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Question Number 156137 by cortano last updated on 08/Oct/21

Answered by mr W last updated on 09/Oct/21

Commented by mr W last updated on 09/Oct/21

$$AB=\sqrt{a^2+c^2}$$$$BC=\sqrt{b^2+d^2}$$$$AC=\sqrt{{(a+b)}^2+{(c+d)}^2}$$$$sinceAB+BC⩽AC$$$$\Rightarrow \sqrt{a^2+c^2}+\sqrt{b^2+d^2}⩽\sqrt{{(a+b)}^2+{(c+d)}^2}$$$$‘‘{equal}^{″}holdswhen\frac{a}{c}=\frac{b}{d}$$

Commented by mr W last updated on 09/Oct/21

$$applyingabove...$$$$\sqrt{x^2+4x+13}+\sqrt{x^2-8x+41}$$$$=\sqrt{{(x+2)}^2+3^2}+\sqrt{{(4-x)}^2+5^2}$$$$⩾\sqrt{{(x+2+4-x)}^2+{(3+5)}^2}=\sqrt{6^2+8^2}=10$$$$\Rightarrow min=10$$$$minimumoccurswhen\frac{x+2}{3}=\frac{4-x}{5},$$$$i.e.x=\frac{1}{4}$$

Commented by cortano last updated on 10/Oct/21

$$\mathrm{the}\mathrm{points}\mathrm{A},\mathrm{B}\mathrm{and}\mathrm{C}\mathrm{are}\mathrm{colinier}$$

Commented by cortano last updated on 10/Oct/21

$$\mathrm{if}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{{\mathrm{x}}^2+4\mathrm{x}+8}+\sqrt{{\mathrm{x}}^2-6\mathrm{x}+34}$$$$\min \mathrm{f}\left(\mathrm{x}\right)=?$$$$\mathrm{solution}:$$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{{(\mathrm{x}+2)}^2+4}+\sqrt{{(3-\mathrm{x})}^2+25}$$$$⩾\sqrt{{(\mathrm{x}+2+3-\mathrm{x})}^2+{(2+5)}^2}$$$$=\sqrt{25+49}=\sqrt{74}$$$$\mathrm{occurs}\mathrm{when}\frac{\mathrm{x}+2}{2}=\frac{3-\mathrm{x}}{5}$$$$\mathrm{i}.\mathrm{e}5\mathrm{x}+10=6-2\mathrm{x},\mathrm{x}=\frac{4}{7}$$

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