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L
ocal
E
xtrema
Previously we have taken the partial derivative of a function
f
(
x
,
y
). But those partial
derivatives were themselves functions and so we can take their partial derivatives.
Local Maximums and Local Minimums of Functions
f
(
x
,
y
) has a local maximum at the point (
x
0
,
y
0
) if
f
(
x
0
,
y
0
)
¥
f
(
x
,
y
) for
all points (
x
,
y
) near (
x
0
,
y
0
)
f
(
x
,
y
) has a local minimum at the point (
x
0
,
y
0
) if
f
(
x
0
,
y
0
)
§
f
(
x
,
y
) for
all points (
x
,
y
) near (
x
0
,
y
0
)
Recall that the gradient vector points in a direction where the function increases.
Suppose that the point (
x
0
,
y
0
) is a local maximum of the function
f
(
x
,
y
) which is not on
the boundary of the domain. If the vector
“
f
(
x
0
,
y
0
) is defined and nonzero, then we can
increase
f
(
x
,
y
) by moving in the direction of
“
f
(
x
0
,
y
0
).
But we just said that the point (
x
0
,
y
0
) is a local maximum, thus there is no direction
that we can travel in to make the function larger. Hence, we must have that
“
f
(
x
0
,
y
0
) =
0
.
Similarly, if (
x
0
,
y
0
) were a local minimum, it would follow that
“
f
(
x
0
,
y
0
) =
0
as well.
Using our results from above, we also have a useful relation between local maxima
and local minima to firstorder partial derivatives.
Local Extrema Theorem
If
f
(
x
,
y
) has a local maximum or a local minimum at (
x
0
,
y
0
) and the
firstorder partial derivatives of
f
(
x
,
y
) exist there, then
f
x
(
x
0
,
y
0
) = 0 and
f
y
(
x
0
,
y
0
) = 0.
This leads us to the definition of a critical point for a function. As is the case with one
variable calculus, we allow for the possibility that a point is neither a local maximum nor
a local minimum.
Critical Points of a Function
A point (
x
0
,
y
0
) is said to be a critical point of the function
f
(
x
,
y
) if the
gradient,
“
f
(
x
0
,
y
0
), is either
0
or undefined.
1
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View Full DocumentExample 1:
Find and analyze the critical point(s) of
f
(
x
,
y
) =
x
2
+
xy
+
y
2
.
Solution:
f
x
(
x
,
y
) = 2
x
+
y
and
f
y
(
x
,
y
) =
x
+ 2
y
. Setting these equations equal to 0, we have two
equations:
2
x
+
y
= 0
and
x
+ 2
y
= 0
We can solve for
x
and
y
by multiplying the second equation by 2 and adding the
equations. Doing this, we have that 3
y
= 0, and so
y
= 0. Plugging that into the first
equation, we see that
x
= 0. Thus, the only critical point is (0, 0).
Notice that
f
(0, 0) = 0. Thus, if
f
(
x
,
y
) is always positive or zero near (0, 0), then we
would have that (0, 0) is a local minimum. If
f
(
x
,
y
) is always negative or zero near (0, 0),
then we would have a local maximum.
Looking at a sketch of the graph (Figure 1 below), it would appear that (0, 0, 0) is a
local minimum.
Figure 1: Sketch of
f
(
x
,
y
) =
x
2
+
xy
+
y
2
We can confirm our suspicion by completing the square on the function. That is, we
have that
2
22
13
(, )
24
2
f xy x x
y y
x
y
y
. Notice that this is a sum of two
squares, so it is always greater than or equal to zero. Thus, the critical point is a local
(and in fact, global) minimum.
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 Spring '07
 Hohnhold
 Calculus, Derivative

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