Discussion
Let \(S = \{(x, y, f(x, y)) \in \mathbf{R}^{3} \mid (x, y) \in U\}\) where U is open in \(\mathbf{R}^{2}\) and \(f : U \to \mathbf{R}\) is smooth. If \(f\) admits a local maximum at \(p\) and a local minimum at \(q\), then the Gaussian curvatures \(k_{1}\) and \(k_{2}\) to S at the points \((p, f(p))\) and \((q, f(q))\) respectively satisfy :
Answer: A
The curvatures satisfy \(k_{1}, k_{2} \ge 0\).
For a graph \(z = f(x, y)\), the Gaussian curvature at a critical point (where \(f_{x} = f_{y} = 0\)) is
\[ K = f_{xx}f_{yy} - f_{xy}^{2} \]
the determinant of the Hessian. At a local maximum the Hessian is negative semi-definite and at a local minimum positive semi-definite; in both cases its two eigenvalues have the same sign (or one is 0), so the determinant is \(\ge 0\). Both surfaces bend the same way in every direction, like a cap or a bowl.
A negative curvature would mean a saddle, which is neither a maximum nor a minimum, so the options with \(\le 0\) are wrong.
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