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PARTIAL DERIVATIVESPARTIAL DERIVATIVES
15
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PARTIAL DERIVATIVES
As we saw in Chapter 4, one of the main
uses of ordinary derivatives is in finding
maximum and minimum values.
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PARTIAL DERIVATIVES
15.7Maximum and
Minimum Values
In this section, we will learn how to:
Use partial derivatives to locate
maxima and minima of functions of two variables.
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MAXIMUM & MINIMUM VALUES
Look at the hills and valleys in the graph
of f shown here.
Fig. 15.7.1, p. 959
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ABSOLUTE MAXIMUM
There are two points (a, b) where f has a local
maximum—that is, where f(a, b) is larger than
nearby values of f(x, y).
The larger of these two values is the absolute maximum.
Fig. 15.7.1, p. 959
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ABSOLUTE MINIMUM
Likewise, f has two local minima—where
f(a, b) is smaller than nearby values.
The smaller of these two values is the absolute minimum.
Fig. 15.7.1, p. 959
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LOCAL MAX. & LOCAL MAX. VAL.
A function of two variables has a local
maximum at (a, b) if f(x, y) ≤ f(a, b) when
(x, y) is near (a, b).
This means that f(x, y) ≤ f(a, b) for all points
(x, y) in some disk with center (a, b).
The number f(a, b) is called a local maximum value.
Definition 1
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LOCAL MIN. & LOCAL MIN. VALUE
If f(x, y) ≥ f(a, b) when (x, y) is near (a, b),
then f has a local minimum at (a, b).
f(a, b) is a local minimum value.
Definition 1
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ABSOLUTE MAXIMUM & MINIMUM
If the inequalities in Definition 1 hold for
all points (x, y) in the domain of f, then f has
an absolute maximum (or absolute minimum)
at (a, b).
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LOCAL MAXIMUM & MINIMUM
If f has a local maximum or minimum at (a, b)
and the first-order partial derivatives of f exist
there, then
fx(a, b) = 0 and fy(a, b) = 0
Theorem 2
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LOCAL MAXIMUM & MINIMUM
Let g(x) = f(x, b).
If f has a local maximum (or minimum) at (a, b), then g has a local maximum (or minimum) at a.
So, g’(a) = 0 by Fermat’s Theorem.
Proof
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LOCAL MAXIMUM & MINIMUM
However, g’(a) = fx(a, b)
See Equation 1 in Section 15.3
So, fx(a, b) = 0.
Proof
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LOCAL MAXIMUM & MINIMUM
Similarly, by applying Fermat’s
Theorem to the function G(y) = f(a, y),
we obtain:
fy(a, b) = 0
Proof
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LOCAL MAXIMUM & MINIMUM
If we put fx(a, b) = 0 and fy(a, b) = 0 in
the equation of a tangent plane (Equation 2
in Section 15.4), we get:
z = z0
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THEOREM 2—GEOMETRIC INTERPRETATION
Thus, the geometric interpretation
of Theorem 2 is:
If the graph of f has a tangent plane at a local maximum or minimum, then the tangent plane must be horizontal.
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CRITICAL POINT
A point (a, b) is called a critical point
(or stationary point) of f if either:
fx(a, b) = 0 and fy(a, b) = 0
One of these partial derivatives does not exist.
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CRITICAL POINTS
Theorem 2 says that, if f has a local
maximum or minimum at (a, b), then
(a, b) is a critical point of f.
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CRITICAL POINTS
However, as in single-variable calculus,
not all critical points give rise to maxima
or minima.
At a critical point, a function could have a local maximum or a local minimum or neither.
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LOCAL MINIMUM
Let f(x, y) = x2 + y2 – 2x – 6y + 14
Then, fx(x, y) = 2x – 2
fy(x, y) = 2y – 6
These partial derivatives are equal to 0 when x = 1 and y = 3.
So, the only critical point is (1, 3).
Example 1
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LOCAL MINIMUM
By completing the square, we find:
f(x, y) = 4 + (x – 1)2 + (y – 3)2
Since (x – 1)2 ≥ 0 and (y – 3)2 ≥ 0, we have f(x, y) ≥ 4 for all values of x and y.
So, f(1, 3) = 4 is a local minimum.
