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Question Number 132853 by EDWIN88 last updated on 17/Feb/21

$$\mathrm{Find}\mathrm{the}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{paraboloid}$$$$\mathrm{z}={\mathrm{x}}^2+{\mathrm{y}}^2\mathrm{which}\mathrm{is}\mathrm{closest}\mathrm{to}\mathrm{the}\mathrm{point}$$$$(3,-6,4)$$

Answered by MJS_new last updated on 17/Feb/21

$$y=-2x\left[\mathrm{horizontal}\mathrm{sections}\mathrm{are}\mathrm{circles}\right]$$$$\left(\begin{array}{c}x\\ y\\ x^2+y^2\end{array}\right)=\left(\begin{array}{c}x\\ -2x\\ 5x^2\end{array}\right)$$$$\mid \left(\begin{array}{c}x\\ -2x\\ 5x^2\end{array}\right)-\left(\begin{array}{c}3\\ -6\\ 4\end{array}\right)\mid =\sqrt{25x^4-35x^2-30x+62}$$$$\frac{d}{dx}\left[\sqrt{25x^4-35x^2-30x+62}\right]=0$$$$\frac{5(10x^3-7x-3)}{\sqrt{25x^4-35x^2-30x+62}}=0$$$$\frac{5(x-1)(10x^2+10x+3)}{\sqrt{25x^4-35x^2-30x+62}}=0$$$$x=1$$$$\Rightarrow y=-2\wedge z=5$$$$\mathrm{minimum}\mathrm{distance}\mathrm{is}\sqrt{21}$$

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