Lagrange Multipler and Max/Min point of intersection

1. Oct 11, 2009

calorimetry

1. The problem statement, all variables and given/known data
The plane 4x − 3y + 8z = 5 intersects the cone z^2 = x^2 + y^2 in an ellipse.
Use LaGrange Multipliers to find the highest and lowest points on the ellipse.

2. Relevant equations
Lagrange Multiplier

3. The attempt at a solution
I guess I lack an understanding of Lagrange multiplier to begin solving this problem. Normally, I would be given some kinda of function express in term of x, y, and z and the plane would be the constraint for the function. Then I would just solve using Lagrange multiplier.

Here what is my function and what is my constrain?

I was thinking of using the cone as my function, but then I would have 0 = x^2 + y^2 - z^2
which is a surface and not a function of x, y, and z or is it?
And the plane is the constrain, but then I wouldn't be just looking for the max and min of the ellipse intersection anymore, it would be the max and min of the cone with the plane as the constrain.

I'm guessing both the plane and the cone are constrains in this problem because their intersection is the actual constrain, then what is my function to maximize/minimize?

Thanks in advance for any help/explanation.

2. Oct 12, 2009

lanedance

how about trying the following (pretty much taken from what you described)

maximise
f(x,y,z) = z
subject to the constraints
g(x,y,z) = 4x − 3y + 8z - 5 = 0
h(x,y,z) = x^2 + y^2 - z^2 = 0

then use two lagrange multipliers...

$$F(x,y,z) = f + \lambda.g + \gamma.h$$
then
$$\nablaF(x,y,z) = \nabla f + \lambda.\nabla g + \gamma.\nabla h$$

gives 5 unknowns & 5 equations, though little messy not sure if there's an easier way, maybe using the z in the plane equation...

Last edited: Oct 12, 2009
3. Oct 12, 2009

calorimetry

Thanks lanedance, this is what I was thinking, but I don't know why I didn't realize that the function I am trying to maximize is f(x, y, z)=z. It should work.

4. Oct 13, 2009

xiseeux

I don't get it... if you use f(x,y,z)=z to maximize, wouldn't you get fx(x,y,z)=0 and fy(x,y,z)=0... Then would you set fx(x,y,z) = fy(x,y,z)?

5. Oct 13, 2009

lanedance

true that fx(x,y,z)=0 and fy(x,y,z)=0, but then you use the lagrange multiplier method to get a set of equations to solve from
$$\nablaF(x,y,z) = \nabla f + \lambda.\nabla g + \gamma.\nabla h = 0$$
and the constraints

6. Oct 13, 2009

HallsofIvy

You would if there were no constraints. And then, because the derivatives are always 0, you would get no "max" or "min"- f(z)= z has no maximum or minimum values.

But here the point is constrained to lie on two given figures. And you use "Lagrange multipliers" for problems like that.

If you want to maximize/minimize a function f(x,y,z) subject to the constraint g(x,y,z)= 0 then you must have $\nabla f= \lambda \nabla g$ for some "multiplier" $\lambda$.