06/Jun/24
y = 3 cos^2 α + 2 cos α find: max(y) = ?
$$\mathrm{y}\:=\:\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\alpha\:+\:\mathrm{2}\:\mathrm{cos}\:\alpha \\ $$$$\mathrm{find}:\:\:\:\mathrm{max}\left(\mathrm{y}\right)\:=\:? \\ $$
Answered by A5T last updated on 07/Jun/24
y≤3×1+2×1=5⇒max(y)=5(Equality at α=0]
$${y}\leqslant\mathrm{3}×\mathrm{1}+\mathrm{2}×\mathrm{1}=\mathrm{5}\Rightarrow{max}\left({y}\right)=\mathrm{5}\left({Equality}\:{at}\:\alpha=\mathrm{0}\right] \\ $$
Commented by A5T last updated on 07/Jun/24
[(√3)cosα+(1/( (√3)))]^2 =3cos^2 α+2cosα+(1/3) ⇒y=3cos^2 α+2cosα=[(√3)cosα+(1/( (√3)))]^2 −(1/3) ⇒max(y)=[(√3)×1+(1/( (√3)))]^2 −(1/3)=((16−1)/( 3))=5 ⇒min(y)=((−1)/3) [Equality at cosα=((−1)/3)]
$$\left[\sqrt{\mathrm{3}}{cos}\alpha+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right]^{\mathrm{2}} =\mathrm{3}{cos}^{\mathrm{2}} \alpha+\mathrm{2}{cos}\alpha+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{y}=\mathrm{3}{cos}^{\mathrm{2}} \alpha+\mathrm{2}{cos}\alpha=\left[\sqrt{\mathrm{3}}{cos}\alpha+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right]^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{max}\left({y}\right)=\left[\sqrt{\mathrm{3}}×\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right]^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{16}−\mathrm{1}}{\:\mathrm{3}}=\mathrm{5} \\ $$$$\Rightarrow{min}\left({y}\right)=\frac{−\mathrm{1}}{\mathrm{3}}\:\left[{Equality}\:{at}\:{cos}\alpha=\frac{−\mathrm{1}}{\mathrm{3}}\right] \\ $$
Commented by hardmath last updated on 07/Jun/24
thamk you dear professor min(y) = 1 ?
$$\mathrm{thamk}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$$$\mathrm{min}\left(\mathrm{y}\right)\:=\:\mathrm{1}\:? \\ $$
Answered by mr W last updated on 07/Jun/24
y=3 cos^2 α+2 cos α y′=−6 cos α sin α−2 sin α =−2 sin α (3 cos α+1)=^! 0 ⇒sin α=0 ⇒cos α=±1 ⇒3 cos α+1=0 ⇒cos α=−(1/3) with cos α=1: y=3+2=5 → max with cos α=−1: y=3−2=1 with cos α=−(1/3): y=(3/9)−(2/3)=−(1/3) →min
$${y}=\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\alpha+\mathrm{2}\:\mathrm{cos}\:\alpha \\ $$$${y}'=−\mathrm{6}\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\alpha−\mathrm{2}\:\mathrm{sin}\:\alpha \\ $$$$\:\:\:=−\mathrm{2}\:\mathrm{sin}\:\alpha\:\left(\mathrm{3}\:\mathrm{cos}\:\alpha+\mathrm{1}\right)\overset{!} {=}\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\mathrm{0}\:\Rightarrow\mathrm{cos}\:\alpha=\pm\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}\:\mathrm{cos}\:\alpha+\mathrm{1}=\mathrm{0}\:\Rightarrow\mathrm{cos}\:\alpha=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${with}\:\mathrm{cos}\:\alpha=\mathrm{1}:\:{y}=\mathrm{3}+\mathrm{2}=\mathrm{5}\:\:\rightarrow\:{max} \\ $$$${with}\:\mathrm{cos}\:\alpha=−\mathrm{1}:\:{y}=\mathrm{3}−\mathrm{2}=\mathrm{1} \\ $$$${with}\:\mathrm{cos}\:\alpha=−\frac{\mathrm{1}}{\mathrm{3}}:\:{y}=\frac{\mathrm{3}}{\mathrm{9}}−\frac{\mathrm{2}}{\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{3}}\:\rightarrow{min} \\ $$
Commented by hardmath last updated on 07/Jun/24
thank you dear professors
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors} \\ $$

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