Lagrange Multipliers (Question)

**VkL** · March 9th 2010, 07:36 PM

I have to find the points on the sphere $x^2+y^2+z^2=36$ that are closest to and furthest from the point $(1,2,2)$

This is where I am up to. And I will show where I am stuck.
$\nabla f(x,y,z) = 2x+2y+2z$

What do I do now? Do I use the distance formula somehow?

Any help is appreciated!

thanks in advance!

**nehme007** · March 9th 2010, 08:03 PM

Whichever point maximizes (minimizes) the square of the distance to the point (1,2,2) also maximizes (minimizes) the distance to the point. So the function you want to optimize is $g(x,y,z) = (x-1)^2 + (y-2)^2 + (z-2)^2$ . You need to find points on the sphere where the gradient of f is a scalar multiple of the gradient of g. What you have written down is not the gradient of f, by the way. The gradient of f is the vector <2x, 2y, 2z>. The gradient of g is <2(x-1), 2(y-2), 2(z-2)>. Let one gradient be a scalar multiple of the other. This gives you 3 equations (one for each component of the vector). The fourth equation is the equation of the sphere. You have four variables (including the scalar multiple, which is usually denoted by $\lambda$ ). I'll let you work it from here.

**VkL** · March 9th 2010, 09:02 PM

$\nabla f(x,y,z)=\lambda( \nabla g(x,y,z))$
where $f(x,y,z) = D^2 = (x-1)^2+(y-2)^2+(z-2)^2$
and where $g(x,y,z)$ = Constraint condition = $x^2+y^2+z^2=36$

$<2(x-1)+2(y-2)+2(z-2)> = \lambda <2x+2y+2z>$

solving for $\lambda$ I got that $\lambda =3/2$
and putting $\lambda$ back into the equations I get

$(-2, -4, -4)$

Is this correct? Is this the answer? If anyone can check that would be amazing! I have a final tomorrow =/

**nehme007** · March 9th 2010, 09:20 PM

You can check it by inputting "maximize (x-1)^2 + (y-2)^2 + (z-2)^2 subject to x^2 + y^2 + z^2 = 36" at wolframalpha.com. Replace maximize with minimize to get the other answer.

**VkL** · March 9th 2010, 10:29 PM

^WOW Thank you so much!

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