# Lagrange multipliers, guidance needed

1. Mar 6, 2015

### ilyas.h

1. The problem statement, all variables and given/known data

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

consider k_f:

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.

3. The attempt at a solution

is this the langrange equation that I'd use?

$$L(x,y,\lambda )=k_{f}(x,y)-\lambda k_{g}(x,y)$$

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

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.

2. Mar 7, 2015

### Ray Vickson

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. Mar 7, 2015

### ilyas.h

my k_f(x,y) is:

$$\frac{4}{(1+4x^{2}+4y^{2})^2}$$

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:

$$\frac{24}{(1+36x^{2}+16y^{2})^2}$$

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

Last edited: Mar 7, 2015
4. Mar 7, 2015

### Ray Vickson

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. Mar 7, 2015

### ilyas.h

ok, then the Lagrange equation we are interested in is:

$$L(x,y, \lambda)=\frac{4}{(1+4x^{2}+4y^{2})^2}- \lambda(3x^{2} + 2y^{2}-1)$$
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. Mar 7, 2015

### Ray Vickson

OK, so don't simplify it; just solve it as-is. It does work, but takes a bit more effort.

7. Mar 8, 2015

### ilyas.h

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

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

$\frac{\partial L}{\partial \lambda}= 1 - 3x^{2}-2y^{2}=0$

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

$(\frac{-32}{3}(1+4x^{2}+4y^{2})^{-3}) = \lambda = ((-24)(1+4x^{2}+4y^{2})^{-3})$

8. Mar 8, 2015

### Ray Vickson

You can write
$$0= L_x = -Rx - 6x \lambda , \; \text{and} \; 0 = L_y = -Ry - 4 y \lambda$$
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. Mar 8, 2015

### ilyas.h

1. $\frac{\partial L}{\partial x}= (-64x(1+4x^{2}+4y^{2})^{-3}) - 6x\lambda = 0$

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

3. $\frac{\partial L}{\partial \lambda}= 1 - 3x^{2}-2y^{2}=0$

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

$\frac{32}{3}xy(1+4x^{2}+4y^{2})^{3} = -24xy(1+4x^{2}+4y^{2})^{3}$

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. Mar 8, 2015

### Ray Vickson

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?

Last edited: Mar 8, 2015
11. Mar 8, 2015

### ilyas.h

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:

$\frac{32}{3}xy = -24xy$

$1 - 3x^{2}-2y^{2}=0$

going along the right lines?

and thanks for the help so far.

12. Mar 8, 2015

### Ray Vickson

13. Mar 8, 2015

### ilyas.h

ok, so then x must equal y which must equal zero.

Or perhaps my calculations are wrong...

14. Mar 8, 2015

### ilyas.h

...............

15. Mar 8, 2015