# Lagrange Multipliers

• March 25th 2010, 11:33 AM
davesface
Lagrange Multipliers
The plane $x+y+z=4$ intersects the paraboloid $z=x^2+y^2$ in an ellipse. Find the points on the ellipse nearest to and farthest from the origin.

So is the goal here to just minimize and then maximize the distance formula $f(x,y,z)=\sqrt{x^2+y^2+z^2}$ with the constraints of $g_1:x+y+z=4$ and $g_2:z=x^2+y^2$?
• March 25th 2010, 12:11 PM
Soroban
Hello, davesface!

Quote:

The plane $x+y+z\:=\:4$ intersects the paraboloid $z\:=\:x^2+y^2$ in an ellipse.
Find the points on the ellipse nearest to and farthest from the origin.

So is the goal to minimize and maximize the distance formula $f(x,y,z)\:=\:\sqrt{x^2+y^2+z^2}$
with the constraints of $g_1:x+y+z\:=\:4$ and $g_2:z\:=\:x^2+y^2$ ? . . . . Yes!

We have: . $F(x,y,z,\lambda,\mu) \;=\;\left(x^2+y^2+z^2\right)^{\frac{1}{2}} + \lambda(x+y+z-4) + \mu\left(z-x^2-y^2\right)$

Find the five partial derivatives, equate to zero, and solve the system.

. . $\begin{array}{ccccc}\dfrac{\partial F}{\partial x} &=& \dfrac{x}{\sqrt{x^2+y^2+z^2}} + \lambda - 2\mu x &=& 0 \\ \\

\dfrac{\partial F}{\partial y} &=& \dfrac{y}{\sqrt{x^2+y^2+z^2}} + \lambda - 2\mu y &=&0 \\ \\

\dfrac{\partial F}{\partial z} &=& \dfrac{z}{\sqrt{x^21+y^2+z^2}} + \lambda + \mu &=&0 \end{array}$

. . . . $\begin{array}{ccccc}\dfrac{\partial F}{\partial \lambda} &=& x + y + z - 4 &=& 0 \\ \\

\dfrac{\partial F}{\partial \mu} &=& z - x^2 - y^2 &=& 0 \end{array}$

I'll wait in the car . . .
.
• March 25th 2010, 01:16 PM
davesface
I haven't ever seen someone take $\frac{\delta F}{d \lambda}$ or $\frac{\delta F}{d \mu}$ (and since those are constants, do those derivatives make sense?). According to my notes, it's set up as $\nabla f= \lambda \nabla g_1+ \mu \nabla g_2$, which I think for this problem would be:

$\left [ \begin{array}{cc} \frac{x}{\sqrt{x^2+y^2+z^2}} \\ \frac{y}{\sqrt{x^2+y^2+z^2}}\\ \frac{z}{\sqrt{x^2+y^2+z^2}} \end{array} \right ]= \lambda \left [ \begin{array}{cc} 1 \\ 1\\1 \end{array} \right ] + \mu \left [ \begin{array}{cc} 2x \\ 2y \\-1 \end{array} \right ]$