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13.7.pdfMaximum and Minimum Values In this
13.7.pdfMaximum and Minimum Values In this
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13.7.pdfMaximum and Minimum Values In this
Page 3
Figure 2: If a relative maximum of
f
occurs at (
a, b
), then the ﬁrst partial derivatives
f
x
(
a, b
) and
f
y
(
a, b
) are both zero.
Figure 3(a) shows the graph of a function
f
with a relative maximum at a point (
a, b
) lying inside the
domain of
f
. As you can see, the tangent plane to the surface
z
=
f
(
x, y
) at the point (
a, b, f
(
a, b
))
is horizontal. Next, Figure 3(b) shows the graph of a function with a relative maximum at a point
(
a, b
). Note that both
f
x
(
a, b
) and
f
y
(
a, b
) do not exist because the surface
z
=
f
(
x, y
) has a point
(
a, b, f
(
a, b
)) that looks like a jagged mountain peak.
All of these points are
critical points
of a
function of two variables.
Figure 3: At a relative extremum of
f
, either
f
x
=
f
y
= 0 or one or both partial derivatives do not
exist.
Deﬁnition 2: Critical Points of a Function
Let
f
be a function of two variables deﬁned on an open region containing the point (
a, b
).
We call (
a, b
)a
critical point
of
f
if
(a)
f
x
and/or
f
y
do not exist at (
a, b
), or
(b) both
f
x
(
a, b
) = 0 and
f
y
(
a, b
) = 0.
Remark:
Theorem 1 tells us that the relative extremum of a function
f
deﬁned on an open region
can occur only at a critical point of
f
. That is, the critical points of
f
are
candidates
for relative
extrema.
3
Page 4
Example 1.
Let
f
(
x, y
)=
x
2
+
y
2

4
x

6
y
+ 17. Find the critical point of
f
, and show that
f
has a relative minimum at that point.
Figure 4 shows the graph of the two functions studied in Example 1.
Figure 4: The graph of
f
(
x, y
)=
x
2
+
y
2

4
x

6
y
+ 17 in Example 1.
4
Page 5
As in the case of a function of a single variable, a critical point of a function of two variables is
only a candidate for a relative extremum of the function. A critical point need not give rise to a
relative extremum, as the following example shows.
Example 2.
Show that the point (0
,
0) is a critical point of
f
(
x, y
)=
y
2

x
2
but that it does not
give rise to a relative extremum of
f
.
Proof.
The partial derivatives of
f
,
f
x
(
x, y
)=

2
x
and
f
y
(
x, y
)=2
y,
are continuous everywhere. Since
f
x
and
f
y
are both equal to zero at (0
,
0), we conclude that (0
,
0)
is a critical point of
f
and that it is the only candidate for a relative extremum of
f
.
But notice that for points on the
x
axis, we have
y
= 0, so
f
(
x, y
)=

x
2
<
0 if
x
6
= 0; and for
points on the
y
axis, we have
x
= 0, so
f
(
x, y
)=
y
2
>
0 if
y
6
= 0. Therefore, every open disk in
the
xy
plane centered at (0
,
0) contains points where the function is positive and points where it
is negative. The point (0
,
0
,
0) is called a
saddle point
. The graph of
f
is shown in Figure 5(a).
Figure 5(b) displays the level curves (they are hyperbolas) of
f
, and shows the function decreasing
and increasing in an alternating fashion among the four groupings of hyperbolas.
Figure 5: (a) The origin is a saddle point of the function
f
(
x, y
)=
y
2

x
2
. There are no relative
extreme values (Example 2). (b) Level curves for the function
f
in Example 2.
Deﬁnition 3: Saddle Point of a Function
A diﬀerentiable function
f
(
x, y
) has a
saddle point
at a critical point (
a, b
) if in every open
disk centered at (
a, b
) there are domain points (
x, y
) where
f
(
x, y
)
>f
(
a, b
) and domain
points (
x, y
) where
f
(
x, y
)
<f
(
a, b
). The corresponding point (
a, b, f
(
a, b
)) on the surface
z
=
f
(
x, y
) is called a saddle point of the surface.
5
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