In fact, it is the absolute minimum of f.
Example 1
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LOCAL MINIMUM
This can be confirmed geometrically from
the graph of f, which is the elliptic paraboloid
with vertex (1, 3, 4).
Example 1
Fig. 15.7.2, p. 959
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EXTREME VALUES
Find the extreme values of
f(x, y) = y2 – x2
Since fx = –2x and fy = –2y, the only critical point is (0, 0).
Example 2
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EXTREME VALUES
Notice that, for points on the x-axis,
we have y = 0.
So, f(x, y) = –x2 < 0 (if x ≠ 0).
For points on the y-axis, we have x = 0.
So, f(x, y) = y2 > 0 (if y ≠ 0).
Example 2
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EXTREME VALUES
Thus, every disk with center (0, 0) contains
points where f takes positive values as well
as points where f takes negative values.
So, f(0, 0) = 0 can’t be an extreme value for f.
Hence, f has no extreme value.
Example 2
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MAXIMUM & MINIMUM VALUES
Example 2 illustrates the fact that
a function need not have a maximum or
minimum value at a critical point.
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MAXIMUM & MINIMUM VALUES
The figure shows how this is possible.
The graph of f is the hyperbolic paraboloid z = y2 – x2.
It has
a horizontal tangent plane (z = 0) at the origin.
Fig. 15.7.3, p. 960
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MAXIMUM & MINIMUM VALUES
You can see that f(0, 0) = 0 is:
A maximum in the direction of the x-axis. A minimum in the direction of the y-axis.
Fig. 15.7.3, p. 960
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SADDLE POINT
Near the origin, the graph has the shape
of a saddle. So, (0, 0) is called a saddle point of f.
Fig. 15.7.3, p. 960
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EXTREME VALUE AT CRITICAL POINT
We need to be able to determine whether
or not a function has an extreme value at
a critical point.
The following test is analogous to the Second
Derivative Test for functions of one variable.
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SECOND DERIVATIVES TEST
Suppose that:
The second partial derivatives of f are continuous on a disk with center (a, b).
fx(a, b) = 0 and fy(a, b) = 0 [that is, (a, b) is a critical point of f].
Theorem 3
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SECOND DERIVATIVES TEST
Let D = D(a, b)
= fxx(a, b) fyy(a, b) – [fxy(a, b)]2
a) If D > 0 and fxx(a, b) > 0, f(a, b) is a local minimum.
b) If D > 0 and fxx(a, b) < 0, f(a, b) is a local maximum.
c) If D < 0, f(a, b) is not a local maximum or minimum.
Theorem 3
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SECOND DERIVATIVES TEST
In case c,
The point (a, b) is called a saddle point of f .
The graph of f crosses its tangent plane at (a, b).
Note 1
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SECOND DERIVATIVES TEST
If D = 0, the test gives no
information:
f could have a local maximum or local minimum at (a, b), or (a, b) could be a saddle point of f.
Note 2
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SECOND DERIVATIVES TEST
To remember the formula for D,
it’s helpful to write it as a determinant:
Note 3
2( )xx xy
xx yy xyyx yy
f fD f f f
f f
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SECOND DERIVATIVES TEST
Find the local maximum and minimum
values and saddle points of
f(x, y) = x4 + y4 – 4xy + 1
Example 3
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SECOND DERIVATIVES TEST
We first locate the critical points:
fx = 4x3 – 4y
fy = 4y3 – 4x
Example 3
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SECOND DERIVATIVES TEST
Setting these partial derivatives equal to 0,
we obtain:
x3 – y = 0
y3 – x = 0
To solve these equations, we substitute y = x3 from the first equation into the second one.
Example 3
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SECOND DERIVATIVES TEST
This gives:
Example 3
9
8
4 4
2 2 4
0
( 1)
( 1)( 1)
( 1)( 1)( 1)
x x
x x
x x x
x x x x
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SECOND DERIVATIVES TEST
So, there are three real roots:
x = 0, 1, –1
The three critical points are:
(0, 0), (1, 1), (–1, –1)
Example 3
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SECOND DERIVATIVES TEST
Next, we calculate the second partial
derivatives and D(x, y):
fxx = 12x2 fxy = – 4 fyy = 12y2
D(x, y) = fxx fyy – (fxy)2
= 144x2y2 – 16
Example 3
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SECOND DERIVATIVES TEST
As D(0, 0) = –16 < 0, it follows from case c
of the Second Derivatives Test that the origin
is a saddle point.
