Understanding Lagrange Multipliers: Solving for Max and Min Values

In summary, to find the maximum and minimum values of f(x, y, z) subject to the constraint g(x, y, z) = k, you need to find all values of x, y, z, and λ that satisfy the vector equation ∇f(x, y, z) = λ ∇g(x, y, z). This can be done by setting corresponding coordinates equal and solving for the variables. Once you have these values, evaluate f at each point and the largest value will be the maximum value of f while the smallest value will be the minimum value of f. This method, known as Lagrange multipliers, is used to find extreme values of functions
  • #1
Dx
Find max and min value…f(x,y,z) =3x+2y+z; x2 + y2+z2 = 1

If g(x,y,z) = x2 + y2+z2 = 1 then what do I do next?

I need help to further solve for this please? I am horrible at math and don't understand lagrange multipiers so can anyone better explain it to me and help me solve for difficult problem.

Dx
 
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  • #2
Well, my text has the following box for lagrange multipliers (interesting theorems / procedures are placed in boxes in the text):

To find the maximum and minimum values of f(x, y, z) subject to the constraint g(x, y, z) = k (assuming these extreme values exist):

(a) Find all values of x, y, z, and λ such that:

[nab]f(x, y, z) = λ [nab]g(x, y, z)

and g(x, y, z) = k

(b) Evaluate f at all the points (x, y, z) that arise from step (a). The largest of these values is the maximum value of f; the smallest is the minimum value of f.

I imagine your text has something similar. Do you understand how to start step (a)?
 
  • #3
Originally posted by Hurkyl
Well, my text has the following box for lagrange multipliers (interesting theorems / procedures are placed in boxes in the text):



I imagine your text has something similar. Do you understand how to start step (a)?

No, I don't understand this what-so-ever. I see one example in my book that wants to find the points of a rectangular hyperbola and its using partial Dx but I don't know how to perform step a could you please help me.

Dx
 
  • #4
∇f(x, y, z) means the gradient of f with respect to its three variables. The gradient is defined as the row vector:

&nabla;f(x, y, z) = < &part;f(x, y, z)/&part;x, &part;f(x, y, z)/&part;y, &part;f(x, y, z)/&part;z >
I.E. the first coordinate is the partial derivative with respect to (WRT) the first variable, the second coordinate is the partial derivative WRT the second variable, and so on for as many variabes as the function has.


The equation in step (a) is, then, a vector equation, which we solve by setting corresponding coordinates equal:

&nabla;f(x, y, z) = &lambda; &nabla;g(x, y, z)

is

< &part;f(x, y, z)/&part;x, &part;f(x, y, z)/&part;y, &part;f(x, y, z)/&part;z > = &lambda; < &part;g(x, y, z)/&part;x, &part;g(x, y, z)/&part;y, &part;g(x, y, z)/&part;z >

is

&part;f(x, y, z)/&part;x = &lambda; &part;g(x, y, z)/&part;x
&part;f(x, y, z)/&part;y = &lambda; &part;g(x, y, z)/&part;y
&part;f(x, y, z)/&part;z = &lambda; &part;g(x, y, z)/&part;z



(BTW, if you also want me to explain the "why" behind this method, ask and I'll do so... but the "why" may be quite difficult to understand)
 

What are Lagrange multipliers?

Lagrange multipliers are a mathematical technique used to find the extreme values (maxima and minima) of a function subject to constraints.

How do Lagrange multipliers work?

Lagrange multipliers use the concept of partial derivatives to find the extreme values of a function. The constraints are incorporated into the function using a Lagrange multiplier, which acts as a scaling factor for the constraints.

When are Lagrange multipliers used?

Lagrange multipliers are used when optimizing a function subject to constraints. This could include problems in economics, physics, engineering, and other fields where there are limitations or restrictions on the variables involved.

What is the formula for Lagrange multipliers?

The formula for Lagrange multipliers is:
L(x,y,z,λ) = f(x,y,z) - λ(g(x,y,z))
where L is the Lagrangian function, f is the objective function, g is the constraint function, and λ is the Lagrange multiplier.

What are the limitations of Lagrange multipliers?

While Lagrange multipliers are a powerful tool for solving constrained optimization problems, they have some limitations. They can only be used for problems with continuous and differentiable functions, and they may not always give the global optimum solution. Additionally, they can become computationally expensive for problems with a large number of constraints.

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