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Question Number 153513 by yeti123 last updated on 08/Sep/21

((x/5) + (y/3))((5/x) + (3/y)) = 139, ∀x,y ∈ R_(>0)   find maximum and minimum of  ((x + y)/( (√(xy))))

$$\left(\frac{{x}}{\mathrm{5}}\:+\:\frac{{y}}{\mathrm{3}}\right)\left(\frac{\mathrm{5}}{{x}}\:+\:\frac{\mathrm{3}}{{y}}\right)\:=\:\mathrm{139},\:\forall{x},{y}\:\in\:\mathbb{R}_{>\mathrm{0}} \\ $$ $$\mathrm{find}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum}\:\mathrm{of}\:\:\frac{{x}\:+\:{y}}{\:\sqrt{{xy}}} \\ $$

Answered by liberty last updated on 08/Sep/21

⇒((x+y)/( (√(xy)))) = (√(x/y))+(√(y/x))=(√((5λ)/3))+(√(3/(5λ)))  from condition  2+((3x)/(5y))+((5y)/(3x)) =139  ((3x)/(5y))+((5y)/(3x))=137 set ((3x)/(5y))=λ⇒(x/y)=((5λ)/3)  λ+(1/λ)=137⇒λ^2 −137λ+1=0  λ=((137 ±(√((137+2)(137−2))))/2)  λ=((137±136.98)/2) → { ((λ_1 =136.99)),((λ_2 =0.007)) :}  f(λ_1 )=(√((5×136.99)/3))+(√(3/(5×136.99)))             ≈15.176 (max)  f(λ_2 )=(√((5×0.007)/3))+(√(3/(5×0.07)))            ≈ 9.366 (min)

$$\Rightarrow\frac{{x}+{y}}{\:\sqrt{{xy}}}\:=\:\sqrt{\frac{{x}}{{y}}}+\sqrt{\frac{{y}}{{x}}}=\sqrt{\frac{\mathrm{5}\lambda}{\mathrm{3}}}+\sqrt{\frac{\mathrm{3}}{\mathrm{5}\lambda}} \\ $$ $${from}\:{condition} \\ $$ $$\mathrm{2}+\frac{\mathrm{3}{x}}{\mathrm{5}{y}}+\frac{\mathrm{5}{y}}{\mathrm{3}{x}}\:=\mathrm{139} \\ $$ $$\frac{\mathrm{3}{x}}{\mathrm{5}{y}}+\frac{\mathrm{5}{y}}{\mathrm{3}{x}}=\mathrm{137}\:{set}\:\frac{\mathrm{3}{x}}{\mathrm{5}{y}}=\lambda\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{5}\lambda}{\mathrm{3}} \\ $$ $$\lambda+\frac{\mathrm{1}}{\lambda}=\mathrm{137}\Rightarrow\lambda^{\mathrm{2}} −\mathrm{137}\lambda+\mathrm{1}=\mathrm{0} \\ $$ $$\lambda=\frac{\mathrm{137}\:\pm\sqrt{\left(\mathrm{137}+\mathrm{2}\right)\left(\mathrm{137}−\mathrm{2}\right)}}{\mathrm{2}} \\ $$ $$\lambda=\frac{\mathrm{137}\pm\mathrm{136}.\mathrm{98}}{\mathrm{2}}\:\rightarrow\begin{cases}{\lambda_{\mathrm{1}} =\mathrm{136}.\mathrm{99}}\\{\lambda_{\mathrm{2}} =\mathrm{0}.\mathrm{007}}\end{cases} \\ $$ $${f}\left(\lambda_{\mathrm{1}} \right)=\sqrt{\frac{\mathrm{5}×\mathrm{136}.\mathrm{99}}{\mathrm{3}}}+\sqrt{\frac{\mathrm{3}}{\mathrm{5}×\mathrm{136}.\mathrm{99}}} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\approx\mathrm{15}.\mathrm{176}\:\left({max}\right) \\ $$ $${f}\left(\lambda_{\mathrm{2}} \right)=\sqrt{\frac{\mathrm{5}×\mathrm{0}.\mathrm{007}}{\mathrm{3}}}+\sqrt{\frac{\mathrm{3}}{\mathrm{5}×\mathrm{0}.\mathrm{07}}} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\approx\:\mathrm{9}.\mathrm{366}\:\left({min}\right) \\ $$