That is, f has no local maximum or minimum at (0, 0).
Example 3
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SECOND DERIVATIVES TEST
As D(1, 1) = 128 > 0 and fxx(1, 1) = 12 > 0,
we see from case a of the test that f(1, 1) = –1
is a local minimum.
Similarly, we have D(–1, –1) = 128 > 0
and fxx(–1, –1) = 12 > 0.
So f(–1, –1) = –1 is also a local minimum.
Example 3
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SECOND DERIVATIVES TEST
The graph of f is shown here.
Example 3
Fig. 15.7.4, p. 961
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CONTOUR MAP
A contour map of the function in
Example 3 is shown here.
Fig. 15.7.5, p. 961
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CONTOUR MAP
The level curves near (1, 1) and (–1, –1)
are oval in shape.
They indicate that:
As we move away from (1, 1) or (–1, –1) in any direction, the values of f are increasing.
Fig. 15.7.5, p. 961
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CONTOUR MAP
The level curves near (0, 0) resemble
hyperbolas.
They reveal that:
As we move away from the origin (where the value of f is 1), the values of f decrease in some directions but increase in other directions.
Fig. 15.7.5, p. 961
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CONTOUR MAP
Thus, the map suggests the presence of
the minima and saddle point that we found
in Example 3.
Fig. 15.7.5, p. 961
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MAXIMUM & MINIMUM VALUES
Find and classify the critical points of
the function
f(x, y) = 10x2y – 5x2 – 4y2 – x4 – 2y4
Also, find the highest point on the graph of f.
Example 4
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MAXIMUM & MINIMUM VALUES
The first-order partial derivatives
are:
fx = 20xy – 10x – 4x3
fy = 10x2 – 8y – 8y3
Example 4
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MAXIMUM & MINIMUM VALUES
So, to find the critical points, we need to
solve the equations
2x(10y – 5 – 2x2) = 0
5x2 – 4y – 4y3 = 0
E. g. 4—Eqns. 4 & 5
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MAXIMUM & MINIMUM VALUES
From Equation 4, we see that
either:
x = 0
10y – 5 – 2x2 = 0
Example 4
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MAXIMUM & MINIMUM VALUES
In the first case (x = 0), Equation 5
becomes:
–4y(1 + y2) = 0
So, y = 0, and we have the critical point (0, 0).
Example 4
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53
MAXIMUM & MINIMUM VALUES
In the second case (10y – 5 – 2x2 = 0),
we get:
x2 = 5y – 2.5
Putting this in Equation 5, we have:
25y – 12.5 – 4y – 4y3 = 0
E. g. 4—Equation 6
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54
MAXIMUM & MINIMUM VALUES
So, we have to solve the cubic
equation
4y3 – 21y + 12.5 = 0
E. g. 4—Equation 7
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MAXIMUM & MINIMUM VALUES
Using a graphing calculator or computer
to graph the function
g(y) = 4y3 – 21y + 12.5
we see Equation 7
has three real
roots.
Example 4
Fig. 15.7.6, p. 962
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MAXIMUM & MINIMUM VALUES
Zooming in, we can find the roots to four
decimal places:
y ≈ –2.5452 y ≈ 0.6468 y ≈ 1.8984
Alternatively, we could have used Newton’s method or a rootfinder to locate these roots.
Example 4
Fig. 15.7.6, p. 962
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MAXIMUM & MINIMUM VALUES
From Equation 6, the corresponding x-values
are given by:
If y ≈ –2.5452, x has no corresponding real values.
If y ≈ 0.6468, x ≈ ± 0.8567
If y ≈ 1.8984, x ≈ ± 2.6442
5 2.5x y
Example 4
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MAXIMUM & MINIMUM VALUES
So, we have a total of five critical points,
which are analyzed in the chart.
All quantities are rounded to two decimal places.
Example 4
p. 962
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MAXIMUM & MINIMUM VALUES
These figures give two views of the graph
of f. We see that the surface opens downward.
