Find the point(s) on the surface z^2 = xy + 1 which are closest to the point (7, 8, 0).

Question:

Find the point(s) on the surface {eq}\; z^2 = xy + 1 \; {/eq} which are closest to the point {eq}(7, \; 8, \; 0) {/eq}.

Minimizing distance:

The distance between a point {eq}(x, y, z) {/eq} and the point {eq}(a, \; b, \; c) {/eq} is the function: {eq}D(x,y,z)= \sqrt{ (x-a)^2 + (y - b)^2 + (z- c)^2 } {/eq}

To minimize the distance subject to a constraint {eq}g(x,y,z)=0 {/eq} we can consider the minimum of the square of the distance instead

{eq}f(x,y,z)=(x-a)^2 + (y - b)^2 + (z- c)^2 {/eq}.

Sometimes we can use the method of Lagrange multipliers and sometimes we can substitute the constraint to the objective function (the square of the distance) and obtain a function in only two variables and use the second partial derivatives test to obtain a max or min.

Answer and Explanation:

The given surface is: {eq}z^2 = xy + 1 {/eq}

The distance between any point of that surface and the point {eq}(7, \; 8, \; 0) {/eq} is:

{eq}\sqrt{...

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from PSAT Prep: Tutoring Solution

Chapter 10 / Lesson 13
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