Lagrange Multipliers to Find Extreme Values of a Multi-Variable Function

In summary, to find the extrema of the function f(x,y) = 3x^{2} + y^{2} with the constraint x^{2} + y^{2} = 1, we can use the Lagrange multiplier method by setting up the system of equations 6x = 2xλ and 2y = 2yλ, with the constraint x^{2} + y^{2} = 1. Solving for x and y, we get x = y = 0 or x = y = ±1, which correspond to the constrained min and max values.
  • #1
ChrisPls
2
0

Homework Statement



I need to find the extrema of [itex]f(x,y) = 3x^{2} + y^{2}[/itex] given the constraint [itex]x^{2} + y^{2} = 1[/itex]

Homework Equations



I'm not sure what goes here. I've been trying to solve it with this:

∇f(x,y) = λ∇g(x,y)

The Attempt at a Solution



[itex]f(x,y) = 3x^{2} + y^{2}[/itex]
[itex]g(x,y) = x^{2} + y^{2} = 1[/itex]

[itex]∇f(x,y) = <6x, 2y>[/itex]
[itex]∇g(x,y) = <2x, 2y>[/itex]

With the previous equation, we get a system of equations:

[itex]6x = 2xλ[/itex]

[itex]2y = 2yλ[/itex]

and the constraint

[itex]x^{2} + y^{2} = 1[/itex]

Solving for x and y, I get x=y=0. But that fails the final constraint. I've used the above procedure for other Lagrange problems without much issue, but this one's stumping me.

So I tried finding a solution on Wolfram: Maxes at (-1, 0) and (1, 0), and Mins at (0, -1) and (0, 1). Okay, nice and simple looking solutions. Right?

But I just can't figure this out. What am I missing?

Thanks
 
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  • #2
ChrisPls said:

Homework Statement



I need to find the extrema of [itex]f(x,y) = 3x^{2} + y^{2}[/itex] given the constraint [itex]x^{2} + y^{2} = 1[/itex]

Homework Equations



I'm not sure what goes here. I've been trying to solve it with this:

∇f(x,y) = λ∇g(x,y)

The Attempt at a Solution



[itex]f(x,y) = 3x^{2} + y^{2}[/itex]
[itex]g(x,y) = x^{2} + y^{2} = 1[/itex]

[itex]∇f(x,y) = <6x, 2y>[/itex]
[itex]∇g(x,y) = <2x, 2y>[/itex]

With the previous equation, we get a system of equations:

[itex]6x = 2xλ[/itex]

[itex]2y = 2yλ[/itex]

and the constraint

[itex]x^{2} + y^{2} = 1[/itex]

Solving for x and y, I get x=y=0. But that fails the final constraint. I've used the above procedure for other Lagrange problems without much issue, but this one's stumping me.

So I tried finding a solution on Wolfram: Maxes at (-1, 0) and (1, 0), and Mins at (0, -1) and (0, 1). Okay, nice and simple looking solutions. Right?

But I just can't figure this out. What am I missing?

Thanks

[tex] 6x = 2x \lambda \Longrightarrow x = 0 \text{ or } \lambda = 3\\
2y = 2y \lambda \Longrightarrow y = 0 \text{ or } \lambda = 1.[/tex]
Either x or y must = 0, and then the other one = ± 1. One value of λ corresponds to the constrained min and the other to the constrained max.

RGV
 
  • #3
Ooh, I see how that works now. Thank you very much!
 

1. What are Lagrange multipliers?

Lagrange multipliers are a mathematical tool used to find extreme values (maximum or minimum) of a multi-variable function subject to a constraint. It involves creating a new function using the original function and the constraint equation, and then solving for the critical points of this new function.

2. When should Lagrange multipliers be used?

Lagrange multipliers should be used when trying to find extreme values of a function while also satisfying a given constraint. This could be in optimization problems, where the constraint represents a limitation on the variables.

3. How do Lagrange multipliers work?

Lagrange multipliers work by finding the critical points of a new function created by combining the original function and the constraint equation. These critical points correspond to the extreme values of the original function subject to the constraint.

4. What is the equation for Lagrange multipliers?

The equation for Lagrange multipliers is ∇f(x,y) = λ∇g(x,y), where f(x,y) is the original function, g(x,y) is the constraint equation, and λ is the Lagrange multiplier. This equation is used to find the critical points of the new function formed by combining f(x,y) and g(x,y).

5. What are the steps for using Lagrange multipliers to find extreme values?

The steps for using Lagrange multipliers to find extreme values are:
1. Write the original function and the constraint equation.
2. Create a new function by combining the two equations and adding a Lagrange multiplier.
3. Take the partial derivatives of the new function with respect to each variable.
4. Set the partial derivatives equal to zero and solve for the variables and the Lagrange multiplier.
5. Plug the values found in step 4 into the original function to get the extreme values subject to the constraint.

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