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Question Number 169826 by cortano1 last updated on 10/May/22
Given f(x)=(√(sin x)) +(√(3 cos x)) x∈ (0, (π/2)) Find max f(x).
$$\:\:{Given}\:{f}\left({x}\right)=\sqrt{\mathrm{sin}\:{x}}\:+\sqrt{\mathrm{3}\:\mathrm{cos}\:{x}}\: \\ $$$$\:\:{x}\in\:\left(\mathrm{0},\:\frac{\pi}{\mathrm{2}}\right) \\ $$$$\:{Find}\:{max}\:{f}\left({x}\right).\: \\ $$
Answered by mr W last updated on 10/May/22
f′(x)=((cos x)/(2(√(sin x))))−((3sin x)/(2(√(3 cos x))))=0 (tan x)^(3/2) =(1/( (√3))) ⇒tan x=(1/( (3)^(1/3) )) sin x=(1/( (√(1+(9)^(1/3) )))), cos x=((3)^(1/3) /( (√(1+(9)^(1/3) )))) f(x)_(max) =((1+(9)^(1/3) )/( ((1+(9)^(1/3) ))^(1/4) ))=(1+(9)^(1/3) )^(3/4) ✓
$${f}'\left({x}\right)=\frac{\mathrm{cos}\:{x}}{\mathrm{2}\sqrt{\mathrm{sin}\:{x}}}−\frac{\mathrm{3sin}\:{x}}{\mathrm{2}\sqrt{\mathrm{3}\:\mathrm{cos}\:{x}}}=\mathrm{0} \\ $$$$\left(\mathrm{tan}\:{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}}} \\ $$$$\mathrm{sin}\:{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{9}}}},\:\mathrm{cos}\:{x}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}}}{\:\sqrt{\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{9}}}} \\ $$$${f}\left({x}\right)_{{max}} =\frac{\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{9}}}{\:\sqrt[{\mathrm{4}}]{\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{9}}}}=\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{9}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \:\checkmark \\ $$

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