# Complete the square ( Potentially )?

1. Dec 14, 2012

### Zondrina

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

There was a question on my exam a few days ago. Using Lagrange to find the max/min on a region. We only had to answer a certain amount of questions and I never got to this one. I'm working on it now though out of curiosity.

R = { (x,y) | x2 + xy + y2 ≤ 3 }

f(x,y) = 4x3 - 3xy2 - 4y3

2. Relevant equations

F = f + λg

3. The attempt at a solution

So I first thought we should check the interior of the ellipse, so I did and found (0,0,0) was the only critical obtained.

Then I formed my Lagrange equation and took all the required derivatives as usual. Then I formed my system of equations which actually appears unsolvable due to the large mix of variables per equation.

I'm glad I didn't attempt this one as I'm stumped. I figured I might want to try completing the square on the ellipse, but I'm not sure.

2. Dec 14, 2012

### Dick

You just have to work through it. What did you get for the system of equations? It looks to me like the derivative equations factor pretty nicely. I don't think completing the square will help particularly.

3. Dec 14, 2012

### Zondrina

$12x^2 - 3y^2 + 2xλ + yλ = 0$

$-6xy - 12y^2 + 2yλ + xλ = 0$

$x^2 + xy + y^2 - 3 = 0$

4. Dec 14, 2012

### Dick

(2y+x) is a factor of the first equation, (y+2x) is a factor of the second equation. Factor them out. Then start thinking about possibilities.

5. Dec 14, 2012

### Zondrina

So my first equation factors to (2x + y)(6x - 3y + λ) and my second to (x + 2y)(-6y + λ).

Solving the first gives x = -y/2 or λ = 3(y - 2x) ( Lots of solutions, but whatever right ). I'm pretty sure I can solve this now and then obtain my points to test.

Just wondering, without wolfram how I may have seen this factoring occur. It seems... not so out there? Is there a trick I might not know?

Thank you once again for your help.

6. Dec 14, 2012

### Dick

Nah, probably no trick you don't know. Except to figure if they are going to give you this kind of problem on a test they probably cooked it so it could be done fairly easily. And when you set something equal to zero you always try to factor it. So they probably cooked it to factor.

7. Dec 14, 2012

### haruspex

If you assume it will factor, it's usually not hard to figure out what the factors must be (of quadratics anyway).
In your first equation the y terms are -3y2+λy. The only way that can factor nicely is in the form (... +y)(... -3y+λ). λ only occurs again as 2xλ, we have (2x ... +y)(... -3y+λ). Introducing that 2x also resulted in -6xy, but there's no xy in the equation, so we have to introduce an x term in the other bracket to cancel it: (2x ... +y)(6x ... -3y+λ). Since that produces the desired x2 term, we're done.

8. Dec 14, 2012

### Dick

Another way is to figure if you try to solve each equation for λ you ought to get something not too bad if it's doable. So whatever is multiplying the λ probably divides the other stuff. That's how I guessed it. Assuming it's going to be easy is a good guessing tool.

9. Dec 15, 2012

### HallsofIvy

Staff Emeritus
From $\nabla f+ \lambda\nabla g= 0$, we can get $\nabla f= -\lambda\nabla g$. That gives a series of such equations for the different components. I find it often simplest to eliminate $\lambda$ first (since it is not part of the solution) by dividing one equation by another.
Here those equations are:
$12x^2 −3y^2= -\lambda(2x+y)$
$−6xy−12y^2= -\lambda(2y+x)$
and you can eliminate $\lambda$ by dividing the first equation by the second
$$\frac{12x^2- 3y^2}{-6xy- 12y^2}= \frac{2x+y}{2y+x}$$
$$\frac{4x^2- y^2}{-2xy- 4y^2}= \frac{2x+y}{2y+x}$$
and get rid of the fractions by multiplying both sides by the two denominators
$$(4x^2- y^2)(2y+ x)= (2x+y)(-2x- 4y^2$$
$$4x^3+ 8x^2y- xy^2- 2y^3= -4x^2- 2xy- xy^2- 4y^3$$