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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname{\mathcal{L}}(x_1,x_2,\ldots, x_n,\lambda)} .

Suppose we now want to find optimum values for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname{\mathcal{L}}(x,y,\lambda)= 2x^2+y^2+\lambda (1-x-y)}

The gradient for this new function is

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial \mathcal{L}}{\partial x}(x,y,\lambda)= 4x+\lambda (-1)=0}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H(\mathcal{L})= \begin{bmatrix} 4 & 0 & -1 \\ 0& 2 & -1 \\ -1&-1&0 \end{bmatrix} }

Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 $f(x,y)=2/3$ .