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Question Number 211529 by MrGaster last updated on 12/Sep/24

$$\mathrm{known}:\mathit{x}+\mathbf{y}=3,\mathrm{ask}:$$$$\mathbf{min}(\sqrt{{\mathit{x}}^2+1}+\sqrt{\mathbf{y}2-4})=?$$

Answered by MrGaster last updated on 12/Sep/24

$$1.\mathit{f}\left(\mathit{x}\right)=\sqrt{{\mathit{x}}^2+1}+\sqrt{{(3+\mathit{x})}^2-4}$$$${\mathit{f}}^{}\left(\mathit{x}\right)=\frac{\mathit{x}}{\sqrt{{\mathit{x}}^2+1}}-\frac{3-\mathit{x}}{\sqrt{{(3-\mathit{x})}^2-4}}$$$${\mathit{f}}^{}\left(\mathit{x}\right)=0$$$$\mathrm{Check}\mathrm{endpoint}:$$$$\mathit{x}=0,\mathbf{y}=3$$$$\mathit{f}\left(0\right)=1+\sqrt{5}$$$$\mathit{x}=1,\mathbf{y}=2$$$$\mathit{f}\left(1\right)=\sqrt{2}$$$$\mathbf{min}\mathit{f}\left(\mathit{x}\right)=\sqrt{2}$$

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