Example 4
Fig. 15.7.7, p. 962 Fig. 15.7.8, p. 962
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MAXIMUM & MINIMUM VALUES
This can also be seen from the expression
for f(x, y): The dominant terms are –x4 – 2y4
when |x| and |y| are large.
Example 4
Fig. 15.7.7, p. 962 Fig. 15.7.8, p. 962
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MAXIMUM & MINIMUM VALUES
Comparing the values of f at its local
maximum points, we see that the absolute
maximum value of f is:
f(± 2.64, 1.90) ≈ 8.50
That is, the highest points on the graph of f are:
(± 2.64, 1.90, 8.50)
Example 4
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MAXIMUM & MINIMUM VALUES
The five critical points of the function f
in Example 4 are shown in red in this contour
map of f.
Example 4
Fig. 15.7.9, p. 963
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63
MAXIMUM & MINIMUM VALUES
Find the shortest distance from the point
(1, 0, –2) to the plane x + 2y + z = 4.
The distance from any point (x, y, z) to the point (1, 0, –2) is:
Example 5
2 2 2( 1) ( 2)d x y z
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MAXIMUM & MINIMUM VALUES
However, if (x, y, z) lies on the plane
x + 2y + z = 4, then z = 4 – x – 2y.
Thus, we have:
2 2 2( 1) (6 2 )d x y x y
Example 5
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MAXIMUM & MINIMUM VALUES
We can minimize d by minimizing
the simpler expression
2
2 2 2
( , )
( 1) (6 2 )
d f x y
x y x y
Example 5
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66
MAXIMUM & MINIMUM VALUES
By solving the equations
we find that the only critical point is .
2( 1) 2(6 2 ) 4 4 14 0
2 4(6 2 ) 4 10 24 0
x
y
f x x y x y
f y x y x y
5116 3( , )
Example 5
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MAXIMUM & MINIMUM VALUES
Since fxx = 4, fxy = 4, and fyy = 10,
we have:
D(x, y) = fxx fyy – (fxy)2 = 24 > 0 and fxx > 0
So, by the Second Derivatives Test, f has a local minimum at .
5116 3( , )
Example 5
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MAXIMUM & MINIMUM VALUES
Intuitively, we can see that this
local minimum is actually an absolute
minimum:
There must be a point on the given plane that is closest to (1, 0, –2).
Example 5
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MAXIMUM & MINIMUM VALUES
If x = and y = , then
The shortest distance from (1, 0, –2) to the plane x + 2y + z = 4 is .
116
53
2 2 2
2 2 25 5 5 56 3 6 6
( 1) (6 2 )
6
d x y x y
Example 5
56 6
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MAXIMUM & MINIMUM VALUES
A rectangular box without a lid is to be made
from 12 m2 of cardboard.
Find the maximum volume of such a box.
Example 6
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MAXIMUM & MINIMUM VALUES
Let the length, width, and height of the box
(in meters) be x, y, and z.
Then, its volume is: V = xyz
Example 6
Fig. 15.7.10, p. 963
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MAXIMUM & MINIMUM VALUES
We can express V as a function of just
two variables x and y by using the fact that
the area of the four sides and the bottom
of the box is:
2xz + 2yz + xy = 12
Example 6
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MAXIMUM & MINIMUM VALUES
Solving this equation for z, we get:
z = (12 – xy)/[2(x + y)]
So, the expression for V becomes:
Example 6
2 212 12
2( ) 2( )
xy xy x yV xy
x y x y
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MAXIMUM & MINIMUM VALUES
We compute the partial derivatives:
Example 6
2 2
2
2 2
2
(12 2 )
2( )
(12 2 )
2( )
V y xy x
x x y
V x xy y
y x y
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MAXIMUM & MINIMUM VALUES
If V is a maximum, then
∂V/∂x = ∂V/∂y = 0
However, x = 0 or y = 0 gives V = 0.
Example 6
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MAXIMUM & MINIMUM VALUES
So, we must solve:
12 – 2xy – x2 = 0
12 – 2xy – y2 = 0
These imply that x2 = y2 and so x = y.
Note that x and y must both be positive in this problem.
