# Thread: Maxima and Minima using Lagrange Multipliers

1. ## Maxima and Minima using Lagrange Multipliers

Find the local maxima and minima of the following problem by introducing two Lagrange multipliers:

$\displaystyle f(x_1,x_2,x_3) = x_1 + 2x_2 + 2x_3$

subject to

$\displaystyle x_1^2 + x_2^2 + 4x_3^2 = 2$ and $\displaystyle x_1^2 + (x_2 - 1)^2 + 4_3^2 = 3$

2. Alright. There are some parts of this problem that should be pretty straightforward. Have you introduced said multipliers? Have you written the equations. It'd be awesome if you can write down the vector equations. (you can use \nabla for $\displaystyle \nabla$, unles you know it of course)

3. Okay, I am stuck. Here is what I have now though....

Define Constraints:
$\displaystyle g_1(x_1, x_2, x_3) = x_1^2 + x_2^2 + 4x_3^2 = 1$ , $\displaystyle g_2(x_1, x_2, x_3) = x_1^2 + (x_2 - 1)^2 + 4x_3^2 = 3$

Then

$\displaystyle L(x) = x_1 + 2x_2 + 2x_3 + \lambda_1(x_1^2 + x_2^2 + 4x_3^2) + \lambda_2(x_1^2 + (x_2 - 1)^2 + 4x_3^2)$

Critical Points satisfy

$\displaystyle \frac{\partial L}{\partial x_1} = 1 + 2 \lambda_1 x_1 + 2 \lambda_2 x_1 = 0$
$\displaystyle \frac{\partial L}{\partial x_2} = 2 + 2 \lambda_1 x_2 + 2 \lambda_2 (2x_2 - 2) = 0$
$\displaystyle \frac{\partial L}{\partial x_3} = 2 + 8 \lambda_1 x_3 + 8 \lambda_2 x_3 = 0$

But im having trouble finding the values for $\displaystyle x_1, x_2, x_3$ and then finding $\displaystyle \lambda_1 , \lambda_2$