# Math Help - LaGrange Multipliers

1. ## 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!

2. 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 $x+y=1$.

3. 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.