Chapter 14 - Section 1 - Functions of Several Variables

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:

Solve $ f^{'} = 0 $
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:

Solve $ \nabla \mathbf{F} = 0 $.
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 $.

Example Set One

One

Two

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.

Example Set Two

One

$n$ 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 $$ F = f - \displaystyle \sum_{i=1}^m \lambda_i (g_i - k_i) $$

Example Set Three

One

→ Chapter 15: Multiple Integrals

Department of Mathematics and Statistics, Binghamton University people:grads:brown:brown_teaching:math323:chapter_14

Chapter 14 - Section 1 - Functions of Several Variables