Lagrange Multiplier Method
目录The End of Inequality: "\(\textsf{Lagrange Multiplier Method}\)"
Prerequisite knowledge - partial derivatives
In a nutshell: principal element derivative.
For a function \(F(x,y)\), its partial derivative with respect to \(x\) = taking \(y\) as a constant to find the derivative of \(x\).
For example: \(F(x,y) = x^2 + 4yx + y^2\), its partial derivative with respect to \(x\) = \(2x + 4y\).
Usage
Given \(F(x,y) = 0\), find the \(G(x,y)\) maximum (note that for \(F(x,y) = k\) you need to make \(F(x,y) = F(x,y) - k\)).
Let \(H(x,y,\lambda) = G(x,y) + \lambda F(x,y)\) and find the partial derivatives of \(x,y,\lambda\) respectively \(x(x,y,\lambda),y(x,y,\lambda),\lambda(x,y,\lambda)\).
Let \(x(x,y,\lambda) = 0, y(x,y,\lambda) = 0, \lambda(x,y,\lambda) = 0\) respectively, we get three equations, according to these 3 equations solve \(x_0,y_0,\lambda_0\), then \(G(x_0,y_0)\) is the *extreme value of the function *.
Then some questions \(x,y\) may have restrictions, so throw them aside first and then count the values according to the restrictions at the end.
OK, so now use it for a question:
\(\theta \in \R\), find the maximum and minimum values of \(\sin\theta(1 - \cos\theta)\).
Let \(\sin\theta = y,\cos\theta = x\), then the question becomes \(F(x,y) = x^2 + y^2 - 1 = 0\), \(G(x,y) = y(1 - x) = y - xy\), and find the maximum value of \(G(x,y)\).
Let \(H(x,y,\lambda) = y - xy + \lambda(x^2 + y^2 - 1)\) and list:
- \(x(x,y,\lambda) = -y + 2\lambda x = 0\) - \(y(x,y,\lambda) = 1 - x + 2\lambda y = 0\) - \(\lambda(x,y,\lambda) = x^2 + y^2 - 1 = 0\)
\(x - y = \dfrac{1}{2\lambda + 1}\)
\(x + y = -\dfrac{1}{2\lambda-1}\)
\((\dfrac{1}{2\lambda + 1})^2 + (\dfrac{1}{2\lambda-1})^2 = 2\)
\(\rm \lambda = 0\ or\ \lambda = \pm \frac{\sqrt 3}{2}\)
Finally bring in and work out \((x,y)=(-\frac{1}{2},\frac{\sqrt 3}{2})\ \text{or} \ (-\frac{1}{2},-\frac{\sqrt 3}{2}) \text{ or } (0,0)\).
Maximum value is \(\frac{3\sqrt 3}{4}\), minimum value is \(-\frac{3\sqrt 3}{4}\)
Ex
What about the three parameters \((x,y,z)\)?
This requires two known equations, then set two parameters \(\lambda,\mu\) and solve in the same way.
More parameters in the same way.
Summary
- There is nothing to do, pulling the multiplication of broken moves.
For some oddly tricky or computationally intensive problems, you can use the Lagrange multiplier method.
标签：partial,theta,Multiplier,find,Lagrange,frac,Method,lambda From： https://www.cnblogs.com/into-qwq/p/lagrange-multiplier-method.html