If we are given a function \(f(x,y,z)\) that we want to optimize (find maximums, minimums or both) subject to a constraint \(g(x,y,z) = 0\), we set up the gradient equation \( \nabla f = \lambda \nabla g \). [ For a version of the equations that do not use the gradient, see below. ]

The variable \( \lambda \) is just a number (independent of \(x\), \(y\) and \(z\)) called a Lagrange multiplier. We introduce \( \lambda \) during the course of solving this problem but it will not appear in our answer.

Remember that the gradient is a vector. So, the above equation gives us three equations ( one for each variable ) and four unknowns ( \(x\), \(y\), \(z\) and \( \lambda \) ). Using also the constraint equation \(g(x,y,z) = 0\), we can now (theoretically) solve for all four unknowns.

This next video clip explains this technique in more detail.

So, why does this work? It seems kind of strange that introducing another variable would enable us to optimize a function. Here is a great video clip that explains this. His use of graphs is very good to visualize what is going on.

Okay, so now you know how to handle one constraint, if you are given two constraints, you just add another lagrange multiplier. We usually use the Greek letter mu, \( \mu \). The equation then looks like \( \nabla f = \lambda \nabla g + \mu \nabla h \) where the function we want to optimize is \(f(x,y,z)\) and the constraint equations are \(g(x,y,z) = 0\) and \(h(x,y,z) = 0\). Here is a video explaining this in more detail, including an example.

An alternate version of the equations using the Lagrange Method that do not use the gradient is given below. We present equations with three variables and two constraints. As you would expect, you can use the same ideas with two variables (drop \(z\)) and one constraint (drop \( h(x,y,z) \)).