Answer to Question #104434 in Calculus for Emmanuel

Question

How to solve a multivariable function using Lagrange multipliers

Answer 1

When you want to maximize (or minimize) a multivariable function $f(x, y, \dots)$  subject to the constraint that another multivariable function equals a constant, ${g(x, y, \dots) = c}$

follow these steps:

Step 1: Introduce a new variable ${\lambda}$, and define a new function $\mathcal{L}$  as follows:

$\mathcal{L}(x, y, \dots, {\lambda}) = {f(x, y, \dots)} - {\lambda} ({g(x, y, \dots)-c})$

This function $\mathcal{L}$  is called the "Lagrangian", and the new variable ${\lambda}$  is referred to as a "Lagrange multiplier"

Step 2: Set the gradient of $\mathcal{L}$  equal to the zero vector.

$\nabla \mathcal{L}(x, y, \dots, {\lambda}) = \textbf{0} \quad \leftarrow \small{\gray{\text{Zero vector}}}$

In other words, find the critical points of $\mathcal{L}$

Step 3: Consider each solution, which will look something like $(x_0, y_0, \dots, {\lambda}_0)$. Plug each one into f. Or rather, first remove the ${\lambda}_0$ then plug it into f, since f does not have ${\lambda}$  as an input. Whichever one gives the greatest (or smallest) value is the maximum (or minimum) point your are seeking.