Lagrange multiplier - find maximum of $F(x,y,z) = 2x+y-z$ on the surface: $4x^2 + 2y^2 + z^2 = 40$

Question

I need someone to walk me through a 3 variable lagrange problem, since I haven't been able to find a reliable source to teach me, please.

Here it is:

Find the maximum of the function $F(x,y,z) = 2x+y-z$ on the surface: $4x^2 + 2y^2 + z^2 = 40$

Answer 1

This is straightforward: your Lagrangian is $$ L(x,y,z,\lambda) = 2x+y-z -\lambda(4x^2 + 2y^2+z^2 - 40). $$ Now just write that all the partial derivatives are 0, and you are done:

$$ 0 = \frac{\partial L}{\partial x} = 2-8\lambda x\implies \lambda = \frac 1{4x}; \\ 0 = \frac{\partial L}{\partial y} = 1-4\lambda y\implies \lambda = \frac 1{4y}; \\ 0 = \frac{\partial L}{\partial z} = -1-2\lambda z\implies \lambda = -\frac 1{2z}; $$

Hence $$z=-2x=-2y. \\ 40 = 4x^2 + 2y^2 + z^2 = 4x^2 + 2x^2 + 4x^2 = 10 x^2; \\ x=y=2; z=-4\text{ or } x=y=-2; z=4.$$ The maximum is for the first solution ($L$ is bigger).

You can check with Wolfram Alpha.