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Previous: Taylor's formula.
Definition 6.1
Let
f be a function of two variables
x and
y defined on an open domain
in the plane. A point (
x_{0},
y_{0}) is a
critical point for
f if
f is differentiable at (
x_{0},
y_{0}) and if
f_{x}(
x_{0},
y_{0}) =
f_{y}(
x_{0},
y_{0})=0.
Example 6.2
Let
f(
x,
y)=
xy. Then
f_{x}=
y and
f_{y}=
x. We have:
The function
f has a single critical point, namely the origin.
Example 6.3
Let
f(
x,
y)=
x^{2}y+
x
y. Then
f_{x}=2
xy+1 and
f_{y}=
x^{2}1. We have:
This system of equations has two solutions, i.e. the function
f has two critical points, namely (1,1/2) and (1, 1/2).
Theorem 6.4 (second derivative test)
Let
f be a function of two variables
x and
y defined on an open domain
in the plane. Suppose that (
x_{0},
y_{0}) is a critical point for
f and that
f has second order partial derivatives at
((
x_{0},
y_{0}). Then:
 1.
 If
f_{xx} < 0 and
f_{xx}f_{yy}f_{xy}^{2} >0 at
((x_{0},y_{0}), then f has a maximum at
((x_{0},y_{0}).
 2.
 If
f_{xx} > 0 and
f_{xx}f_{yy}f_{xy}^{2} >0 at
((x_{0},y_{0}), then f has a minimum at
((x_{0},y_{0}).
 3.
 If
f_{xx}f_{yy}f_{xy}^{2} < 0 at
((x_{0},y_{0}), then f has a saddle point at
((x_{0},y_{0}).
This theorem is proved by use of Taylor's formula (see section 1).
Example 6.5
Take
f(
xy)=
xy, as in Example
6.2. We have:
f_{xx}= 0,
f_{xy}=1 and
f_{yy}=0. Thus
f_{xx}f_{yy}
f_{xy}^{2} = 1 <0 and the function
f has a saddle point at the origin (see Figure
1(a)).
Figure 1:
Surfaces with a saddle point.

Example 6.6
Let
f(
x,
y)=
x^{2}+
y^{2}. Then
f_{x}=2
x and
f_{y}=2
y and the function has a single critical point (0,0). Now
f_{xx}= 2,
f_{xy}=0 and
f_{yy}=2. The second derivative test (i.e. Thm
6.4) gives:
f_{xx}f_{yy}
f_{xy}^{2} =4 >0 and
f_{xx}=2 >0, i.e. the function
f has a minimum at (0,0) (see Fig.
2).
Figure 2:
The graph of a function with a minimum.

Example 6.7
Let
f(
x,
y)=
x^{2}y^{3}. Then
f_{x}=2
xy^{3} and
f_{y}=3
x^{2}y^{2}. The function has a critical point at (0,0), but the second derivative test is unconclusive here, as
f_{xx}=
y^{3},
f_{xy}=6
xy^{2} and
f_{yy}=6
x^{2}y, and all three are equal to 0 at (0,0) (see Figure
1(b)). Only ``hand work'' can bring us a conclusion (which conclusion?).
Example 6.8
Let
f(
x,
y)=
x^{3}+
xy+
y^{3}. You can check that the function has a saddle point at the origin and a local maximum at (1,1), equal to 1. See Figure
3.
Figure 3:
z=x^{3}+xy+y^{3}.

Next: Directional derivative and Gradient
Up: Applications of the derivative.
Previous: Taylor's formula.
Noah DanaPicard
20010530