Max/Min f subject to g: Lagrange Multipliers

In summary, the equations can all be solved for x, y, and z when the condition that x2+y2+z2=1 is met.
  • #1
terminal.velo
4
0

Homework Statement



Find max/min of f subject to constraint: x^2+y^2+z^1 = 1


Homework Equations




f(x,y,z) = 1/4*x^2 + 1/9*y^2 + z^2
g(x,y,z) = x^2 + y^2 + z^2 - 1

The Attempt at a Solution



L = 1/4*x^2 + 1/9*y^2 + z^2 - λ(x^2 + y^2 + z^2 - 1)
Lx = 2/4*x - λ*x*2
Ly = 2/9*y - λ*2*y
Lz = 2*z - λ*2*z
 
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  • #2
terminal.velo said:

Homework Statement



Find max/min of f subject to constraint: x^2+y^2+z^1 = 1


Homework Equations




f(x,y,z) = 1/4*x^2 + 1/9*y^2 + z^2
g(x,y,z) = x^2 + y^2 + z^2 - 1

The Attempt at a Solution



L = 1/4*x^2 + 1/9*y^2 + z^2 - λ(x^2 + y^2 + z^2 - 1)
Lx = 2/4*x - λ*x*2
Ly = 2/9*y - λ*2*y
Lz = 2*z - λ*2*z

Correct so far. What should be the partial derivatives at a minimum/maximum?

ehild
 
  • #3
I don't know, the lecturer told us to solve the rest of the problem at home.
 
  • #4
Well, check your notes, what is the condition that a minimum or maximum exist?

ehild
 
  • #5
All 3 derivates of L (Lx, Ly, Lz) have to equal zero, to find the λ, then the x/y/z for critical points of f(x,y,z)

Lx = 2/4*x - λ*x*2 = 0
Ly = 2/9*y - λ*2*y = 0
Lz = 2*z - λ*2*z = 0
 
  • #6
OK, factorize the left hand sides.

ehild
 
  • #7
2x(1/4 - λ) = 0
2y(1/9 - λ) = 0
2z(1 - λ) = 0

EDIT: I'm in a bit of a funk, apologies! (lectures from 8am 'til 3pm)
 
  • #8
terminal.velo said:
2x(1/4 - λ) = 0
2y(1/9 - λ) = 0
2z(1 - λ) = 0

EDIT: I'm in a bit of a funk, apologies! (lectures from 8am 'til 3pm)

So you need some rest, without Maths... :smile:
Anyway: you can omit the factors 2. All your equations are products, equal to zero. That means, one of the two factors must be zero in each equation. More than one factor containing lambda can not be zero, it would mean contradiction. Can all x, y,z equal to zero? Remember, you have the condition that x2+y2+z2=1.

ehild
 

What is the purpose of using Lagrange multipliers in this context?

Lagrange multipliers are used in optimization problems to find the maximum or minimum value of a function subject to a constraint. They help to simplify the problem and make it easier to solve by converting it into a system of equations.

How do you set up the equations for solving a max/min problem using Lagrange multipliers?

To set up the equations, you first need to define the objective function (f) and the constraint function (g). Then, you create a new function called the Lagrangian by adding a Lagrange multiplier (λ) multiplied by the constraint function to the objective function. Finally, you take the partial derivatives of the Lagrangian with respect to all the variables and set them equal to 0.

Can Lagrange multipliers be used for both single and multiple constraints?

Yes, Lagrange multipliers can be used for both single and multiple constraints. In the case of multiple constraints, you will have multiple Lagrange multipliers, one for each constraint.

What are the limitations of using Lagrange multipliers for optimization problems?

Lagrange multipliers can only be used when the objective and constraint functions are continuous and differentiable. They are also only applicable for finding local maxima and minima, not global ones. Additionally, they may not be suitable for problems with a large number of variables or constraints as they can become computationally expensive.

Are there any real-world applications of using Lagrange multipliers?

Yes, Lagrange multipliers have various real-world applications, such as in economics, engineering, and physics. They can be used to optimize production processes, minimize costs, and find optimal solutions for various systems and processes.

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