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Question Number 111560 by dw last updated on 04/Sep/20

Commented by dw last updated on 04/Sep/20

$$Stepbystepsolution$$

Answered by Her_Majesty last updated on 04/Sep/20

$$\frac{d}{dz}[\sqrt{x^2+1}+\sqrt{{(x-y)}^2+25}+\sqrt{{(z-y)}^2+4}+\sqrt{{(9-z)}^2+16}]=$$$$=\frac{z-9}{\sqrt{{(9-z)}^2+16}}+\frac{z-y}{\sqrt{{(z-y)}^2+4}}=0$$$$\frac{{(z-9)}^2}{{(9-z)}^2+16}=\frac{{(z-y)}^2}{{(z-y)}^2+4}\left({}_{becauseofsquaring!}^{bewareoffalsesolutions}\right)$$$$z^2-(\frac{8y}{3}-6)z+\frac{4y^2}{3}-27=0$$$$z=2y-9(false!)\vee z=\frac{2y}{3}+3$$$$\Rightarrow$$$$\sqrt{x^2+1}+\sqrt{{(x-y)}^2+25}+\sqrt{{(y-9)}^2+36}$$$$\frac{d}{dy}[\sqrt{x^2+1}+\sqrt{{(x-y)}^2+25}+\sqrt{{(y-9)}^2+36}]=$$$$=\frac{y-x}{\sqrt{{(x-y)}^2+25}}+\frac{y-9}{\sqrt{{(y-9)}^2+36}}=0$$$$\frac{{(y-x)}^2}{{(x-y)}^2+25}=\frac{{(y-9)}^2}{{(y-9)}^2+36}\left({}_{becauseofsquaring!}^{bewareoffalsesolutions}\right)$$$$y^2-\frac{72x-450}{11}y+\frac{36x^2-2025}{11}=0$$$$y=6x-45(false!)\vee y=\frac{6x+45}{11}$$$$\Rightarrow$$$$\sqrt{x^2+1}+\sqrt{{(x-9)}^2+121}$$$$\frac{d}{dx}[\sqrt{x^2+1}+\sqrt{{(x-9)}^2+121}]=$$$$=\frac{x}{\sqrt{x^2+1}}+\frac{x-9}{\sqrt{{(x-9)}^2+121}}=0$$$$\frac{x^2}{x^2+1}=\frac{{(x-9)}^2}{{(x-9)}^2+121}\left({}_{becauseofsquaring!}^{bewareoffalsesolutions}\right)$$$$x^2+\frac{3}{30}x-\frac{27}{40}=0$$$$x=-\frac{9}{10}(false!)\vee x=\frac{3}{4}$$$$\Rightarrow y=\frac{9}{2}\wedge z=6$$$$\Rightarrow answeris15$$

Commented by dw last updated on 05/Sep/20

$$ThankyiuSir$$

Answered by 1549442205PVT last updated on 04/Sep/20

$$\mathrm{Applying}\mathrm{the}\mathrm{inequality}:$$$$\sqrt{{\mathrm{a}}_1^2+{\mathrm{b}}_1^2}+\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_2^2}+...+\sqrt{{\mathrm{a}}_{\mathrm{n}}^2+{\mathrm{b}}_{\mathrm{n}}^2}$$$$⩾\sqrt{{({\mathrm{a}}_1+{\mathrm{a}}_2+...+{\mathrm{a}}_{\mathrm{n}})}^2+{({\mathrm{b}}_1+{\mathrm{b}}_2+...+{\mathrm{b}}_{\mathrm{n}})}^2}$$$$\mathrm{we}\mathrm{have}:$$$$\mathrm{P}=\sqrt{{\mathrm{x}}^2+1}+\sqrt{{(\mathrm{x}-\mathrm{y})}^2+25}+\sqrt{{(\mathrm{z}-\mathrm{y})}^2+4}+\sqrt{{(9-\mathrm{z})}^2+16}$$$$=\sqrt{{\mathrm{x}}^2+1}+\sqrt{{(\mathrm{y}-\mathrm{x})}^2+25}+\sqrt{{(\mathrm{z}-\mathrm{y})}^2+4}+\sqrt{{(9-\mathrm{z})}^2+16}$$$$⩾\sqrt{{(\mathrm{x}+\mathrm{y}-\mathrm{x}+\mathrm{z}-\mathrm{y}+9-\mathrm{z})}^2+{(1+5+2+4)}^2}$$$$=\sqrt{9^2+{12}^2}=15$$$$\mathrm{The}\mathrm{equality}\mathrm{ocurrs}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}$$$$\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-\mathrm{x}}{5}=\frac{\mathrm{z}-\mathrm{y}}{2}=\frac{9-\mathrm{z}}{4}=\frac{\mathrm{x}+\mathrm{y}-\mathrm{x}+\mathrm{z}-\mathrm{y}+9-\mathrm{z}}{1+5+2+4}=\frac{9}{12}=\frac{3}{4}$$$$\Rightarrow \mathrm{x}=0.75;\mathrm{y}=4.5;\mathrm{z}=6$$$$\mathrm{Thus},\mathrm{P}\mathrm{has}\mathrm{smallest}\mathrm{value}\mathrm{equal}\mathrm{to}$$$$15\mathbf{when}(\mathbf{x},\mathbf{y},\mathbf{z})=(\frac{3}{4};\frac{9}{2};6)$$

Commented by dw last updated on 05/Sep/20

$$thankyouSir$$

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