Lagrange Multiplier Method

Lagrange Multiplier Method

目录

• Lagrange Multiplier Method
  • Prerequisite knowledge - partial derivatives
  • Usage
  • Ex
  • Summary

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#

• When there is nothing can do, pull the multiplication of broken moves.

For some oddly tricky or computationally intensive problems, you can use the Lagrange multiplier method.