1. ## 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?

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

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