# LaGrange Multipliers

• Nov 7th 2008, 07:48 AM
JonathanEyoon
LaGrange Multipliers
This was a question on my test yesterday and I was confused on how to do this. Could someone solve this one out for me?

Find a point on the sphere x^2 + y^2 + z^2 = 4 that is the farthest away from point (1,-1,1)

• Nov 7th 2008, 08:00 AM
Moo
Hello,
Quote:

Originally Posted by JonathanEyoon
This was a question on my test yesterday and I was confused on how to do this. Could someone solve this one out for me?

Find a point on the sphere x^2 + y^2 + z^2 = 4 that is the farthest away from point (1,-1,1)

Find an equation translating the "farthest away from point (1,-1,1)"

That is you want to maximize the distance from a point on the sphere to this point.
Let $\displaystyle (x,y,z) \in S$

Let $\displaystyle f ~:~ \mathbb{R}^3 \to \mathbb{R}$
$\displaystyle f(x,y,z)=x^2+y^2+z^2-4$

Define $\displaystyle \varphi ~:~ \mathbb{R}^3 \to \mathbb{R}$
$\displaystyle \varphi(x,y,z)=\sqrt{(x-1)^2+(y+1)^2+(z-1)^2}$

f represents the constraint. That is if $\displaystyle (x,y,z) \in S$, $\displaystyle f(x,y,z)=0$

Let $\displaystyle \lambda$ be a Lagrange multiplier for this problem.
Define $\displaystyle h(x,y,z,\lambda)=f(x,y,z)+\lambda \varphi(x,y,z)$

Solve for $\displaystyle \lambda$ in $\displaystyle \nabla_{x,y,z,\lambda} h(x,y,z,\lambda)=0$, that is to say :

$\displaystyle \left\{\begin{array}{lllllll} \frac{\partial h}{\partial x}=0 \\ \\ \frac{\partial h}{\partial y}=0 \\ \\ \frac{\partial h}{\partial z}=0 \\ \\ \frac{\partial h}{\partial \lambda}=0 \end{array} \right.$
• Nov 7th 2008, 08:24 AM
JonathanEyoon
Uhhhmmm.. you lost me after the first line (Crying)
• Nov 7th 2008, 10:18 AM
HallsofIvy
Quote:

Originally Posted by JonathanEyoon
Uhhhmmm.. you lost me after the first line (Crying)

If you don't even know the formula for distance (the first line), why are you even attempting a problem like this?