Lagrange multipliers, guidance needed

In summary: I should emphasize, it should tell you that you have made a mistake, somewhere, if you really were to get that far, but you have not yet even written down the correct equation for the critical points. Try again.)For any real ##x,y## we have ##1 + 4 x^2 + 4 y^2 \geq 1 > 0##, so you are allowed to divide it out on both sides, to end up with \frac{32}{3} xy = - 24 xy What does that tell... you? (I should emphasize, it should tell you that you have made a mistake, somewhere, if you really were to get that far, but you have not
  • #1
ilyas.h
60
0

Homework Statement



f(x,y) is function who's mixed 2nd order PDE's are equal.

consider k_f:

Untitled.png


determine the points on the graph of the parabloid f(x,y) = x^2 + y^2 above the ellipse 3x^2 + 2y^2 = 1 at which k_f is maximised and minimised.

The Attempt at a Solution



is this the langrange equation that I'd use?

[tex]L(x,y,\lambda )=k_{f}(x,y)-\lambda k_{g}(x,y)[/tex]

or maybe it's just:
[tex]L(x,y,\lambda )=k_{f}(x,y)-\lambda g(x,y)[/tex]

im not sure if I should put the g function (3x^2 + 2y^2 = 1) into the k_f. If I do need to do this, then the algebra is very ugly.
 
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  • #2
ilyas.h said:

Homework Statement



f(x,y) is function who's mixed 2nd order PDE's are equal.

consider k_f:

Untitled.png


determine the points on the graph of the parabloid f(x,y) = x^2 + y^2 above the ellipse 3x^2 + 2y^2 = 1 at which k_f is maximised and minimised.

The Attempt at a Solution



is this the langrange equation that I'd use?

[tex]L(x,y,\lambda )=k_{f}(x,y)-\lambda k_{g}(x,y)[/tex]

or maybe it's just:
[tex]L(x,y,\lambda )=k_{f}(x,y)-\lambda g(x,y)[/tex]

im not sure if I should put the g function (3x^2 + 2y^2 = 1) into the k_f. If I do need to do this, then the algebra is very ugly.

Have you made an attempt to simplify ##k_f(x,y)## as much as possible before using it in a Lagrangian or anywhere else? Far from ugly, the algebra will be very simple and straightforward.
 
  • #3
Ray Vickson said:
Have you made an attempt to simplify ##k_f(x,y)## as much as possible before using it in a Lagrangian or anywhere else? Far from ugly, the algebra will be very simple and straightforward.

my k_f(x,y) is:

[tex]\frac{4}{(1+4x^{2}+4y^{2})^2}[/tex]

How would you simplify this/the problem? I tried simplifying with wolfram alpha, it doesn't return anything useful.

my k_g(x,y) is:

[tex]\frac{24}{(1+36x^{2}+16y^{2})^2}[/tex]

please, someone help. It's a bit of an emergency.
 
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  • #4
ilyas.h said:
my k_f(x,y) is:

[tex]\frac{4}{(1+4x^{2}+4y^{2})^2}[/tex]

How would you simplify this/the problem? I tried simplifying with wolfram alpha, it doesn't return anything useful.

my k_g(x,y) is:

[tex]\frac{24}{(1+36x^{2}+16y^{2})^2}[/tex]

please, someone help. It's a bit of an emergency.

Why do you care about ##k_g##? It has nothing at all to do with the question as originally posed. Also: knowing the form of ##k_f## should allow you to simplify the problem significantly, to get a problem that is much easier than the original one. I cannot say more, without essentially doing the problem for you
 
  • #5
Ray Vickson said:
Why do you care about ##k_g##? It has nothing at all to do with the question as originally posed. Also: knowing the form of ##k_f## should allow you to simplify the problem significantly, to get a problem that is much easier than the original one. I cannot say more, without essentially doing the problem for you
ok, then the Lagrange equation we are interested in is:

[tex]L(x,y, \lambda)=\frac{4}{(1+4x^{2}+4y^{2})^2}- \lambda(3x^{2} + 2y^{2}-1)[/tex]
what do you mean by form of k_f(x,y)? could you come up with a mini-alternative example to what you mean, I just don't see how you can simplify it, unless this has something unique to do with the problem as opposed to general algebraic manipulation.
 
  • #6
ilyas.h said:
ok, then the Lagrange equation we are interested in is:

[tex]L(x,y, \lambda)=\frac{4}{(1+4x^{2}+4y^{2})^2}- \lambda(3x^{2} + 2y^{2}-1)[/tex]
what do you mean by form of k_f(x,y)? could you come up with a mini-alternative example to what you mean, I just don't see how you can simplify it, unless this has something unique to do with the problem as opposed to general algebraic manipulation.

OK, so don't simplify it; just solve it as-is. It does work, but takes a bit more effort.
 
  • #7
Ray Vickson said:
OK, so don't simplify it; just solve it as-is. It does work, but takes a bit more effort.

