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Question Number 111699 by bemath last updated on 04/Sep/20

find minimum value of function f(x) = ((27)/(2x^2 )) + ((96)/(27)) x^2

$${find}\:{minimum}\:{value}\:{of}\:{function} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{27}}{\mathrm{2}{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{96}}{\mathrm{27}}\:{x}^{\mathrm{2}} \\ $$

Answered by john santu last updated on 04/Sep/20

Answered by ajfour last updated on 04/Sep/20

f(x)=u+v uv=c=48 f(x)∣_(min) =mn((√((u−v)^2 +4uv)) ) = 2(√(uv)) =2(√(48)) =8(√3) .

$${f}\left({x}\right)={u}+{v} \\ $$$${uv}={c}=\mathrm{48} \\ $$$${f}\left({x}\right)\mid_{{min}} ={mn}\left(\sqrt{\left({u}−{v}\right)^{\mathrm{2}} +\mathrm{4}{uv}}\:\right)\:=\:\mathrm{2}\sqrt{{uv}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\sqrt{\mathrm{48}}\:=\mathrm{8}\sqrt{\mathrm{3}}\:. \\ $$

Commented by bemath last updated on 04/Sep/20

how you got uv = c = 48

$${how}\:{you}\:{got}\:{uv}\:=\:{c}\:=\:\mathrm{48} \\ $$

Commented by ajfour last updated on 04/Sep/20

f(x)=u+v uv=((27)/(2x^2 ))×((96x^2 )/(27)) = 48

$$\:\:\:{f}\left({x}\right)={u}+{v} \\ $$$${uv}=\frac{\mathrm{27}}{\mathrm{2}{x}^{\mathrm{2}} }×\frac{\mathrm{96}{x}^{\mathrm{2}} }{\mathrm{27}}\:=\:\mathrm{48} \\ $$

Answered by 1549442205PVT last updated on 04/Sep/20

Applying Cauchy′s inequality for positive numbers we have f(x)=((27)/(2x^2 )) + ((96)/(27)) x^2 ≥2(√(((27)/(2x^2 )).((96x^2 )/(27)))) =2(√(48))=8(√3) The equality ocurrs if and only if ((27)/(2x^2 ))=((96x^2 )/(27))⇔x^4 =((27^2 )/(192))⇔x^2 =((27)/(8(√3))) ⇔∣x∣=((3(√3))/( (√(8(√3)))))⇔x=±((3(√3))/( (√(8(√3))))) Thus,f(x) has smallest value equal to 8(√(3 )) when x=±((3(√3))/( (√(8(√3)))))

$$\mathrm{Applying}\:\mathrm{Cauchy}'\mathrm{s}\:\mathrm{inequality}\:\mathrm{for} \\ $$$$\mathrm{positive}\:\mathrm{numbers}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{27}}{\mathrm{2}{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{96}}{\mathrm{27}}\:{x}^{\mathrm{2}} \geqslant\mathrm{2}\sqrt{\frac{\mathrm{27}}{\mathrm{2x}^{\mathrm{2}} }.\frac{\mathrm{96x}^{\mathrm{2}} }{\mathrm{27}}} \\ $$$$=\mathrm{2}\sqrt{\mathrm{48}}=\mathrm{8}\sqrt{\mathrm{3}} \\ $$$$\mathrm{The}\:\mathrm{equality}\:\mathrm{ocurrs}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\: \\ $$$$\frac{\mathrm{27}}{\mathrm{2x}^{\mathrm{2}} }=\frac{\mathrm{96x}^{\mathrm{2}} }{\mathrm{27}}\Leftrightarrow\mathrm{x}^{\mathrm{4}} =\frac{\mathrm{27}^{\mathrm{2}} }{\mathrm{192}}\Leftrightarrow\mathrm{x}^{\mathrm{2}} =\frac{\mathrm{27}}{\mathrm{8}\sqrt{\mathrm{3}}} \\ $$$$\Leftrightarrow\mid\mathrm{x}\mid=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{8}\sqrt{\mathrm{3}}}}\Leftrightarrow\mathrm{x}=\pm\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{8}\sqrt{\mathrm{3}}}} \\ $$$$\mathrm{Thus},\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{has}\:\mathrm{smallest}\:\mathrm{value}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{8}\sqrt{\mathrm{3}\:}\:\mathrm{when}\:\mathrm{x}=\pm\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{8}\sqrt{\mathrm{3}}}} \\ $$

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