Question-204205

Question Number 204205 by universe last updated on 08/Feb/24

Answered by ajfour last updated on 08/Feb/24

x=p^2 , y=q^2 p+q=p^2 +q^2 =t t^2 =t+2pq 2p^2 +2pq=2pt 2p^2 +t^2 −t=2pt 4p(dp/dt)+2t−1=2p+2t(dp/dt) (dp/dt)=0 ⇒ t=p+(1/2) 2p^2 +p^2 =p(2p+1)+(1/4) ⇒ p^2 −p−(1/4)=0 p_(max) =(1/2)±(√(1/2)) we take x_(max) =p_m ^2 =p_m +(1/4) x_(max) =(3/4)+(1/( (√2)))

$${x}={p}^{\mathrm{2}} \:,\:{y}={q}^{\mathrm{2}} \\ $$$${p}+{q}={p}^{\mathrm{2}} +{q}^{\mathrm{2}} ={t} \\ $$$${t}^{\mathrm{2}} ={t}+\mathrm{2}{pq} \\ $$$$\mathrm{2}{p}^{\mathrm{2}} +\mathrm{2}{pq}=\mathrm{2}{pt} \\ $$$$\mathrm{2}{p}^{\mathrm{2}} +{t}^{\mathrm{2}} −{t}=\mathrm{2}{pt} \\ $$$$\mathrm{4}{p}\frac{{dp}}{{dt}}+\mathrm{2}{t}−\mathrm{1}=\mathrm{2}{p}+\mathrm{2}{t}\frac{{dp}}{{dt}} \\ $$$$\:\:\frac{{dp}}{{dt}}=\mathrm{0}\:\:\:\Rightarrow\:\:{t}={p}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}{p}^{\mathrm{2}} +{p}^{\mathrm{2}} ={p}\left(\mathrm{2}{p}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\:\:{p}^{\mathrm{2}} −{p}−\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$${p}_{{max}} =\frac{\mathrm{1}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${we}\:{take}\:\:\:{x}_{{max}} ={p}_{{m}} ^{\mathrm{2}} ={p}_{{m}} +\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:{x}_{{max}} =\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$ \\ $$$$ \\ $$

Commented by ajfour last updated on 08/Feb/24

Answered by a.lgnaoui last updated on 08/Feb/24

(x−(1/2))^2 −(1/4)=(√y) −y (x>0 y>0) [ x=(1/2)+(√((1/4)−(y−(√y) ))) ] (dx/dy)=(((1/(2(√y)))−1)/(2(√((1/4)+(√y) −y))))=((1−2(√y))/(4(√((y/4)+y−y^2 )))) =0⇒ 2(√y) =1 y=(1/4) x−(√x) −(1/4) =0 ⇒ (√x) =((1+(√2))/2) soit x_(max) =((3+2(√2))/4)

$$\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}=\sqrt{\mathrm{y}}\:−\mathrm{y}\:\:\:\:\:\:\:\left(\mathrm{x}>\mathrm{0}\:\:\:\:\mathrm{y}>\mathrm{0}\right) \\ $$$$\:\:\:\:\left[\:\:\:\:\boldsymbol{\mathrm{x}}=\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\left(\mathrm{y}−\sqrt{\mathrm{y}}\:\right)}\:\:\:\right] \\ $$$$\:\: \\ $$$$\:\:\:\frac{\boldsymbol{\mathrm{dx}}}{\boldsymbol{\mathrm{dy}}}=\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\boldsymbol{\mathrm{y}}}}−\mathrm{1}}{\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\sqrt{\boldsymbol{\mathrm{y}}}\:−\boldsymbol{\mathrm{y}}}}=\frac{\mathrm{1}−\mathrm{2}\sqrt{\boldsymbol{\mathrm{y}}}}{\mathrm{4}\sqrt{\frac{\boldsymbol{\mathrm{y}}}{\mathrm{4}}+\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{y}}^{\mathrm{2}} }} \\ $$$$ \\ $$$$\:=\mathrm{0}\Rightarrow\:\:\:\mathrm{2}\sqrt{\mathrm{y}}\:=\mathrm{1}\:\:\:\:\:\mathrm{y}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\boldsymbol{\mathrm{x}}−\sqrt{\boldsymbol{\mathrm{x}}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:\:=\mathrm{0}\:\:\: \\ $$$$\Rightarrow\:\:\:\:\:\:\:\sqrt{\mathrm{x}}\:=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\:\:\:\:\:\mathrm{soit}\:\: \\ $$$$\:\:\:\:\mathrm{x}_{\mathrm{max}} =\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\:\: \\ $$$$\: \\ $$

Answered by mr W last updated on 08/Feb/24

((√y))^2 −(√y)+x−(√x)=0 such that (√y) ∈R exists, Δ=1^2 −4(x−(√x))≥0 ((√x))^2 −(√x)−(1/4)≤0 ((1−(√(1^2 +4×(1/4))))/2)≤(√x)≤((1+(√(1^2 +4×(1/4))))/2) 0≤(√x)≤((1+(√2))/2) 0≤x≤(((1+(√2))/2))^2 =((3+2(√2))/4) i.e. x_(max) =((3+2(√2))/4) ✓

$$\left(\sqrt{{y}}\right)^{\mathrm{2}} −\sqrt{{y}}+{x}−\sqrt{{x}}=\mathrm{0} \\ $$$${such}\:{that}\:\sqrt{{y}}\:\in{R}\:{exists}, \\ $$$$\Delta=\mathrm{1}^{\mathrm{2}} −\mathrm{4}\left({x}−\sqrt{{x}}\right)\geqslant\mathrm{0} \\ $$$$\left(\sqrt{{x}}\right)^{\mathrm{2}} −\sqrt{{x}}−\frac{\mathrm{1}}{\mathrm{4}}\leqslant\mathrm{0} \\ $$$$\frac{\mathrm{1}−\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{4}×\frac{\mathrm{1}}{\mathrm{4}}}}{\mathrm{2}}\leqslant\sqrt{{x}}\leqslant\frac{\mathrm{1}+\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{4}×\frac{\mathrm{1}}{\mathrm{4}}}}{\mathrm{2}} \\ $$$$\mathrm{0}\leqslant\sqrt{{x}}\leqslant\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{0}\leqslant{x}\leqslant\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${i}.{e}.\:{x}_{{max}} =\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{4}}\:\checkmark \\ $$

Answered by MM42 last updated on 08/Feb/24

((√x)−(1/2))^2 =(1/2)−((√y)−(1/2))^2 ⇒((√x)−(1/2))^2 ≤(1/2)⇒(√x)−(1/2)≤((√2)/2) ⇒(√x)≤((1+(√2))/2)⇒max_x =((3+2(√2))/4) ✓

$$\left(\sqrt{{x}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}−\left(\sqrt{{y}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\sqrt{{x}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\sqrt{{x}}−\frac{\mathrm{1}}{\mathrm{2}}\leqslant\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{{x}}\leqslant\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}\Rightarrow{max}_{{x}} =\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{4}}\:\:\checkmark \\ $$$$ \\ $$

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