Find-the-point-on-the-paraboloid-z-x-2-y-2-which-is-closest-to-the-point-3-6-4-

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

Find the point on the paraboloid

z = x^{2} + y^{2} which is closest to the point

(3, - 6, 4)

Answered by MJS_new last updated on 17/Feb/21

y = - 2 x [horizontal sections are circles]

(\begin{matrix} x \\ y \\ x^{2} + y^{2} \end{matrix}) = (\begin{matrix} x \\ - 2 x \\ 5 x^{2} \end{matrix})

∣ (\begin{matrix} x \\ - 2 x \\ 5 x^{2} \end{matrix}) - (\begin{matrix} 3 \\ - 6 \\ 4 \end{matrix}) ∣= \sqrt{25 x^{4} - 35 x^{2} - 30 x + 62}

\frac{d}{d x} [\sqrt{25 x^{4} - 35 x^{2} - 30 x + 62}] = 0

\frac{5 (10 x^{3} - 7 x - 3)}{\sqrt{25 x^{4} - 35 x^{2} - 30 x + 62}} = 0

\frac{5 (x - 1) (10 x^{2} + 10 x + 3)}{\sqrt{25 x^{4} - 35 x^{2} - 30 x + 62}} = 0

x = 1

\Rightarrow y = - 2 \land z = 5

minimum distance is \sqrt{21}

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