Question #104434

How to solve a multivariable function using Lagrange multipliers

Expert's answer

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

follow these steps:

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

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


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

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

∇L(x,y,…,λ)=0←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 L\mathcal{L}

Step 3: Consider each solution, which will look something like (x0,y0,…,λ0)(x_0, y_0, \dots, {\lambda}_0). Plug each one into f. Or rather, first remove the λ0{\lambda}_0 then plug it into f, since f does not have λ{\lambda}  as an input. Whichever one gives the greatest (or smallest) value is the maximum (or minimum) point your are seeking.



LATEST TUTORIALS
APPROVED BY CLIENTS