# LaGrange Multipliers

• Nov 1st 2008, 04:26 PM
JonathanEyoon
LaGrange Multipliers
Grrr... This problem has been stressing me out for the past half hour

Find the extreme values of f(x,y) = xy on the ellipse ((x^2)/8) + ((y^2)/2) = 1

i've gone ahead and figured out the gradient f and g and ended up with (we'll call L lambda since I don't know how to insert that

y = (Lx)/4

x = Ly

((x^2)/8) + ((y^2)/2) = 1

I've gotten it down to this system of equations but i'm having trouble solving for this. Any help is appreciate.
• Nov 1st 2008, 05:00 PM
Soroban
Hello, Jonathan!

Sorry, your set-up is all wrong . . .

Quote:

Find the extreme values of $\displaystyle f(x,y) = xy$ on the ellipse $\displaystyle \frac{x^2}{8} + \frac{y^2}{2} \: =\:1$
We want the extreme values of $\displaystyle f(x,y)$ with the constraint: .$\displaystyle x^2+4y^2-8\:=\:0$

Our function is: .$\displaystyle F(x,y) \;=\;xy + \lambda(x^2+4y^2-8)$

Then: .$\displaystyle \begin{array}{ccccc}\frac{\partial F}{\partial x} &=& y + 2x\lambda &=& 0 \\ \\[-3mm] \frac{\partial F}{\partial y} &=& x + 8y\lambda &=&0 \\ \\[-3mm] \frac{\partial F}{\partial\lambda} &=& x^2+4y^2-8 &=& 0 \end{array}$

Now solve that system . . .

• Nov 1st 2008, 05:11 PM
JonathanEyoon
Wow... just wow... you made it so much easier.

I just figured it out using my same setup but this would of been so much easier. This is what I did

Eq. 1 = y = (Lx)/4

Eq. 2 = x = Ly

Eq. 3 = ((x^2)/8) + ((y^2)/2) = 1

I substituted Eq.2 into Eq.1 for the x value and ended up with L = +2 and -2

Taking the value of lambda of +2 I put it in Eq.2 and got x = 2y

I then subbed 2y for the x value in Eq.3 and ended up with a y value of +1 and -1. Finding x is simple and got +2 and -2

I ended with extreme values of 2 and -2 for my final answer. All in all though, your way seems WAYYY easier and i'm gonna try it right now. Thanks!