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Question Number 171727 by cortano1 last updated on 20/Jun/22

Answered by mahdipoor last updated on 20/Jun/22

$$y⩾\frac{x^2+y^2}{2}\Rightarrow 1⩾x^2+{(y-1)}^2\Rightarrow -1⩽x⩽1$$$$forx=x_0,max(x_0+y)=x_0+max\left(y\right)=$$$$x_0+1+\sqrt{1-x_0^2}=f\left(x_0\right)$$$$\frac{df}{dx}=1-\frac{x_0}{\sqrt{1-x_0^2}}=0or\nexists \Rightarrow x_0=\pm 1,\sqrt{\frac{1}{2}}$$$$\Rightarrow max(x+y)=f\left(\sqrt{\frac{1}{2}}\right)=2\sqrt{\frac{1}{2}}+1$$

Answered by mr W last updated on 20/Jun/22

$$y⩾\frac{x^2+y^2}{2}=\frac{{(x+y)}^2}{2}-xy$$$$(x+y+1-y)y⩾\frac{{(x+y)}^2}{2}$$$$letk=x+y$$$$(k+1)y-y^2⩾\frac{k^2}{2}$$$$y^2-(k+1)y-\frac{k^2}{2}⩽0$$$$\mathrm{\Delta }={(k+1)}^2-4\times \frac{k^2}{2}⩾0$$$$k^2-2k-1⩽0$$$${(k-1)}^2⩽2$$$$-\sqrt{2}+1⩽k⩽\sqrt{2}+1$$$${(x+y)}_{min}=k_{min}=-\sqrt{2}+1$$$${(x+y)}_{max}=k_{max}=\sqrt{2}+1$$

Answered by mr W last updated on 20/Jun/22

$$y⩾\frac{x^2+y^2}{2}$$$$x^2+y^2-2y⩽0$$$$x^2+{(y-1)}^2⩽1^2\to circle$$$$\Rightarrow x=\cos \theta ,y=1+\sin \theta$$$$x+y=1+\sin \theta +\cos \theta =1+\sqrt{2}\sin (\theta +\frac{\pi }{4})$$$${(x+y)}_{min}=1-\sqrt{2}$$$${(x+y)}_{max}=1+\sqrt{2}$$

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