# Lagrange multiplier with inequality and point constraint?

1. May 2, 2010

### mintygreen

Find an equation of the largest sphere that passes through the point (-1,1,4) and is such that each of the points (x,y,z) inside the sphere satisfies the condition

x^2 + y^2 + z^2 < 136 + 2(x + 2y + 3z)

I know this problem requires Lagrange multipliers. I assume that x^2 + y^2 + z^2 is the f(x,y,z) that I want to maximize and I think that 2x + 4y + 6z = -136 is my g(x,y,z)=k? Beyond that, I can't figure out how to work with the inequality or incorporate the sphere passing through the points... Thanks for any help!

2. May 2, 2010

### gabbagabbahey

Let's start with the basics....what is the equation of a sphere centered at $(a,b,c)$ with radius $r$? If $(-1,1,4)$ lies on that sphere, what can you say about $a$, $b$, $c$ and $r$?

3. May 3, 2010

### mintygreen

Ok, so the forumula is (x-a)^2 + (y-b)^2 + (z-c)^2 = r^2. If (-1,1,4) lies on the sphere, can we say that what r must equal at least what we get for r when we substitute -1,1,and 4 for the values of x,y, and z in this equation?

4. May 3, 2010

### gabbagabbahey

Good.

Sort of.. all you can really say is that $(-1-a)^2+(1-b)^2+(4-c)^2=r^2$...which gives you one constraint.

Now, what quantity are you trying to maximize when you want as large a sphere as possible?

5. May 3, 2010

### mathymath

(-1-a)^2 + (1-b)^2 + (Z-c)^2 = R^2 can be simplified to

a^2 + b^2 +c^2 +2a-2b-8c = r ^2 -18

Since, (a,b,c) lies inside the circle,

a^2 + b^2 +c^2 < 136 +2a+4b+6c

=> a^2 + b^2 +c^2-2a-4b-6c < 136

So, to maximize r don't we need to maximize (a, b, c)????

6. May 4, 2010

### yaffa

how do you maximize a^2 + b^2 +c^2-2a-4b-6c < 136 given a^2 + b^2 +c^2 +2a-2b-8c = r ^2 -18?

7. May 5, 2010

### yaffa

you're trying to maximize the radius, but how do you do that with the inequality?

8. May 6, 2010

### mintygreen

OK, so I guess that solves the problem of incorporating (-1,1,4).

Can we actually say we want to maximize this equation, because we want to maximize the radius of the sphere?
a^2 + b^2 +c^2 +2a-2b-8c = r ^2 -18

Does 2x + 4y + 6z = -136 still work as a constraint, and if so, how do I incorporate the fact that it must be greater than our maximized sphere? Or, can we simply set it equal because the problem specifies that all points INSIDE the sphere satisfies the condition?

9. May 6, 2010

### gabbagabbahey

It's true that (a,b,c) lies inside the sphere, but that certainly won't be the only point that does. In order to get the most information possible out of the inequality constraint, you just leave x, y and z as-is.

And no, you don't need to maximize a,b and c...you only need to find the values of a,b and c that maximize the radius.

First, are you also yaffa, mathymath and/or letsgobuffalo? If so, you will find that using a single account will cause less confusion and get you quicker help (I held off responding further until you posted again), and having multiple accounts is against forum rules.

Anyways, you want to maximize the radius (you might as well maximize $r^2$ to make things easier) of the sphere $r^2=(x-a)^2+(y-b)^2+(z-c)^2$ and you have two constraints:

(1) [tex](-1-a)^2 + (1-b)^2 + (4-c)^2 = r^2=(x-a)^2+(y-b)^2+(z-c)^2[/itex]

(2) $x^2 + y^2 + z^2 < 136 + 2(x + 2y + 3z)$

So, start by defining $f(x,y,z)\equiv r^2=(x-a)^2+(y-b)^2+(z-c)^2$ and then rewrite your constraints so they are in the form $g(x,y,z)=0$ and $h(x,y,z)<0$ and then use the method covered in your text.

10. May 6, 2010

### mintygreen

No, this is my only account. I posted again because I figured you were holding off posting until I responded.

11. May 7, 2010

### mintygreen

First of all, thank you so much for the help. I'm still struggling even though I think I set up the equation correctly, I think its just figuring out how to work everything out algebraically.
f(x,y,z)' = Lg(x,y,z)' + Mh(x,y,z)'
2(x-a)=L2(x-a) + M(2x -2)
2(y-b)=L2(x-b) + M(2y-2)
2(z-c)=L2(z-c) + M(2z-3)
2ax + 2a + 2by -2 - 2cy + 2cz - x^2 - z^2 + 2= 0
x^2 - 2x + y^2 -2y + z^2 -3z <136

I've been trying to work it all out, but there are SO many variables, and nothing is reducing easily...