If I understand well, you want to use the Lagrange multiplier method right. you should notice that the constraint in this methods has to be an equation, not an inequality like here (unless there are some new variation which I don't know). But still i don't understand how did you get the first 3 equations
As I said, you can still solve the problem geometrically. Think of a circle and a line which distance from the origin of you axes in is parametrized the [tex]c[\math]
As Tah said, you cannot use the Lagrange multiplier method on an inequality constraint like this.
What you can do is find x, y, z such that . IF those are inside the ball, that is, if then that might be a max or min. Here, of course, f(x,y,z)= x+ 2y- z which is linear so which is never 0. That means the max and min values must be ON the sphere, not inside it.
Now, there are two things you could do: use the Lagrange multiplier method on as you have done. Yes, that gives , , and . One way I like to handle equations like these, since we really don't want to know is to divide one equation by another. Since and , y/x= 2 and y= 2x. Since, also, , z/x= -1 so z= -x. Put y= 2x and z= -x into and solve for x.
The other thing you could do is write f(x,y,z) ON the surface of the sphere. In spherical coordinates, , , . On the surface of this sphere, so , , . f(x,y,z)= x+ 2y- z= . Now minimize that.