Difference between revisions of "Lagrange Multipliers"

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The method of Lagrange multipliers solves the constrained optimization problem by transforming it into a non-constrained optimization problem of the form:
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* <math>\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))</math>
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Then finding the gradient and Hessian as was done above will determine any optimum values of <math>\operatorname{\mathcal{L}}(x_1,x_2,\ldots, x_n,\lambda)</math>.
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Suppose we now want to find optimum values for <math>f(x,y)=2x^2+y^2</math> subject to <math>x+y=1</math> from [2].
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Then the Lagrangian method will result in a non-constrained function.
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* <math>\operatorname{\mathcal{L}}(x,y,\lambda)= 2x^2+y^2+\lambda (1-x-y)</math>
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The gradient for this new function is
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* <math>\frac{\partial \mathcal{L}}{\partial x}(x,y,\lambda)= 4x+\lambda (-1)=0</math>
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* <math>\frac{\partial \mathcal{L}}{\partial y}(x,y,\lambda)= 2y+\lambda (-1)=0</math>
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* <math>\frac{\partial \mathcal{L}}{\partial \lambda}(x,y,\lambda)=1-x-y=0</math>
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Finding the stationary points of the above equations can be obtained from their matrix from.
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: <math> \begin{bmatrix}
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4 & 0 & -1 \\
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0& 2 & -1 \\
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-1 & -1 & 0
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\end{bmatrix} \begin{bmatrix}
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x\\
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y \\
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\lambda \end{bmatrix}= \begin{bmatrix}
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0\\
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0\\
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-1
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\end{bmatrix}
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</math>
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This results in <math>x=1/3, y=2/3, \lambda=4/3</math>.
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Next we can use the Hessian as before to determine the type of this stationary point.
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: <math> H(\mathcal{L})=
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\begin{bmatrix}
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4 & 0 & -1 \\
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0& 2 & -1 \\
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-1&-1&0
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\end{bmatrix}
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</math>
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Since <math> H(\mathcal{L}) >0 </math> then the solution <math>(1/3,2/3,4/3)</math> minimizes <math>f(x,y)=2x^2+y^2</math> subject to <math>x+y=1</math> with <math>f(x,y)=2/3</math>.
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==Resources==
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===Videos===
 
*[https://www.youtube.com/watch?v=ry9cgNx1QV8&list=RDCMUCFe6jenM1Bc54qtBsIJGRZQ&index=5 LaGrange Multipliers - Finding Maximum or Minimum Values ]Video by patrickJMT
 
*[https://www.youtube.com/watch?v=ry9cgNx1QV8&list=RDCMUCFe6jenM1Bc54qtBsIJGRZQ&index=5 LaGrange Multipliers - Finding Maximum or Minimum Values ]Video by patrickJMT
  

Revision as of 12:22, 6 October 2021

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

Then finding the gradient and Hessian as was done above will determine any optimum values of .

Suppose we now want to find optimum values for subject to 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}
  • 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 \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 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 (1/3,2/3,4/3)} minimizes 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 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+y=1} with 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)=2/3} .


Resources

Videos

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