0. Lagrange multipliers technique¶
First of all, let's wrapp you head around lagrange multipliers technique. To make it simple, we're gonna deal with only 2 dimention in this article.
Let's say you wanna maximize multivariate function $f(x,y)$. subject to the constraint that another multivariate function equals a constant, $g(x,y) = C$
In that situation we can apply lagrange multiplier technique. At first introduce new variable $\lambda$. And define new function $L$ which takes form
$L$ is called "Lagrange function", $\lambda$ is "Lagrange multiplier".
Suppose $(x_0, y_0)$ maximaize (or minimize) $f(x,y)$ subject to $\varphi(x,y) = 0$, the following equality holds,
1. Intuitive understanding¶
From above equality,
$$\begin{eqnarray}\frac{\partial \left\{f(x,y) - \lambda \varphi(x,y)\right\}}{\partial x}&=&0 \\
\frac{\partial f(x)}{\partial x} &=& \lambda \frac{\partial g(x)}{\partial x}\end{eqnarray}$$
In the same way,
$$\begin{eqnarray}\frac{\partial f(x)}{\partial y} &=& \lambda \frac{\partial g(x)}{\partial y}\end{eqnarray}$$
Therefore, $$\nabla f = \lambda \nabla \varphi$$
You can tell gradient of $f(x,y)$ and $g(x,y)$ are parallel at the point of $(x,y)$ which maximaize (or minimize) $f(x,y)$ subject to $g(x,y)$.
2. What the $\nabla f(x,y)$ is ??¶
Let's think about contour of $f(x,y)$ where $f(x,y) = C$. Suppose moving the point along contour by $(\varDelta x, \varDelta y)$. At the time, total derivative can be captured as below,
$$\frac{\partial f}{\partial x}dx + \frac{\partial y}{\partial y} dy = \nabla f(x) \cdot \varDelta x= 0$$Therefore gradient $\nabla f$ is gonna be normal vector to tangent vector of contour.
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