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Question Number 213688 by Abdullahrussell last updated on 13/Nov/24

Answered by mr W last updated on 13/Nov/24

$$at{x}_{min}=-2:$$$$-2+y=4\sqrt{y+3}$$$$y+3-4\sqrt{y+3}-5=0$$$$(\sqrt{y+3}+1)(\sqrt{y+3}-5)=0$$$$\Rightarrow \sqrt{y+3}=5\Rightarrow y=22\Rightarrow x+y=20$$$$at{y}_{min}=-3:$$$$x-3=4\sqrt{x+2}$$$$x+2-4\sqrt{x+2}-5=0$$$$(\sqrt{x+2}+1)(\sqrt{x+2}-5)=0$$$$\Rightarrow \sqrt{x+2}=5\Rightarrow x=23\Rightarrow x+y=20$$$$\Rightarrow {(x+y)}_{min}=20✓$$

Answered by mr W last updated on 13/Nov/24

$$letp=x+2,q=y+3$$$$p+q-5=4(\sqrt{p}+\sqrt{q})$$$${(\sqrt{p}+\sqrt{q})}^2-2\sqrt{pq}-5=4(\sqrt{p}+\sqrt{q})$$$${(\sqrt{p}+\sqrt{q})}^2-4(\sqrt{p}+\sqrt{q})-5=2\sqrt{pq}⩽p+q=4(\sqrt{p}+\sqrt{q})+5$$$${(\sqrt{p}+\sqrt{q})}^2-8(\sqrt{p}+\sqrt{q})-10⩽0$$$$\Rightarrow \sqrt{p}+\sqrt{q}⩽4+\sqrt{26}$$$$\Rightarrow x+y=4(\sqrt{p}+\sqrt{q})⩽4(4+\sqrt{26})$$$$\Rightarrow {(x+y)}_{max}=4(4+\sqrt{26})✓$$

Commented by mr W last updated on 14/Nov/24

$$alternatively$$$$letu=\sqrt{x+2},v=\sqrt{y+3}$$$$x+y=u^2+v^2-5=4(u+v)$$$$letu+v=k$$$$u^2+{(k-u)}^2-5=4k$$$$2u^2-2ku+k^2-4k-5=0$$$$\mathrm{\Delta }=k^2-2(k^2-4k-5)⩾0$$$$k^2-8k-10⩽0$$$$\Rightarrow k⩽4+\sqrt{26}$$$$x+y=4k⩽4(4+\sqrt{26})$$$$\Rightarrow {(x+y)}_{max}=4(4+\sqrt{26})✓$$

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