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Question Number 211529 by MrGaster last updated on 12/Sep/24
known:x+y=3,ask:  min((√(x^2 +1))+(√(y2−4)))=?
$$\mathrm{known}:\boldsymbol{{x}}+\boldsymbol{\mathrm{y}}=\mathrm{3},\mathrm{ask}: \\ $$$$\boldsymbol{\mathrm{min}}\left(\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}+\sqrt{\boldsymbol{\mathrm{y}}\mathrm{2}−\mathrm{4}}\right)=? \\ $$
Answered by MrGaster last updated on 12/Sep/24
1.f(x)=(√(x^2 +1))+(√((3+x)^2 −4))  f^� (x)=(x/( (√(x^2 +1))))−((3−x)/( (√((3−x)^2 −4))))  f^� (x)=0  Check endpoint:  x=0,y=3  f(0)=1+(√5)  x=1,y=2  f(1)=(√2)  minf(x)=(√2)
$$\mathrm{1}.\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}+\sqrt{\left(\mathrm{3}+\boldsymbol{{x}}\right)^{\mathrm{2}} −\mathrm{4}} \\ $$$$\boldsymbol{{f}}^{} \left(\boldsymbol{{x}}\right)=\frac{\boldsymbol{{x}}}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}}−\frac{\mathrm{3}−\boldsymbol{{x}}}{\:\sqrt{\left(\mathrm{3}−\boldsymbol{{x}}\right)^{\mathrm{2}} −\mathrm{4}}} \\ $$$$\boldsymbol{{f}}^{} \left(\boldsymbol{{x}}\right)=\mathrm{0} \\ $$$$\mathrm{Check}\:\mathrm{endpoint}: \\ $$$$\boldsymbol{{x}}=\mathrm{0},\boldsymbol{\mathrm{y}}=\mathrm{3} \\ $$$$\boldsymbol{{f}}\left(\mathrm{0}\right)=\mathrm{1}+\sqrt{\mathrm{5}} \\ $$$$\boldsymbol{{x}}=\mathrm{1},\boldsymbol{\mathrm{y}}=\mathrm{2} \\ $$$$\boldsymbol{{f}}\left(\mathrm{1}\right)=\sqrt{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{min}{f}}\left(\boldsymbol{{x}}\right)=\sqrt{\mathrm{2}} \\ $$

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