# Calculus Optimization Methods/Lagrange Multipliers

The method of Lagrange multipliers solves the constrained optimization problem by transforming it into a non-constrained optimization problem of the form:

• $\operatorname{\mathcal{L}}(x_1,x_2,\ldots, x_n,\lambda)= \operatorname{f}(x_1,x_2,\ldots, x_n)+\operatorname{\lambda}(k-g(x_1,x_2,\ldots, x_n))$

Then finding the gradient and hessian as was done above will determine any optimum values of $\operatorname{\mathcal{L}}(x_1,x_2,\ldots, x_n,\lambda)$.

Suppose we now want to find optimum values for $f(x,y)=2x^2+y^2$ subject to $x+y=1$ from [2].

Then the Lagrangian method will result in a non-constrained function.

• $\operatorname{\mathcal{L}}(x,y,\lambda)= 2x^2+y^2+\lambda (1-x-y)$

The gradient for this new function is

• $\frac{\partial \mathcal{L}}{\partial x}(x,y,\lambda)= 4x+\lambda (-1)=0$
• $\frac{\partial \mathcal{L}}{\partial y}(x,y,\lambda)= 2y+\lambda (-1)=0$
• $\frac{\partial \mathcal{L}}{\partial \lambda}(x,y,\lambda)=1-x-y=0$

Finding the stationary points of the above equations can be obtained from their matrix from.

$\begin{bmatrix} 4 & 0 & -1 \\ 0& 2 & -1 \\ -1 & -1 & 0 \end{bmatrix} \begin{bmatrix} x\\ y \\ \lambda \end{bmatrix}= \begin{bmatrix} 0\\ 0\\ -1 \end{bmatrix}$

This results in $x=1/3, y=2/3, \lambda=4/3$.

Next we can use the hessian as before to determine the type of this stationary point.

$H(\mathcal{L})= \begin{bmatrix} 4 & 0 & -1 \\ 0& 2 & -1 \\ -1&-1&0 \end{bmatrix}$

Since $H(\mathcal{L}) >0$ then the solution $(1/3,2/3,4/3)$ minimizes $f(x,y)=2x^2+y^2$ subject to $x+y=1$ with $f(x,y)=2/3$.