= Solution
{c}
The constraint intersection is nonempty, closed and bounded, hence <compact>, so extrema exist. The two constraint gradients are $(x/2,2y,2z)$ and $(1,1,-1)$. If the first were $t$ times the second, the plane condition would give $2t+t/2+t/2=0$, so $t=0$ and all coordinates would vanish, contradicting the ellipsoid equation. Their <rank> is therefore two at every feasible point, justifying the <Lagrange multiplier> equations.
These equations are
$$
(1,2,0)=\lambda(x/2,2y,2z)+\mu(1,1,-1),\quad z=x+y.
$$
Eliminating $\mu$ gives $1=\lambda(5x/2+2y)$ and $2=\lambda(2x+4y)$. The first equation makes $\lambda\ne0$; subtracting twice the first from the second forces $x=0$. Hence $z=y$, and the ellipsoid gives $y=z=\pm1/\sqrt2$. Thus
$$
\boxed{g_{\max}=\sqrt2\text{ at }(0,1/\sqrt2,1/\sqrt2),\qquad
g_{\min}=-\sqrt2\text{ at }(0,-1/\sqrt2,-1/\sqrt2).}
$$
All candidates were found, and <compactness> ensures they include the extrema.
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