Example 6
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MAXIMUM & MINIMUM VALUES
If we put x = y in either equation,
we get:
12 – 3x2 = 0
This gives: x = 2y = 2
z = (12 – 2 ∙ 2)/[2(2 + 2)] = 1
Example 6
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MAXIMUM & MINIMUM VALUES
We could use the Second Derivatives
Test to show that this gives a local
maximum of V.
Example 6
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MAXIMUM & MINIMUM VALUES
Alternatively, we could simply argue from
the physical nature of this problem that
there must be an absolute maximum volume,
which has to occur at a critical point of V.
So, it must occur when x = 2, y = 2, z = 1.
Example 6
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MAXIMUM & MINIMUM VALUES
Then,
V = 2 ∙ 2 ∙ 1
= 4
Thus, the maximum volume of the box is 4 m3.
Example 6
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ABSOLUTE MAXIMUM & MINIMUM VALUES
For a function f of one variable,
the Extreme Value Theorem says that:
If f is continuous on a closed interval [a, b], then f has an absolute minimum value and an absolute maximum value.
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According to the Closed Interval Method
in Section 4.1, we found these by evaluating
f at both:
The critical numbers
The endpoints a and b
ABSOLUTE MAXIMUM & MINIMUM VALUES
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There is a similar situation for functions
of two variables.
Just as a closed interval contains its
endpoints, a closed set in is one that
contains all its boundary points.
CLOSED SET
2¡
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A boundary point of D is a point (a, b)
such that every disk with center (a, b)
contains points in D and also points not in D.
BOUNDARY POINT
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For instance, the disk
D = {(x, y) | x2 + y2 ≤ 1}
which consists of all points on and inside
the circle x2 + y2 = 1 is a closed set
because:
It contains all of its boundary points (the points on the circle x2 + y2 = 1).
CLOSED SETS
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86
However, if even one point on the boundary
curve were omitted,
the set would not
be closed.
NON-CLOSED SETS
Fig. 15.7.11, p. 964
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A bounded set in is one that is
contained within some disk.
In other words, it is finite in extent.
BOUNDED SET
2
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Then, in terms of closed and bounded sets,
we can state the following counterpart of
the Extreme Value Theorem (EVT) for
functions of two variables in two dimensions.
CLOSED & BOUNDED SETS
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EVT (TWO-VARIABLE FUNCTIONS)
If f is continuous on a closed, bounded set D
in , then f attains an absolute maximum
value f(x1, y1) and an absolute minimum value
f(x2, y2) at some points (x1, y1) and (x2, y2)
in D.
Theorem 8
2
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90
EVT (TWO-VARIABLE FUNCTIONS)
To find the extreme values guaranteed by
Theorem 8, we note that, by Theorem 2,
if f has an extreme value at (x1, y1), then
(x1, y1) is either:
A critical point of f
A boundary point of D
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91
EVT (TWO-VARIABLE FUNCTIONS)
Thus, we have the following
extension of the Closed Interval
Method.
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CLOSED INTVL. METHOD (EXTN.)
To find the absolute maximum and minimum
values of a continuous function f on a
closed, bounded set D:
1. Find the values of f at the critical points of f in D.
2. Find the extreme values of f on the boundary of D.
3. The largest value from steps 1 and 2 is the absolute maximum value. The smallest is the absolute minimum value.
Method 9
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CLOSED & BOUNDED SETS
Find the absolute maximum and minimum
values of the function f(x, y) = x2 – 2xy + 2y
on the rectangle
D = {(x, y) | 0 ≤ x ≤ 3, 0 ≤ y ≤ 2}
Example 7
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CLOSED & BOUNDED SETS
As f is a polynomial, it is continuous on
the closed, bounded rectangle D.
So, Theorem 8 tells us there is both an absolute maximum and an absolute minimum.
Example 7
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CLOSED & BOUNDED SETS
According to step 1 in Method 9, we first
find the critical points.
These occur when fx = 2x – 2y = 0 fy = –2x + 2 = 0
So, the only critical point is (1, 1).
The value of f there is f(1, 1) = 1.
Example 7
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CLOSED & BOUNDED SETS
In step 2, we look at the values of f on
the boundary of D. This consists of the four line segments
L1, L2, L3, L4
Example 7
Fig. 15.7.12, p. 965
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CLOSED & BOUNDED SETS
On L1, we have y = 0 and
f(x, 0) = x2 0 ≤ x ≤ 3
This is an increasing function of x.
