Local extrema and saddle points
Problem 10.93 · easy
Find and classify the critical points of \( \displaystyle f(x, y) = - x^{2} - 2 y^{2} - 8 y - 8 \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(- x^{2} - 2 y^{2} - 8 y - 8\right)\\\frac{\partial}{\partial y} \left(- x^{2} - 2 y^{2} - 8 y - 8\right)\end{matrix}\right] = \left[\begin{matrix}- 2 x\\- 4 y - 8\end{matrix}\right] \]The partial derivatives.✓ Proved
- \[ \left[\begin{matrix}0\\0\end{matrix}\right] \]Both vanish at (0, -2), the only solution.✓ Proved
- \[ 8 \]D = f_xx f_yy − f_xy².✓ Proved
- D = 8 > 0 with f_xx < 0: a local maximum.
Answer \( \text{local maximum at } (0, -2) \)
Lines: 3 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | Not checked | — | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | f on a small circle around the point is compared with its centre value |
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/critical_points_2var, checked 2026-09-26 with SymPy 1.14.0.