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Question Number 80027 by jagoll last updated on 30/Jan/20

$$findminimum$$$$valueof\sqrt{x^2+4}+\sqrt{x^2-24x+153}$$$$forx⩾0in\mathbb{R}$$

Commented by john santu last updated on 30/Jan/20

$$\mathrm{yes}\mathrm{sir}$$

Commented by john santu last updated on 30/Jan/20

$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2+4}}+\frac{\mathrm{x}-12}{\sqrt{{\mathrm{x}}^2-24\mathrm{x}+153}}=0$$$$\frac{\mathrm{x}}{12-\mathrm{x}}=\sqrt{\frac{{\mathrm{x}}^2+4}{{\mathrm{x}}^2-24\mathrm{x}+153}}$$$$\frac{{\mathrm{x}}^2}{144-24\mathrm{x}+{\mathrm{x}}^2}=\frac{{\mathrm{x}}^2+\mathrm{x}}{{\mathrm{x}}^2-24\mathrm{x}+153}$$$${\mathrm{x}}^4-{24\mathrm{x}}^3+{153\mathrm{x}}^2={\mathrm{x}}^4-{24\mathrm{x}}^3+{144\mathrm{x}}^2$$$$+{4\mathrm{x}}^2-96\mathrm{x}+576$$$${5\mathrm{x}}^2+96\mathrm{x}-576=0$$$$\mathrm{x}=\frac{-96+\sqrt{20736}}{10}=\frac{-96+144}{10}=4.8$$$${\mathrm{f}}_{\min }=\sqrt{{\left(\frac{24}{5}\right)}^2+4}+\sqrt{{(\frac{24}{5}-12)}^2+9}$$$$$$

Commented by mr W last updated on 30/Jan/20

$$withx=4.8wegetexactlyf=13.$$$$i.e.f_{min}=13.$$

Commented by jagoll last updated on 30/Jan/20

$$thanksmrwandjohn$$

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