# Lagrange Multiplier problem

• Oct 29th 2006, 02:09 PM
splash
Lagrange Multiplier problem
Take home exam problem. Someone want to tell me if I'm missing something with this Lagrange multiplier problem?

Problem: Find the maximum and minimum values of f(x,y)=xy on the curve (x+2)^2 + y^2 =1

This is what I've done-

So we have the following equations.
y=lambda*(2x+4)
x=lambda*2y
(x+2)^2 + y^2 =1

That's 3 equations and 3 unknowns but I can't seem to find a solution. I feel like I'm missing something really stupid. Or my brain is hitting a wall to solve those equations. Can someone help?
• Oct 29th 2006, 02:34 PM
ThePerfectHacker
Quote:

Originally Posted by splash
Take home exam problem. Someone want to tell me if I'm missing something with this Lagrange multiplier problem?

Problem: Find the maximum and minimum values of f(x,y)=xy on the curve (x+2)^2 + y^2 =1

This is what I've done-

So we have the following equations.
y=lambda*(2x+4)
x=lambda*2y
(x+2)^2 + y^2 =1

That's 3 equations and 3 unknowns but I can't seem to find a solution. I feel like I'm missing something really stupid. Or my brain is hitting a wall to solve those equations. Can someone help?

Okay, we have...
$\displaystyle f(x,y)=xy$
Thus,
$\displaystyle \nabla f(x,y)=<y,x>$
And,
$\displaystyle c(x,y)=(x+2)^2+y^2$
Thus,
$\displaystyle \nabla c(x,y)=<2x+4,2y>$
This agrees with what you have.

By Lagrange multipliers we have,
$\displaystyle <y,x>=k<2x+4,2y>$
Thus,
$\displaystyle y=(2x+4)k$
$\displaystyle x=2ky$
$\displaystyle (x+2)^2+y^2=1$
From the second equation we have,
$\displaystyle k=\frac{x}{2y}$
Substitute that into the first equation,
$\displaystyle y=\frac{x(2x+4)}{2y}$
Thus,
$\displaystyle y^2=x(x+2)$
Substitute that into third equation,
$\displaystyle (x+2)^2+x(x+2)=1$
Thus,
$\displaystyle x^2+4x+4+x^2+2x=1$
Thus,
$\displaystyle 2x^2+6x+3=0$
Now solve.