
\vspace*{.3in}
\noindent{\bf Supplementary problem 1: }
Since we'll need it in all parts of the problem, let us first note that
$$
f'(x,y,z,w) = 
\left(\begin{array}{cccc}
2x & -1 & 1 & 0 \\
-1 & 2y & 0 & 1
\end{array}\right).
$$
\begin{itemize}
\item In particular, 
$$
f'(1,1,0,0) = \left(\begin{array}{cccc}
2 & -1 & 1 & 0 \\
-1 & 2 & 0 & 1
\end{array}\right).
$$
Since the leftmost $2\times 2$ submatrix of this matrix is invertible
(its determinant is $3$), the implicit function theorem guarantees us
neighborhoods $U\ni (1,1)$, $V\ni (0,0)$ and a $C^1$ function 
$g:V\to U$ such that
$$
\{(x,y,z,w)\in V\times U: f(x,y,z,w) = (0,0)\} =
\{(x,y,z,w)\in V\times U: (x,y) = g(z,w)\}.
$$
That is, we can solve $f(x,y,z,w) = (0,0)$ locally near $(1,1,0,0)$
for $x$ and $y$ in terms of $z$ and $w$ by setting $(x,y) = g(z,w)$.

\item The obvious initial guess for a solution $(x,y)$ of
$f(x,y,.1,-.2)=(0,0)$ is $(x_0,y_0) = (1,1)$.  So let us improve this
guess by replacing $f$ by its linear approximation about $(1,1,.1,-.2)$
setting it equal to zero and solving for $x$ and $y$ while $(z,w)$ is
fixed at $(.1,-.2)$: i.e.
\begin{eqnarray*}
\left(\begin{array}{c}
0\\
0
\end{array}\right)
& = & 
f(1,1,.1,-.2) + f'(1,1,.1,-.2)
\left(\begin{array}{c}
x_1-1\\
y_1-1\\
0 \\
0
\end{array}\right) \\
& = & 
\left(\begin{array}{c}
.1\\
-.2
\end{array}\right)
+
\left(\begin{array}{cccc}
2 & -1 & 1 & 0 \\
-1 & 2 & 0 & 1
\end{array}\right)\left(\begin{array}{c}
x_1-1\\
y_1-1\\
0 \\
0
\end{array}\right) \\
& = &
\left(\begin{array}{c}
2x_1 - y_1 - .9\\
2y_1 - x_1 - 1.2
\end{array}\right)
\end{eqnarray*}
which implies that $(x_1,y_1) = (1,1.1)$.  Note that this actually does
improve on our starting guess: we had $f(x_0,y_0,.1,-.2) = (.1,-.2)$
whereas $f(x_1,y_1,.1,-.2) = (0,.01)$.

\item
To get an even better guess $(x_2,y_2)$, I repeat the above, using the
linear approximation of $f$ about $(x_1,y_1,.1,-.2)$.
\begin{eqnarray*}
\left(\begin{array}{c}
0\\
0
\end{array}\right)
& = & 
f(1,1.1,.1,-.2) + f'(1,1.1,.1,-.2)
\left(\begin{array}{c}
x_2-1\\
y_2-1.1\\
0 \\
0
\end{array}\right) \\
& = & 
\left(\begin{array}{c}
0\\
.01
\end{array}\right)
+
\left(\begin{array}{cccc}
2 & -1 & 1 & 0 \\
-1 & 2.2 & 0 & 1
\end{array}\right)
\left(\begin{array}{c}
x_2-1\\
y_2-1.1\\
0 \\
0
\end{array}\right) \\
& = &
\left(\begin{array}{c}
2x_2 - y_2 - .9\\
2.2y_2 - x_2 - 1.41
\end{array}\right)
\end{eqnarray*}
which implies that $(x_2,y_2) = (.9970588\dots,1.0941176\dots)$.
Plugging this guess into $f$ gives 
$$
f(x_2,y_2,.1,-.2) \approx (8.7\times 10^{-6},3.5\times 10^{-5}).
$$
Not bad, if I do say so.
\end{itemize}