Lagrange Multipliers

We are going to take a look at another way of optimizing a function subject to given constraint(s). The constraint(s) may be the equation(s) that describe the boundary of a region although in this section we won’t concentrate on those types of problems since this method just requires a general constraint and doesn’t really care where the constraint came from.

We want to optimize (i.e. find the minimum and maximum value of) a function, $f\left(x,y,z\right)$, subject to the constraint $g\left(x,y,z\right)=k$. Again, the constraint may be the equation that describes the boundary of a region or it may not be. The process is actually fairly simple, although the work can still be a little overwhelming at times.

Notice that the system of equations from the method actually has four equations, we just wrote the system in a simpler form. To see this let’s take the first equation and put in the definition of the gradient vector to see what we get.

$$⟨f_x,f_y,f_z⟩=\lambda ⟨g_x,g_y,g_z⟩=⟨\lambda g_x,\lambda g_y,\lambda g_z⟩$$

In order for these two vectors to be equal the individual components must also be equal. So, we actually have three equations here.

$$f_x=\lambda g_x{f}_y=\lambda g_y{f}_z=\lambda g_z$$

These three equations along with the constraint, $g\left(x,y,z\right)=c$, give four equations with four unknowns $x$, $y$, $z$, and $\lambda$.

Note as well that if we only have functions of two variables then we won’t have the third component of the gradient and so will only have three equations in three unknowns $x$, $y$, and $\lambda$.

As a final note we also need to be careful with the fact that in some cases minimums and maximums won’t exist even though the method will seem to imply that they do. In every problem we’ll need to make sure that minimums and maximums will exist before we start the problem.

To see a physical justification for the formulas above. Let’s consider the minimum and maximum value of $f\left(x,y\right)=8x^2-2y$ subject to the constraint $x^2+y^2=1$ as an exercise.