Can't solve lagrange multiplier problem I've been working on this particular problem for some time and can't seem to see what I'm doing wrong. The solution provided says the extrema are $\pm 7^{1/2}/8$. If someone could be so kind as to show me where my error lies, it would be much appreciated.
The questions posed is as follows:


Use lagrange multipliers to find the extrema of $f(x,y,z) = x + y + z$ subject to the constraint $x^2 + 2y^2 + 4z^2 = 1/16$.



I've attached a photo of my work 
Thanks so much!!
 A: So you have
$\lambda2x = \lambda 4y = \lambda 8z = 1\\
y = \frac x2\\
z = \frac x4$
$x^2 + 2 y^2 + 4z^2 = \frac 1{16}\\
x^2 + 2 (\frac x2)^2 + 4(\frac x4) = \frac {1}{16}$
And this is where you made your mistake.  Carrying it through.
$x^2 + \frac {x^2}{2} + \frac {x^2}{4} = \frac {1}{16}\\
16x^2 + 8x^2 + 4x^2 = 1\\
28 x^2 = 1\\
x = \frac {1}{2\sqrt{7}}\\
y = \frac {1}{4\sqrt{7}}\\
z = \frac {1}{8\sqrt{7}}$
$x+y+z = \frac {7}{8\sqrt {7}}\\
x+y+z = \frac {\sqrt 7}{8}$
A: Okay so define $g(x,y,z)=x^2+2y^2+4z^2$. We then have that
$$\nabla f=\langle 1,1,1\rangle$$
$$\text{and}$$
$$\nabla g=\langle 2x,4y,8z\rangle$$
So we are then solving the system of equations
$$\begin{array}{c}
2x=\lambda \\
4y=\lambda \\
8z=\lambda \\
x^2+2y^2+4z^2=1/16
\end{array}$$
Substituting $x=\lambda/2$, $y=\lambda/4$, and $z=\lambda/8$ into our constraint equation we have that
$$\lambda^2\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{1}{16}\Rightarrow \lambda=\pm\frac{1}{\sqrt{7}}$$
We then have have the coordinates of two possible extrema:
$$\pm\left(\frac{1}{2\sqrt{7}},\frac{1}{4\sqrt{7}},\frac{1}{8\sqrt{7}}\right)$$
Evaluating $f(x,y,z)$ at these extrema gives
$$f_{\text{extrema}}=\pm\left(\frac{1}{2\sqrt{7}}+\frac{1}{4\sqrt{7}}+\frac{1}{8\sqrt{7}}\right)=\pm\frac{7}{8\sqrt{7}}=\pm\frac{\sqrt{7}}{8}$$
