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Question Number 205338 by cortano12 last updated on 17/Mar/24

Answered by mr W last updated on 17/Mar/24

$$x⩾-2,y⩾-3$$$$x+2-4\sqrt{x+2}+4+y+3-4\sqrt{y+3}+4=13$$$${(\sqrt{x+2}-2)}^2+{(\sqrt{y+3}-2)}^2={\left(\sqrt{13}\right)}^2$$$$\sqrt{x+2}-2=\sqrt{13}\cos \theta$$$$\sqrt{y+3}-2=\sqrt{13}\sin \theta$$$$\Rightarrow x={(2+\sqrt{13}\cos \theta )}^2-2=2+4\sqrt{13}\cos \theta +13{\cos }^2\theta$$$$\Rightarrow y={(2+\sqrt{13}\sin \theta )}^2-3=1+4\sqrt{13}\sin \theta +13{\sin }^2\theta$$$$x+y=16+4\sqrt{13}(\cos \theta +\sin \theta )$$$$x+y=16+4\sqrt{26}\sin (\theta +\frac{\pi }{4})$$$$\Rightarrow {(x+y)}_{max}=16+4\sqrt{26}✓$$$$atx=-2:$$$${(\sqrt{y+3}-2)}^2=13-{(-2)}^2=9$$$$\sqrt{y+3}-2=3$$$$y=22$$$$\Rightarrow x+y=-2+22=20$$$$aty=-3:$$$${(\sqrt{x+2}-2)}^2=13-{(-2)}^2=9$$$$\sqrt{x+2}-2=3$$$$x=23$$$$\Rightarrow x+y=23-3=20$$$$\Rightarrow {(x+y)}_{min}=20✓$$

Answered by A5T last updated on 17/Mar/24

$$Anothermethodtoget{(x+y)}_{max}:$$$$\frac{a+b}{2}⩾{\left(\frac{\sqrt{a}+\sqrt{b}}{2}\right)}^2\Rightarrow \frac{x+2+y+3}{2}⩾{\left(\frac{\sqrt{x+2}+\sqrt{y+3}}{2}\right)}^2$$$$\Rightarrow {\left(\frac{x+y}{8}\right)}^2⩽\frac{x+y+5}{2};x+y=p\Rightarrow \frac{p^2}{32}⩽p+5$$$$\Rightarrow p^2-32p-160⩽0\Rightarrow 16-4\sqrt{26}⩽x+y⩽4\sqrt{26}+16$$$$Equalitywhenx+2=y+3\Rightarrow x=1+y$$$$\Rightarrow 1+2y=+8\sqrt{y+3}\Rightarrow y=2\sqrt{26}+\frac{15}{2}\Rightarrow x=2\sqrt{26}+\frac{17}{2}$$$$\Rightarrow (x,y)=(2\sqrt{26}+\frac{17}{2},2\sqrt{26}+\frac{15}{2})atx+y=4\sqrt{26}+16$$

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