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PD.8 Lagrange Multipliers

                        

In this section we present Lagrange's maximizing or minimizing a general function $f(x,y,z)$ subject to a constraint (or side condition) of the form $g(x,y,z)=k.$ This is shown in the picture above.

    It is easier to explain the geometric basis of Lagrange's method for functions of two variables. So we start by trying to find the extreme values of $f(x,y)$ subject to a constraint of the form $g(x,y)=k.$ In other words, we seek the extreme values of $f(x,y)$ when the point $(x,y)$ is restricted to like on the level curve $g(x,y)=k.$ The figure below shows this curve together with several level curves of $f$, These have the equations $f(x,y)=c $. To maximize $f(x,y)$ subject to $g(x,y)=k$ is to find the largest value of c such that the level curve $f(x,y)=c$ intersects $g(x,y)=k.$It appears from the figure below that this happens when these curves just touch each other, that is, when they have a common tangent line (Otherwise, the value of c could be increased further) This means that the normal lines at the point $(x_{0},y_{0})$ where they touch are identical. So the gradient vectors are parallel; that is MATH for some scalar $\lambda $.




 

 

The procedure described above can be expanded into the form $f(x,y,z)$ with the constraint $g(x,y,z)=k.$ When expanded into this form, the final equation looks like this:


MATH

 

The number $\lambda $ in the equation above is called a Lagrange multiplier. The procedure based on this equation is described below:

 

 

Method of Lagrange Multipliers

To find the maximum and minimum values of $f(x,y,z)$ subject to the constraint $g(x,y,z)=k$ (assuming that these extreme values exist):

 

(a) Find all values of $x,y,z,$ and $\lambda $ such that


MATH               and $\nabla g(x,y,z)=k$
 

 

(b) Evaluate $f$ at all the points $(x,y,z)$ that result from step (a). The largest of these values is the maximum value of $f$; the smallest is the minimum value of $f$.

 

 

If we write the vector equation MATH in terms of its components, the equations in step (a) become:


MATH
 

 

This is a system of four equations with four unknowns x, y, z, and $\lambda $, but it is not necessary to find explicit values for $\lambda .$

Functions of two variables are solved in much the same way:


MATHand $\nabla g(x,y)=k$
 

 

These equations can be broken down into:


MATH
 

 

These are then solved, like above, as a system of three equations with three variables.

 

Two Constrains

 

Suppose now that we want to find the maximum and minimum values of $f(x,y,z)$ subject not to one constraints, but to two constraints ( and $h(x,y,z)=c)$. In this case, and by the same line of reasoning that was used with one constraint, would have to set up the following equation:


MATH

 

In this case of Lagrange's method is to look for extreme values by solving five equations in the five unknowns x, y, z, $\lambda ,$ and $\mu .$ These equations are obtained by writing the equation in terms of its components and using the constraint equations:


MATH
MATH
MATH
$g(x,y,z)=k$
$h(x,y,z)=c$
 

 

One simply needs to solve these find equations and then plug the results back into the original function and find the maximum value.

 

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