So, Its minimum
value is: f(0, 0) = 0
Its maximum value is:
f(3, 0) = 9
Example 7
Fig. 15.7.12, p. 965
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98
CLOSED & BOUNDED SETS
On L2, we have x = 3 and
f(3, y) = 9 – 4y 0 ≤ y ≤ 2
This is a decreasing function of y.
So, Its maximum
value is: f(3, 0) = 9
Its minimum value is:
f(3, 2) = 1
Example 7
Fig. 15.7.12, p. 965
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99
CLOSED & BOUNDED SETS
On L3, we have y = 2 and
f(x, 2) = x2 – 4x + 4 0 ≤ x ≤ 3
Example 7
Fig. 15.7.12, p. 965
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100
CLOSED & BOUNDED SETS
By the methods of Chapter 4, or simply by
observing that f(x, 2) = (x – 2)2, we see
that:
The minimum value of the function is f(2, 2) = 0.
The maximum value of the function is f(0, 2) = 4.
Example 7
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101
CLOSED & BOUNDED SETS
Finally, on L4, we have x = 0 and
f(0, y) = 2y 0 ≤ y ≤ 2
with: Maximum
valuef(0, 2) = 4
Minimum value f(0, 0) = 0
Example 7
Fig. 15.7.12, p. 965
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102
CLOSED & BOUNDED SETS
Thus, on the boundary,
The minimum value of f is 0.
The maximum value of f is 9.
Example 7
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103
CLOSED & BOUNDED SETS
In step 3, we compare these values with
the value f(1, 1) = 1 at the critical point.
We conclude that:
The absolute maximum value of f on D is f(3, 0) = 9.
The absolute minimum value of f on D is f(0, 0) = f(2, 2) = 0.
Example 7
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104
CLOSED & BOUNDED SETS
This figure shows the graph of f.
Example 7
Fig. 15.7.13, p. 965
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105
SECOND DERIVATIVES TEST
We close this section by giving a proof
of the first part of the Second Derivatives
Test.
The second part has a similar proof.
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106
SECOND DERIVS. TEST (PART A)
We compute the second-order directional
derivative of f in the direction of u = <h, k>.
The first-order derivative is given by
Theorem 3 in Section 14.6:
Duf = fxh + fyk
Proof
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107
SECOND DERIVS. TEST (PART A)
Applying the theorem a second time,
we have:
(Clairaut’s Theorem)
2
2 2
( )
( ) ( )
( ) ( )
2
xx yx xy yy
xx xy yy
D f D D f
D f h D f kx y
f h f k h f h f k k
f h f hk f k
u u u
u u
Proof
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108
SECOND DERIVS. TEST (PART A)
If we complete the square in that expression,
we obtain:
Proof—Equation 10
2 22 2( )xy
xx xx yy xyxx xx
f kD f f h k f f f
f f
u
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109
SECOND DERIVS. TEST (PART A)
We are given that fxx(a, b) > 0 and D(a, b) > 0.
However, fxx and are
continuous functions.
So, there is a disk B with center (a, b) and radius δ > 0 such that fxx(x, y) > 0 and D(x, y) > 0 whenever (x, y) is in B.
2xx yy xyD f f f
Proof
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110
SECOND DERIVS. TEST (PART A)
Therefore, by looking at Equation 10,
we see that
whenever (x, y) is in B.
Proof
2 ( , ) 0D f x y u
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111
SECOND DERIVS. TEST (PART A)
This means that:
If C is the curve obtained by intersecting the graph of f with the vertical plane through P(a, b, f(a, b)) in the direction of u, then C is concave upward on an interval of length 2δ.
Proof
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112
SECOND DERIVS. TEST (PART A)
This is true in the direction of every
vector u.
So, if we restrict (x, y) to lie in B, the graph of f lies above its horizontal tangent plane at P.
Proof
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113
SECOND DERIVS. TEST (PART A)
Thus, f(x, y) ≥ f(a, b) whenever (x, y)
is in B.
This shows that f(a, b) is a local minimum.
Proof