# LaGrange Multipliers

• Mar 21st 2009, 06:09 PM
TheBerkeleyBoss
LaGrange Multipliers
Hey guys! I just took a Calculus III midterm, and I believe I did fine. However, the last problem was the most difficult, and I am unsure if I will get full credit on it.

It involves LaGrangian Multipliers...I'm going to start the problem with two objective equations both equaling "lambda" (after I've taken partial derivatives of both the objective and constraint equations). I cannot recall the original objective equation.

The constraint equation is:

$x^2 + y^2 \leq 9$

Equation of all partials calculated and equal to each other through the value $\lambda$ :

$x(1-x) = \lambda = y(1-y)$

therefore....

$x(1-x) = y(1-y)$

$-x^2 + x = -y^2 + 1$

multiply both sides by -1....

$x^2 - x = y^2 -y$

I believe that I need to complete the squares...however, I'm not sure...I used the quadratic formula a few steps further, but the values I got do not make any sense. Can anyone help? Thanks! (Clapping)
• Mar 22nd 2009, 02:59 AM
Opalg
Quote:

Originally Posted by TheBerkeleyBoss
....

$x(1-x) = y(1-y)$

$-x^2 + x = -y^2 + 1$

multiply both sides by -1....

$x^2 - x = y^2 -y$

I believe that I need to complete the squares...however, I'm not sure...

It's hard to see where this is going without knowing the whole problem. But I would continue the calculation like this:

If $x^2 - x = y^2 -y$ then $x^2-y^2-(x-y) = 0$. Factorise this as $(x-y)(x+y-1) = 0$. Therefore either $x=y$ or $x+y=1$.
• Mar 22nd 2009, 04:32 AM
HallsofIvy
Quote:

Originally Posted by Opalg
It's hard to see where this is going without knowing the whole problem. But I would continue the calculation like this:

If $x^2 - x = y^2 -y$ then $x^2-y^2-(x-y) = 0$. Factorise this as $(x-y)(x+y-1) = 0$. Therefore either $x=y$ or $x+y=1$.

Or simply go from $x^2- y^- (x-y)= 0$ to $x^2- y^2= x- y$ so $(x-y)(x+ y)= (x-y)$.

If $x- y\ne 0$ divide both sides by it to get x+y= 1. Of course, if x- y= 0, then x= y.