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Question Number 212928 by efronzo1 last updated on 27/Oct/24

Commented by mr W last updated on 27/Oct/24

$$doesthequestionmeanmsximum$$$$orminimum?ifCandDmustlie$$$$betweenBandE,thenthereisonly$$$$min(x+y),butnomax(x+y).ifC$$$$andDmayalsolieatBandE,then$$$$max(x+y)=max(\sqrt{6^2+6^2}+2,6+\sqrt{2^2+6^2})$$$$=2+6\sqrt{2}$$$$ifthequestionmeansmin(x+y),$$$$answerseebelow.$$

Answered by golsendro last updated on 27/Oct/24

$$\mathrm{x}=\sqrt{36+{\mathrm{a}}^2}$$$$\mathrm{y}=\sqrt{4+{(6-\mathrm{a})}^2}$$$$\mathrm{Let}\mathrm{x}+\mathrm{y}=\mathrm{f}\left(\mathrm{a}\right)=\sqrt{36+{\mathrm{a}}^2}+\sqrt{4+{(6-\mathrm{a})}^2}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{a}\right)=\frac{\mathrm{a}}{\sqrt{36+{\mathrm{a}}^2}}-\frac{(6-\mathrm{a})}{\sqrt{40-12\mathrm{a}+{\mathrm{a}}^2}}=0$$$${\mathrm{a}}^2(40-12\mathrm{a}+{\mathrm{a}}^2)=(36+{\mathrm{a}}^2)(36-12\mathrm{a}+{\mathrm{a}}^2)$$$$$$

Answered by mr W last updated on 27/Oct/24

Commented by mr W last updated on 27/Oct/24

$$min(x+y)=A^{\prime }F=\sqrt{6^2+{(6+2)}^2}=10$$

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