# Lagrange multipliers and partial derivatives

• phrygian
In summary, to find the point on 2x + 3y + z - 11 = 0 for which 4x^2 + y^2 + z^2 is a minimum, we can use Lagrange multipliers and set the partial derivatives equal to 0. This will give us a system of equations that can be solved to find the values of x, y, and z.
phrygian

## Homework Statement

Find the point on 2x + 3y + z - 11 = 0 for which 4x^2 +y^2 +z^2 is a minimum

## The Attempt at a Solution

Using lagrange multipliers I find:

F = 4x^2 + y^2 + z^2 + l(2x + 3y + z)

Finding the partial derivatives I get the three equations:

df/dx = 8x + 2l
df/dy= 2y + 3l
df/dz= 2z + l

This is where I am stuck, what are the next steps for solving the system of equations?

Well, first set them equal to 0! They are equivalent to $x= -\lambda$, $2y/3-\lambda$ and $2z= -\lambda$. You can easily eliminate $\lambda$ by setting those equal to each other. And don't forget that you have 2x+ 3y+ z= 11 as a third equation.

## 1. What is the purpose of using Lagrange multipliers?

Lagrange multipliers are used to find the extreme values of a multivariable function subject to one or more constraints. It allows us to incorporate the constraints into the optimization problem and find the maximum or minimum value of the function.

## 2. How are Lagrange multipliers related to partial derivatives?

Lagrange multipliers use partial derivatives to find the critical points of the function subject to the constraints. These critical points correspond to the extreme values of the function.

## 3. Can Lagrange multipliers be used for functions with more than two variables?

Yes, Lagrange multipliers can be used for functions with any number of variables. It is a general method for solving constrained optimization problems.

## 4. Can Lagrange multipliers be used for non-linear functions?

Yes, Lagrange multipliers can be used for both linear and non-linear functions. It is a versatile method that can handle a wide range of functions and constraints.

## 5. Are there any limitations to using Lagrange multipliers?

One limitation of Lagrange multipliers is that it can only find the local extrema of a function, not the global extrema. It is also not suitable for functions with non-differentiable points or discontinuities.

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