Local Extrema Finder — Theory
1. Local extrema in two variables
For a two-variable function \(f(x,y)\), a local maximum occurs when nearby function values are smaller.
A local minimum occurs when nearby function values are larger.
\[
f(a,b)\ge f(x,y)
\quad\text{near }(a,b)
\]
means \((a,b)\) is a local maximum, while
\[
f(a,b)\le f(x,y)
\quad\text{near }(a,b)
\]
means \((a,b)\) is a local minimum.
2. Critical point condition
Interior local extrema usually occur at critical points. A critical point satisfies
\[
f_x(a,b)=0,
\qquad
f_y(a,b)=0.
\]
These equations mean the tangent plane is horizontal at the point.
3. Hessian matrix
The Hessian matrix contains the second partial derivatives of the function:
\[
H(x,y)
=
\begin{pmatrix}
f_{xx} & f_{xy}\\
f_{yx} & f_{yy}
\end{pmatrix}.
\]
The Hessian describes the local curvature of the surface near a critical point.
4. Hessian determinant
For the two-variable second derivative test, compute
\[
D
=
f_{xx}f_{yy}-f_{xy}^{2}.
\]
This value is the determinant of the Hessian matrix when \(f_{xy}=f_{yx}\).
5. Second derivative test
At a critical point \((a,b)\), evaluate \(D\) and \(f_{xx}\).
6. Saddle points
A saddle point is a critical point that is not a local maximum or local minimum.
The surface increases in some directions and decreases in others.
\[
f(x,y)=x^2-y^2
\]
has a saddle point at \((0,0)\).
7. Example: local minimum
Consider
\[
f(x,y)=x^2+y^2-4x-6y.
\]
First compute the partial derivatives:
\[
f_x=2x-4,
\qquad
f_y=2y-6.
\]
Set them equal to zero:
\[
2x-4=0,
\qquad
2y-6=0.
\]
Therefore,
\[
x=2,
\qquad
y=3.
\]
The critical point is \((2,3)\).
8. Classifying the example
Now compute the second partial derivatives:
\[
f_{xx}=2,
\qquad
f_{yy}=2,
\qquad
f_{xy}=0.
\]
Then
\[
D
=
f_{xx}f_{yy}-f_{xy}^{2}
=
2\cdot 2-0^2
=
4.
\]
Since \(D>0\) and \(f_{xx}>0\), the point \((2,3)\) is a local minimum.
9. Inconclusive cases
When \(D=0\), the second derivative test does not give a classification.
This does not mean there is no maximum or minimum. It only means this test is not enough.
\[
f(x,y)=x^4+y^4
\]
has a local minimum at \((0,0)\), but the Hessian determinant test is inconclusive there.
11. Common mistakes
- Only solving one equation: a critical point must satisfy both \(f_x=0\) and \(f_y=0\).
- Forgetting to evaluate at the point: \(D\) must be computed at the critical point.
- Mixing up the signs: \(D>0\) alone is not enough; \(f_{xx}\) decides max or min.
- Calling \(D=0\) a saddle point: \(D=0\) means the test is inconclusive.
- Ignoring the search region: numerical solvers only find points inside the selected search interval.
