### Chapter 14 - Section 8 - Lagrange Multipliers

Section Overview

In this section we learn how to use the Lagrange Technique to locate extreme (maximum or minimum) values of a multivariable function subject to some constraints. As we shall see, the technique outlined is extremely similar to the method we used in Calculus I to locate extreme values of a single variable function on an interval. We will describe the technique three times, once for a three-variable function with one constraint, once for a three-variable function with two constraints, and finally for the case of an $n$-variable function with $m$ constraints. We will assume throughout that our objective function is differentiable and our constraints have non-zero gradient, except for a couple examples in which we will discuss how and why the technique may fail when a constraint gradient is the zero vector.

#### Three variable function, single constraint

Let $w = f(x,y,z)$ denote our objective function and $g(x,y,z) = k$ our constraint; that is, we seek for the extreme values attained by the function $f$ on the level surface $g = k$. Suppose $f$ does have an extreme value at a point $P = (x_0,y_0,z_0)$ on the surface $g = k$. Let $C$ be a curve with differentiable parametrization $\mathbf{r}(t)$ that lies on the surface $g = k$ and passes through the point $P$. Let $t_0$ denote the parameter value corresponding to the point $P$, so $\mathbf{r}(t_0) = \langle x_0 , y_0 , z_0 \rangle$, and let $h(t) = f \circ \mathbf{r}(t)$. Now $h$ has an extreme value at $t_0$, so it has a critical point there. Since $f$ is differentiable at $P$ and $\mathbf{r}$ is differentiable, it follows that the composition $h$ is differentiable at $t_0$, and so the critical point $t_0$ is a root of the derivative. Hence, by the chain rule we have

\begin{aligned} 0 &= h^{'}(t_0) \\ &= \frac{\partial f}{\partial x} (x_0,y_0,z_0) \frac{dx}{dt} (t_0) + \frac{\partial f}{\partial y} (x_0,y_0,z_0) \frac{dy}{dt} (t_0) + \frac{\partial f}{\partial z} (x_0,y_0,z_0) \frac{dz}{dt} (t_0) \\ &= \nabla f(x_0,y_0,z_0) \cdot \mathbf{r}^{'}(t_0) \end{aligned}

Thus the gradient of $f$ at the point $P$ is orthogonal to the line tangent to the curve $C$ at the point $P$. Since this holds for all curves in the surface $g = k$ through the point $P$, it follows that the gradient of $f$ at $P$ is parallel to the direction of the plane tangent to the surface $g = k$ at the point $P$; in other words, at the point $P$, if $\nabla g \neq \mathbf{0}$ then $\nabla f = \lambda \nabla g$ for some number $\lambda$.

Let us recall the technique used in Calculus I to locate extreme values of a differentiable function $y = f(x)$ on an open interval $I$. If $f$ has an extreme value at a point $x_0 \in I$, then $x_0$ is a critical point of $f$, and since we are assuming that $f$ is differentiable on $I$ we must have $f^{'}(x_0) = 0$. We use this result to develop our technique as follows. Assuming that $f$ has extreme values on $I$, we locate them in two steps:

1. Solve $f^{'} = 0$
2. Evaluate $f$ at each solution $c$ found in step 1.; largest value is the maximum, smallest is the minimum.

The Lagrange technique works in a very similar way; assuming extreme values exist, we first find critical points, then evaluate at each to determine the extreme values. There is a slight difference in our Lagrange technique though, in that it is not the critical points of $f$ which we seek for but rather one which involves $f$ together with our constraint $g = k$. Let $\lambda$ denote a fourth variable and set $F = f - \lambda (g - k)$. Critical points of $F$ are found by solving the vector equation $\mathbf{0} = \nabla F$, which yields the following system of four equations with four unknowns:

$$\begin{array}{ll} 0 = F_x = f_x - \lambda g_x \\ 0 = F_y = f_y - \lambda g_y \\ 0 = F_z = f_z - \lambda g_z \\ 0 = F_\lambda = g - k \end{array}$$

We see that the first three equations represent the vector equation $\nabla f = \lambda \nabla g$ and the fourth equation is our constraint $g = k$. From our discussion above, we can conclude that if $f$ has an extreme value at a point $P = (x_0,y_0,z_0)$ on the surface $g=k$ and if $\nabla g \neq \mathbf{0}$ at $P$ then there exists a number $\lambda_0$ such that $\nabla f = \lambda_0 \nabla g$ at $P$ and so $(x_0,y_0,z_0,\lambda_0)$ is a critical point of $F$. Therefore, assuming that $f$ has extreme values on the surface $g=k$ and that the gradient of $g$ is non-zero, we can find the extreme values in two steps:

1. Solve $\nabla \mathbf{F} = 0$.
2. Evaluate $f$ at the projections in $\mathbb{R}^3$ of each solution found in step 1.; largest is maximum, smallest is minimum

Note that in step 1. above, we don't actually need the $\lambda$ part of the solution. You may find it's values in the course of finding $x,y,$ and $z$, but more often it will just be used to help relate the important variables. The points found in step 1. are in four-space, but we need points in three space to evaluate $f$; specifically, we need the $x,y,z$ part. The projection referred to above works as follows:

$$(x,y,z,\lambda) \rightarrow (x,y,z,0) \simeq (x,y,z)$$

In words, we replace our fourth coordinate with zero, then identify the point $(x,y,z,0)$ in $\mathbb{R}^4$ with the point $(x,y,z)$ in $\mathbb{R}^3$.

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#### Three variable function, two constraints

Let $f(x,y,z)$ be our objective function and $g(x,y,z) = k$ and $h(x,y,z) = \ell$ our two constraints. Let $$F = f - \lambda (g - k) - \mu (h - \ell)$$ If $f$ has an extreme value at a point $(x_0,y_0,z_0)$ on the intersection of the level surfaces $g=k$ and $h=\ell$ and both $\nabla g$ and $\nabla h$ are non-zero, then there exist numbers $\lambda_0$ and $\mu_0$ such that $(x_0,y_0,z_0,\lambda_0,\mu_0)$ is a critical point of $F$. Thus, to find extreme values of $f$ subject to the given constraints, we solve $\mathbf{0} = \nabla F$ and check each point.

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#### $n$ variable function, $m$ constraints

As usual we form an auxiliary function $F$ with of objective and all constraints and seek for solutions to the vector equation $\mathbf{0} = \nabla F$. Let $x_1,x_2,\ldots,x_n$ denote the $n$-variables and $g_1 = k_1, g_2 = k_2, \ldots, g_m = k_m$ the $m$ constraints. Then we introduce $m$ variables $\lambda_1,\lambda_2,\ldots,\lambda_m$ and set $$F = f - \displaystyle \sum_{i=1}^m \lambda_i (g_i - k_i)$$

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## Section 14.8 Discussion

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