Commented byyeti123 last updated on 08/Sep/21

ok, thanks

$$\mathrm{ok},\:\mathrm{thanks} \\ $$

Commented byMJS_new last updated on 08/Sep/21

you should round significant figures  λ_1 ≈136.99  λ_2 ≈.0072997  ⇒  f(λ_1 )≈15.176  f(λ_2 )≈9.1765  as you can see your path is ok but your 2^(nd)   result is not...

$$\mathrm{you}\:\mathrm{should}\:\mathrm{round}\:\mathrm{significant}\:\mathrm{figures} \\ $$ $$\lambda_{\mathrm{1}} \approx\mathrm{136}.\mathrm{99} \\ $$ $$\lambda_{\mathrm{2}} \approx.\mathrm{0072997} \\ $$ $$\Rightarrow \\ $$ $${f}\left(\lambda_{\mathrm{1}} \right)\approx\mathrm{15}.\mathrm{176} \\ $$ $${f}\left(\lambda_{\mathrm{2}} \right)\approx\mathrm{9}.\mathrm{1765} \\ $$ $$\mathrm{as}\:\mathrm{you}\:\mathrm{can}\:\mathrm{see}\:\mathrm{your}\:\mathrm{path}\:\mathrm{is}\:\mathrm{ok}\:\mathrm{but}\:\mathrm{your}\:\mathrm{2}^{\mathrm{nd}} \\ $$ $$\mathrm{result}\:\mathrm{is}\:\mathrm{not}... \\ $$

Commented byliberty last updated on 08/Sep/21

Commented byliberty last updated on 08/Sep/21

o yes...

$${o}\:{yes}... \\ $$

Answered by MJS_new last updated on 08/Sep/21

strange question...  ((x/5)+(y/3))((5/x)+(3/y))=139 ⇒ y=((411±9(√(2085)))/(10))x  ⇒ ((x+y)/( (√(xy))))=±3+((4(√(2085)))/(15))  ⇒ there are only 2 values for x>0  min=−3+((4(√(2085)))/(15))  max=3+((4(√(2085)))/(15))

$$\mathrm{strange}\:\mathrm{question}... \\ $$ $$\left(\frac{{x}}{\mathrm{5}}+\frac{{y}}{\mathrm{3}}\right)\left(\frac{\mathrm{5}}{{x}}+\frac{\mathrm{3}}{{y}}\right)=\mathrm{139}\:\Rightarrow\:{y}=\frac{\mathrm{411}\pm\mathrm{9}\sqrt{\mathrm{2085}}}{\mathrm{10}}{x} \\ $$ $$\Rightarrow\:\frac{{x}+{y}}{\:\sqrt{{xy}}}=\pm\mathrm{3}+\frac{\mathrm{4}\sqrt{\mathrm{2085}}}{\mathrm{15}} \\ $$ $$\Rightarrow\:\mathrm{there}\:\mathrm{are}\:\mathrm{only}\:\mathrm{2}\:\mathrm{values}\:\mathrm{for}\:{x}>\mathrm{0} \\ $$ $$\mathrm{min}=−\mathrm{3}+\frac{\mathrm{4}\sqrt{\mathrm{2085}}}{\mathrm{15}} \\ $$ $$\mathrm{max}=\mathrm{3}+\frac{\mathrm{4}\sqrt{\mathrm{2085}}}{\mathrm{15}} \\ $$

Commented byliberty last updated on 08/Sep/21

 3+((4(√(2085)))/(15)) = 15.176

$$\:\mathrm{3}+\frac{\mathrm{4}\sqrt{\mathrm{2085}}}{\mathrm{15}}\:=\:\mathrm{15}.\mathrm{176} \\ $$

Commented byMJS_new last updated on 08/Sep/21

no.  ≈15.176

$$\mathrm{no}. \\ $$ $$\approx\mathrm{15}.\mathrm{176} \\ $$

Commented byliberty last updated on 08/Sep/21

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Commented byyeti123 last updated on 08/Sep/21

ok, I get it. thanks

$$\mathrm{ok},\:\mathrm{I}\:\mathrm{get}\:\mathrm{it}.\:\mathrm{thanks} \\ $$

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