📂Mathematical Physics

Lagrange Multiplier Method

Definition¹

The optimal value of the multivariable function $f(x_{1}, \dots, x_{n})$

Description

Let’s assume the graph of $y = 2 - x^{2}$ is given as shown in the figure above. Let’s denote the distance between the origin and the graph as $d$.

$$ d(x,y) = \sqrt{x^{2} + y^{2}} $$

Then, the problem of finding the point that minimizes the distance $d$ is the same as finding the point where the function $f(x,y) = x^{2} + y^{2}$ takes its minimum value. However, since $x, y$ is a point on the graph, it must satisfy the constraint $x^{2} + y = 2$. Let’s denote this constraint as $\phi$.

$$ \phi (x,y) = x^{2} + y = 2 $$

We know that the point where the derivative becomes $0$ is a candidate for the minimum (maximum) value. Therefore, we need to find the point where the total derivative becomes $0$.

$$ df = \dfrac{\partial f}{\partial x}dx + \dfrac{\partial f}{\partial y}dy = 0 $$

The Lagrange multiplier method is a method to find the point of minimum by solving the equation that adds the product of the constraint $d\phi$ and the multiplier $\lambda$ to the original equation.

$$ \begin{align*} df + \lambda d\phi =&\ \left( \dfrac{\partial f}{\partial x}dx + \dfrac{\partial f}{\partial y}dy \right) + \lambda \left( \dfrac{\partial \phi}{\partial x}dx + \dfrac{\partial \phi}{\partial y}dy \right) \\ =&\ \left( \dfrac{\partial f}{\partial x} + \lambda\dfrac{\partial \phi}{\partial x} \right)dx + \left( \dfrac{\partial f}{\partial y} + \lambda \dfrac{\partial \phi}{\partial y} \right) dy \\ =&\ 0 \end{align*} $$

Then, we get the following two equations.

$$ \dfrac{\partial f}{\partial x} + \lambda\dfrac{\partial \phi}{\partial x} = 0 \\[1em] \dfrac{\partial f}{\partial y} + \lambda \dfrac{\partial \phi}{\partial y} = 0 $$

If we specifically substitute $f$ and $\phi$, it reads as follows.

$$ 2x + \lambda \cdot 2x = 0 \\ 2y + \lambda = 0 $$

From the first equation, we know that it is either $x = 0$ or $\lambda = -1$.

• In the case of $x = 0$

  Since $y= 2- x^{2}$, it follows that $y=2$. (From the second equation, $\lambda = -4$)

• In the case of $\lambda = -1$

  From the second equation, it results in $y = \dfrac{1}{2}$. Hence, $x = \pm \sqrt{\dfrac{3}{2}}$

Listing these cases results in the following.

$$ (-\sqrt{\frac{3}{2}}, \frac{1}{2}), \quad (0, 2), \quad (\sqrt{\frac{3}{2}}, \frac{1}{2}) $$

This is the same as substituting $y=2-x^{2}$ into $f(x,y) = x^{2} + y^{2}$ and finding the point where the derivative becomes $0$. Of course, this is a simple example, so it’s easier and simpler to solve by substitution, but this is not the case for complex problems. For example, to solve the problem “When is the volume of a rectangular parallelepiped inscribed in a three-dimensional ellipsoid the largest?”, $f$ and $\phi$ would be as follows.

$$ f(x,y,z) = 8xyz,\quad \phi (x,y,z) = \dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} + \dfrac{z^{2}}{c^{2}} = 1 $$

In this case, it might be difficult to solve by substitution, but the Lagrange multiplier method makes it easier to solve.