[itex]\frac{\partial L}{\partial x}= (-64x(1+4x^{2}+4y^{2})^{-3}) - 6x\lambda = 0[/itex]

[itex]\frac{\partial L}{\partial y}= (-64y(1+4x^{2}+4y^{2})^{-3}) - 4y\lambda = 0[/itex]

[itex]\frac{\partial L}{\partial \lambda}= 1 - 3x^{2}-2y^{2}=0[/itex]

this leads to a contradiction though, using the first and second equation we get:

[itex](\frac{-32}{3}(1+4x^{2}+4y^{2})^{-3}) = \lambda = ((-24)(1+4x^{2}+4y^{2})^{-3})[/itex]
 
  • #8
ilyas.h said:
[itex]\frac{\partial L}{\partial x}= (-64x(1+4x^{2}+4y^{2})^{-3}) - 6x\lambda = 0[/itex]

[itex]\frac{\partial L}{\partial y}= (-64y(1+4x^{2}+4y^{2})^{-3}) - 4y\lambda = 0[/itex]

[itex]\frac{\partial L}{\partial \lambda}= 1 - 3x^{2}-2y^{2}=0[/itex]

this leads to a contradiction though, using the first and second equation we get:

[itex](\frac{-32}{3}(1+4x^{2}+4y^{2})^{-3}) = \lambda = ((-24)(1+4x^{2}+4y^{2})^{-3})[/itex]

No contradiction; you have just made some fatal errors.

You can write
[tex] 0= L_x = -Rx - 6x \lambda , \; \text{and} \; 0 = L_y = -Ry - 4 y \lambda [/tex]
where ##R = 64/(1 + 4x^2+4 y^2)^3##. You have attempted to convert ##-Rx - 6 x \lambda = 0## into ##-R - 6 \lambda = 0## (similarly for ##y##), and you are NOT allowed to do that without some extra conditions.
 
  • #9
1. [itex]\frac{\partial L}{\partial x}= (-64x(1+4x^{2}+4y^{2})^{-3}) - 6x\lambda = 0[/itex]

2. [itex]\frac{\partial L}{\partial y}= (-64y(1+4x^{2}+4y^{2})^{-3}) - 4y\lambda = 0[/itex]

3. [itex]\frac{\partial L}{\partial \lambda}= 1 - 3x^{2}-2y^{2}=0[/itex]

if you get an equation for lambda from the first and second equation without dividing through by any variables you get that (using [itex]\lambda[/itex]):

[itex]\frac{32}{3}xy(1+4x^{2}+4y^{2})^{3} = -24xy(1+4x^{2}+4y^{2})^{3}[/itex]

how am I meant to get a relationship between x and y from the above equation that I can utilize in the third equation?
 
  • #10
ilyas.h said:
1. [itex]\frac{\partial L}{\partial x}= (-64x(1+4x^{2}+4y^{2})^{-3}) - 6x\lambda = 0[/itex]

2. [itex]\frac{\partial L}{\partial y}= (-64y(1+4x^{2}+4y^{2})^{-3}) - 4y\lambda = 0[/itex]

3. [itex]\frac{\partial L}{\partial \lambda}= 1 - 3x^{2}-2y^{2}=0[/itex]

if you get an equation for lambda from the first and second equation without dividing through by any variables you get that (using [itex]\lambda[/itex]):

[itex]\frac{32}{3}xy(1+4x^{2}+4y^{2})^{3} = -24xy(1+4x^{2}+4y^{2})^{3}[/itex]

how am I meant to get a relationship between x and y from the above equation that I can utilize in the third equation?

For any real ##x,y## we have ##1 + 4 x^2 + 4 y^2 \geq 1 > 0##, so you are allowed to divide it out on both sides, to end up with
[tex] \frac{32}{3} xy = - 24 xy [/tex]
What does that tell you?
 
Last edited:
  • #11
Ray Vickson said:
For any real ##x,y## we have ##1 + 4 x^2 + 4 y^2 \geq 1 > 0##, so you are allowed to divide it out on both sides, to end up with
[tex] \frac{32}{3} xy = - 24 xy [/tex]
What does that tell you?
that x =y = 0, however this cannot be true due to the third equation.

therefore x and y has to take multiple values that satisfy the following relationships:

[itex]\frac{32}{3}xy = -24xy[/itex]

[itex]1 - 3x^{2}-2y^{2}=0[/itex]

going along the right lines?

and thanks for the help so far.
 
  • #12
ilyas.h said:
that x =y = 0, however this cannot be true due to the third equation.

****************************************************************************
That is not what it tells me. Back to the drawing board.


*****************************************************************************

therefore x and y has to take multiple values that satisfy the following relationships:

[itex]\frac{32}{3}xy = -24xy[/itex]

[itex]1 - 3x^{2}-2y^{2}=0[/itex]

going along the right lines?

and thanks for the help so far.
 
  • #13
Ray Vickson said:
...
ok, so then x must equal y which must equal zero.

Or perhaps my calculations are wrong...
 
  • #14
...
 
  • #15
someone please help me!
 

1. What are Lagrange multipliers and why are they used in optimization?

Lagrange multipliers are a mathematical tool used in optimization problems to find the maximum or minimum value of a function subject to constraints. They are used to convert a constrained optimization problem into an unconstrained one, making it easier to solve.

2. How do Lagrange multipliers work?

Lagrange multipliers work by introducing a new variable, called a multiplier, to the objective function. This new variable is then used to incorporate the constraints into the objective function, creating a new function that can be optimized without constraints.

3. What types of problems can be solved using Lagrange multipliers?

Lagrange multipliers can be used to solve both equality and inequality constrained optimization problems. They are commonly used in fields such as economics, physics, and engineering to find optimal solutions to complex problems.

4. What are the limitations of Lagrange multipliers?

One limitation of Lagrange multipliers is that they can only be used for problems with a single objective function and a set of constraints. Additionally, they may not always provide a unique solution, and multiple local optima may exist.

5. How can I use Lagrange multipliers in my research or work?

If you are working in a field that involves optimization problems, such as economics or engineering, you may encounter situations where Lagrange multipliers can be applied. It is important to have a strong understanding of the concept and its applications to effectively use it in your work.

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