# Thread: Lagrange multipliers

1. ## Lagrange multipliers

I am suppose to calculate the max and min of f(x,y,z) = x^2 - 2y + 2z^2 with the constraint function of x^2 + y^2 + z^2 = 1

I am to use Lagrange multipliers for this. When I do I get F sub y equal to -2 so how can this be set equal to zero!!! I am so confused. I also get lambda = 1 and = - y I am suppose to get max of function at 4 and min at -2 but I am really confused here.

2. Originally Posted by Frostking
I am suppose to calculate the max and min of f(x,y,z) = x^2 - 2y + 2z^2 with the constraint function of x^2 + y^2 + z^2 = 1

I am to use Lagrange multipliers for this. When I do I get F sub y equal to -2 so how can this be set equal to zero!!! I am so confused. I also get lambda = 1 and = - y I am suppose to get max of function at 4 and min at -2 but I am really confused here.
Define $F(x,y,z,\lambda)=x^2-2y+2z^2-\lambda (x^2+y^2+z^2-1)$

thus $F_y=-2-2 \lambda y$

take the successive derivatives of F and make them equal 0.

3. ## Lagrange mulitplier problem

I have f sub x as 2x - lambda 2x = 0
f sub z as 4z - lambda 2 z = 0

So I think since I get lambda = -y and also to 1 that y is -1 and since a squared is involved it could be 1 also. BUt when I plug these values in to original equation with x and z = to zero I get max of function at 2 and min at -2 whereas the answer says it should be 4 and -2

Can you advise as to my error???? FrostKing

4. Originally Posted by Frostking

I have f sub x as 2x - lambda 2x = 0
f sub z as 4z - lambda 2 z = 0

So I think since I get lambda = -y and also to 1 that y is -1 and since a squared is involved it could be 1 also. BUt when I plug these values in to original equation with x and z = to zero I get max of function at 2 and min at -2 whereas the answer says it should be 4 and -2

Can you advise as to my error???? FrostKing
I think it is more complicated. And it's y=-1/lambda

You have the following system :
$\left\{\begin{array}{llll} 2x-2\lambda x=0 \\ -2-2\lambda y=0 \\ 4z-2 \lambda z=0 \\ x^2+y^2+z^2-1=0 \end{array} \right.$

$\left\{\begin{array}{llll} x(1-\lambda)=0 \\ y=-1/\lambda \\ z(2-\lambda)=0 \\ x^2+y^2+z^2-1=0 \end{array} \right.$

So from the first one, x=0 or lambda=1.
- If x=0, then from the third equation, we have z=0 or lambda=2
• if z=0, the fourth one will give y^2=1, so y=1 or -1. f(0,1,0)=-2 and f(0,-1,0)=2
• if lambda=2, y=-1/2. From the fourth one, you'll get z=...

- If lambda=1, y=-1. And from the third equation, we have z=0. The fourth equation will give you x=...

can you see the reasoning ?

5. ## Lagrange mulitplier problem comments

So in answer to your question from the fourth equation z = + or 1 0.866 since its square must be 0.75 so the equation is 1=1

And x = 0 for your second question, I think. I did not realize that lambda could be equal to one number and in the same set of equations that it was able to equal something else..Thanks much for your time! Frostking