The Method of Lagrange Multipliers

# The Method of Lagrange Multipliers

Recall from the Lagrangian Functions page that provided that extrema exist for the function $z = f(x, y)$ whose points $(x, y)$ are restricted to lie on the constraint level curve $g(x, y) = 0$, then we simply need to consider the critical points of the corresponding Lagrangian function:

(1)
\begin{align} \quad L(x, y, \lambda) = f(x, y) + \lambda g(x, y) \end{align}

We will now solidify the method of Lagrange multipliers to find existing extrema.

## The Method of Lagrange Multipliers for Two Variable Functions

Let $z = f(x, y)$ be a two variable real-valued function. Then $f$ represents a surface in $\mathbb{R}^3$. Now suppose that we want to maximize or minimize $f(x, y)$ subject to a constraint equation $g(x, y) = C$. We note that we are trying to maximize or minimize $f$ for which we only consider points $(x, y)$ that lie on the level curve $g(x, y) = C$. In order to obtain a maximum value, we must find the value of $k$ such that the level curve $f(x, y) = k$ touches the level curve $g(x, y) = C$ at some point $(x_0, y_0)$. This occurs when both curves $f(x, y) = k$ and $g(x, y) = C$ have a common tangent line at $(x_0, y_0)$. Since both curves $f(x, y) = k$ and $g(x, y) = C$ have a common tangent line at $(x_0, y_0)$, we also have that both curves share normal lines at $(x_0, y_0)$ and thus the gradient vectors of $f$ and $g$ at $(x_0, y_0)$ are parallel, and so for some scalar $\lambda$ we have that:

(2)
\begin{align} \quad \nabla f(x_0, y_0) = \lambda \nabla g(x_0, y_0) \end{align}
 Definition: If $(x_0, y_0)$ produces an extreme value for the function $z = f(x, y)$ subject to the constraint $g(x, y) = C$, then there exists a $\lambda \in \mathbb{R}$ called a Lagrange Multiplier such that $\nabla f(x_0, y_0) = \lambda \nabla g(x_0, y_0)$.

From this, we obtain the following method for finding the maximum and/or minimum values of a function $z = f(x, y)$ subject to a constraint $g(x, y) = C$.

 Theorem 1 (The Method of Lagrange Multipliers for Two Variable Functions): Let $z = f(x, y)$ be a two variable real-valued function subject to the constraint $g(x, y) = C$. Suppose that a maximum/minimum value is achieved and that $\nabla g(x, y) \neq (0, 0)$ on the level curve $g(x, y) = C$. Then the maximum/minimum can be found by finding the values of $x$, $y$, and possibly $\lambda$ such that $\nabla f(x, y) = \lambda \nabla g(x, y)$ AND $g(x, y) = C$ and then evaluating $f$ at these points $(x, y)$ for which the largest is the maximum of $z = f(x, y)$ subject to the constraint $g(x, y) = C$ and the smallest is the minimum of $z = f(x, y)$ subject to the constraint $g(x, y) = C$.

## The Method of Lagrange Multipliers for Three Variable Functions

The method of Lagrange multipliers for a three variable function $w = f(x, y, z)$ whose points $(x, y, z)$ are constrained to the level surface $g(x, y, z) = C$ follow analogously. Provided that an extreme value exists, we would need to examine the critical points of the corresponding Lagrangian function:

(3)
\begin{align} \quad L(x, y, z, \lambda) = f(x, y, z) + \lambda g(x, y, z) \end{align}

That is, getting $\nabla L (x, y, z, \lambda) = (0, 0, 0, 0)$. Equivalently, we would need to solve the following system of equations for $x$, $y$, $z$ and possibly $\lambda$:

(4)
\begin{align} \quad \frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x} \\ \quad \frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y} \\ \quad \frac{\partial f}{\partial z} = \lambda \frac{\partial g}{\partial z} \\ \quad g(x, y, z) = C \end{align}