Note first that since $L(x,y,z,k) = −xyz−k(xy+2z(x+y)−50) = L(y,x,z,k)$, the problem is symmetric in $x,y$.
Let's proceed with brutal arithmetic. We have 4 equations in 4 unknowns:
$$
\begin{split}
0 &= -L_x = yz + k(y + 2z)\\
0 &= -L_y = xz + k(x + 2z)\\
0 &= -L_z = xy + 2k(x + y)\\
50 &= xy+2z(x+y) \quad (\text{from } c=0)
\end{split}
$$
Claim 1. $x \neq 0$ and $y \neq 0$ and $x + y \neq 0$.
Proof.
Note first that if $x=0$, then the second constraint implies $k=0$ or $z=0$. If $x=k=0$, the first constraint implies $yz=0$ and the last constraint implies $2yz=50$, which is impossible.
Similarly, if $x=z=0$, the first constraint implies $ky = 0$. As we proved, $k=0$ is impossible, so must be $x=y=z=0$, which contradicts the last constraint.
To establish the last claim, note that if $x+y =0$, the third constraint yields $xy=0$ and the last one yields $xy=50$, which is a contradiction.
QED
Now, solving the last constraint for $z$ and the $L_z$ constraint for $k$, we get
$$k = \frac{-xy}{2(x+y)} \text{ and } z = \frac{50-xy}{2(x+y)}.$$
Both of these are well-defined since $x\neq0$ and $y \neq 0$. Plug both of these into the first constraint, getting
$$0 = \frac{y(50-xy)}{2(x+y)}
- \frac{xy^2}{2(x+y)}
- \frac{2xy(50-xy)}{2^2(x+y)^2}$$
and now multiply both sides by $2(x+y)^2 \neq 0$ to get
$$0 = y(50-xy)(x+y) - xy^2(x+y) - xy(50-xy).$$
Divide by $y \neq 0$ and bring the last term into the first term and divide by $y$ again:
$$
\begin{split}
0 &= (50-xy)(x+y) - xy(x+y) - x(50-xy)\\
0 &= y(50-xy) - xy(x+y) \\
0 &= 50-xy - x(x+y) = 50 - 2xy - x^2\\
50 &= x^2 + 2xy
\end{split}
$$
and because the problem is symmetric, the symmetric constraint must hold (if you want, you can derive it the same way from the second constraint):
$$
\begin{split}
x^2 + 2xy &= 50 \\
y^2 + 2xy &= 50
\end{split}
$$
Hence subtracting them yields $x^2 = y^2$, so $x = \pm y$, but $x \neq -y$ by Claim 1, so $x = y$. Now $50 = x^2 + 2xy = 3x^2$ implies $x = \pm \sqrt{50/3} = y$, and there are two solutions: $(\sqrt{50},\sqrt{50})$ and $(-\sqrt{50},-\sqrt{50})$.
Since $x,y$ are lengths, we know only one will make sense: $x = y = \sqrt{50/3}$. Now plug this back in to find $z$ and $k$ as desired.
Best Answer
I guess the answer is late, but maybe it can help others with the same problem.
When I needed to understand the KKT conditions, this explanation helped me a lot, with a very simple and complete step by step example. If it is still complicated, I recommend reading this chapter from Mathematial Optimization and Economic Analysis (Luptácik, M.; Springer) on the subject carefully. When you master the basics concepts, you can already go to practically any book or handbook on nonlinear optimization.