Answer:
We want to find the maximum value of the surface \(f(x, y) = 49 - x^2 - y^2\) subject to the constraint \(x + y = 3\). We can use the method of Lagrange multipliers to solve the problem. Let
$$g(x,y) = x+y-3,$$
and consider the function
$$F(x,y,\lambda) = f(x,y) - \lambda g(x,y) = 49 - x^2 - y^2 - \lambda(x+y-3).$$
Then, we need to find the critical points of \(F(x,y,\lambda)\), which satisfy the following system of equations:
\begin{align}
\frac{\partial F}{\partial x} &= -2x - \lambda = 0, \\
\frac{\partial F}{\partial y} &= -2y - \lambda = 0, \\
\frac{\partial F}{\partial \lambda} &= x + y - 3 = 0.
\end{align}
The first two equations yield that \(x = -\frac{\lambda}{2}\) and \(y = -\frac{\lambda}{2}\). Substituting these into the third equation, we get \(-\lambda + (-\lambda) - 3 = 0\), which implies that \(\lambda = -\frac{3}{2}\). Thus, the critical point is
$$(x,y) = \left(\frac{3}{2}, \frac{3}{2}\right).$$
We also need to check the endpoints of the line segment. When \(x = 0\), we have \(y = 3\), and when \(y = 0\), we have \(x = 3\). We evaluate the function \(f(x,y)\) at these three points:
\begin{align*}
f(0,3) &= 40, \\
f(3,0) &= 40, \\
f\left(\frac{3}{2}, \frac{3}{2}\right) &= 42.25 - \frac{27}{4} = \frac{11}{4}.
\end{align*}
Therefore, the maximum value of the surface \(f(x, y) = 49 - x^2 - y^2\) on the line \(x + y = 3\) is \(\boxed{\frac{11}{4}}\), which occurs at the point \(\left(\frac{3}{2}, \frac{3}{2}\right)\).