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:
[ttps://www.youtube.com/watch?v=ry9cgNx1QV8&list=RDCMUCFe6jenM1Bc54qtBsIJGRZQ&index=5 LaGrange Multipliers - Finding Maximum or Minimum Values] -patrickJMT
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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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* [https://en.wikibooks.org/wiki/Calculus_Optimization_Methods/Lagrange_Multipliers Lagrange Multipliers], WikiBooks: Calculus Optimization Methods
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===Videos===
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*[https://www.youtube.com/watch?v=ry9cgNx1QV8&list=RDCMUCFe6jenM1Bc54qtBsIJGRZQ&index=5 LaGrange Multipliers - Finding Maximum or Minimum Values ]Video by patrickJMT
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*[https://www.youtube.com/watch?v=x6j6yFzTUgU Lagrange Multipliers Practice Problems] Video by ames Hamblin 2017 
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*[https://www.youtube.com/watch?v=HyqBcD_e_Uw Lagrange multipliers | MIT 18.02SC Multivariable Calculus, Fall 2010] Video by MIT OpenCourseWare
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*[https://www.youtube.com/watch?v=qXhcpqslNUU Lagrange Multipliers - Two Constraints -patrickJMT 2009 ] Video by patrickJMT 2009
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*[https://www.youtube.com/watch?v=nDuS5uQ7-lo Lagrange multipliers (3 variables) | MIT 18.02SC Multivariable Calculus, Fall 2010] Video by MIT OpenCourseWare
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==Licensing==
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Content obtained and/or adapted from:
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* [https://en.wikibooks.org/wiki/Calculus_Optimization_Methods/Lagrange_Multipliers Lagrange Multipliers, WikiBooks: Calculus Optimization Methods] under a CC BY-SA license

Latest revision as of 15:52, 2 November 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.

The gradient for this new function is

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

This results in .

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

Since then the solution minimizes subject to with .


Resources

Videos

Licensing

Content obtained and/or adapted from: