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

Answered by ajfour last updated on 08/Feb/24

$$x=p^2,y=q^2$$$$p+q=p^2+q^2=t$$$$t^2=t+2pq$$$$2p^2+2pq=2pt$$$$2p^2+t^2-t=2pt$$$$4p\frac{dp}{dt}+2t-1=2p+2t\frac{dp}{dt}$$$$\frac{dp}{dt}=0\Rightarrow t=p+\frac{1}{2}$$$$2p^2+p^2=p(2p+1)+\frac{1}{4}$$$$\Rightarrow p^2-p-\frac{1}{4}=0$$$$p_{max}=\frac{1}{2}\pm \sqrt{\frac{1}{2}}$$$$wetake{x}_{max}=p_m^2=p_m+\frac{1}{4}$$$$x_{max}=\frac{3}{4}+\frac{1}{\sqrt{2}}$$$$$$$$$$

Commented by ajfour last updated on 08/Feb/24

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

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

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

$${\left(\sqrt{y}\right)}^2-\sqrt{y}+x-\sqrt{x}=0$$$$suchthat\sqrt{y}\in Rexists,$$$$\mathrm{\Delta }=1^2-4(x-\sqrt{x})⩾0$$$${\left(\sqrt{x}\right)}^2-\sqrt{x}-\frac{1}{4}⩽0$$$$\frac{1-\sqrt{1^2+4\times \frac{1}{4}}}{2}⩽\sqrt{x}⩽\frac{1+\sqrt{1^2+4\times \frac{1}{4}}}{2}$$$$0⩽\sqrt{x}⩽\frac{1+\sqrt{2}}{2}$$$$0⩽x⩽{\left(\frac{1+\sqrt{2}}{2}\right)}^2=\frac{3+2\sqrt{2}}{4}$$$$i.e.x_{max}=\frac{3+2\sqrt{2}}{4}✓$$

Answered by MM42 last updated on 08/Feb/24

$${(\sqrt{x}-\frac{1}{2})}^2=\frac{1}{2}-{(\sqrt{y}-\frac{1}{2})}^2$$$$\Rightarrow {(\sqrt{x}-\frac{1}{2})}^2⩽\frac{1}{2}\Rightarrow \sqrt{x}-\frac{1}{2}⩽\frac{\sqrt{2}}{2}$$$$\Rightarrow \sqrt{x}⩽\frac{1+\sqrt{2}}{2}\Rightarrow {max}_x=\frac{3+2\sqrt{2}}{4}✓$$$$$$

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