Find maximum point of $f(x,y,z) = 8x^2 +4yz -16z +600$ with one restriction I need to find the critical points of $$f(x,y,z) = 8x^2 +4yz -16z +600$$ restricted by $4x^2+y^2+4z^2=16$.
I constructed the lagrangian function $$L(x, y, z, \lambda ) = 8x^2 +4yz -16z +600 - \lambda (4x^2+y^2+4z^2-16) $$
but I'm very confused about how to determine those points. I know I need to make a system with all the first derivatives of $L$ equaled to $0$. I did it but every time I try to solve it I get different solutions. How can I get the points?
Thanks.

Using Lagrange Multipliers, we wish to find the points $(x, y, z)$ such that $\nabla f(x, y, z) = \lambda \nabla g(x, y, z),$ where $g(x, y, z) = 4x^2 + y^2 + 4z^2 = 16$ and $\lambda$ is some constant. Observe that the gradients are given by $\nabla f = \langle f_x, f_y, f_z \rangle = \langle 16x, 4z, 4y - 16 \rangle$ and $\nabla g = \langle 8x, 2y, 8z \rangle,$ hence we must solve the following $4 \times 4$ system of equations. $$\begin{cases} 16x = 8 \lambda x \\ 4z = 2 \lambda y \\ 4y - 16 = 8 \lambda z \\ 4x^2 + y^2 + 4z^2 = 16\end{cases}$$ Using the first equation, we have that $(16 - 8 \lambda)x = 0,$ from which it follows that $\lambda = 2$ or $x = 0.$ Using the second equation, we have that $2z = \lambda y$ so that $4z^2 = \lambda^2 y^2.$ Using the third equation, we have that $4y = 8 \lambda z + 16 = 4 \lambda^2 y + 16$ (by the second equation) so that $(4 \lambda^2 - 4)y + 16 = 0.$ Using the fourth equation, we have that $4x^2 + (\lambda^2 + 1) y^2 = 16$ (by the second equation). $$\begin{cases} 4z^2 = \lambda^2 y^2 \\ (4 \lambda^2 - 4)y + 16 = 0 \\ 4x^2 + (\lambda^2 + 1) y^2 = 16 \end{cases}$$
Given that $\lambda = 2,$ we may solve for $y$ in the second equation; solve for $z$ in the first equation; and solve for $x$ in the third equation. On the other hand, if $x = 0,$ then we are dealing with the functions $f(0, y, z) = 4yz - 16z + 600 = h(y, z)$ and $g(0, y, z) = y^2 + 4z^2 = k(y, z) = 16,$ and we can use Lagrange Multipliers to find the critical points of $h(y, z)$ with respect to the constraint $k(y, z) = 16.$ Explicitly, we have that $\langle 4z, 4y - 16 \rangle = \nabla h(y, z) = \mu \nabla k(y, z) = \mu \langle 2y, 8z \rangle$ so that $$\begin{cases} 4z = 2 \mu y \\ 4y - 16 = 8 \mu z \end{cases}$$ is the relevant system of equations. Considering that $\mu$ is nonzero, we can eliminate it by taking $$16 \mu z^2 = (2z)(8 \mu z) = (2z)(4y - 16) = (\mu y)(4y - 16) = \mu (4y^2 - 16y)$$ and cancelling a factor of $\mu$ from the left- and right-hand sides. We have therefore that $16z^2 = 4y^2 - 16y.$ Use the fact that $16z^2 = 4(16 - y^2)$ to solve for $y.$
