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Absolute Minimums
and Maximums

In this section we are going to extend the work from the previous section
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In this section we are want to optimize a function, that is identify the
absolute minimum and/or the absolute maximum of the function, on a given region in 2
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In order to optimize a function in a region we are going to need to get a couple of definitions out
of the way and a fact
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Definitions
2
1
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A region is called open if it
doesn’t include any of its boundary points
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A region in 2 is called bounded if it can be completely contained in a disk
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Let’s think a little more about the definition of closed
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Just what does this mean? Let’s think of a rectangle
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Open

Closed

−5 < x < 3

−5 ≤ x ≤ 3

1< y < 6

1≤ y ≤ 6

In this first case we don’t allow the ranges to include the endpoints (i
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we aren’t including the
edges of the rectangle) and so we aren’t allowing the region to include any points on the edge of
the rectangle
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In the second case we are allowing the region to contain points on the edges and so will contain
its entire boundary and hence will be close
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Extreme Value Theorem
If f ( x, y ) is continuous in some closed, bounded set D in

( x1 , y1 )

2

then there are points in D,

and ( x2 , y2 ) so that f ( x1 , y1 ) is the absolute maximum and f ( x2 , y2 ) is the absolute

minimum of the function in D
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It only tells us that they will exist
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The basic process for finding absolute maximums is pretty much identical to the process that we
used in Calculus I when we looked at finding absolute extrema of functions of single variables
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Here is the process
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Find all the critical points of the function that lie in the region D and determine the function
value at each of these points
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Find all extrema of the function on the boundary
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3
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The main difference between this process and the process that we used in Calculus I is that the
“boundary” in Calculus I was just two points and so there really wasn’t a lot to do in the second
step
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Let’s take a look at an example or two
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Solution
Let’s first get a quick picture of the rectangle for reference purposes
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x = 1, − 1 ≤ y ≤ 1
x = −1, − 1 ≤ y ≤ 1
y = 1, − 1 ≤ x ≤ 1

right side :
left side :
upper side :
lower side :

y = −1, − 1 ≤ x ≤ 1

These will be important in the second step of our process
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To do this
we’ll need the two first order derivatives
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To find the critical points we will need to solve the system,

2 x − 4 xy = 0
8 y − 2x = 0
2

We can solve the second equation for y to get,
2

x
y=
4
Plugging this into the first equation gives us,

⎛ x2 ⎞
2 x − 4 x ⎜ ⎟ = 2 x − x3 = x ( 2 − x 2 ) = 0
⎝ 4 ⎠
This tell us that must have x = 0 or x =± 2 =±1
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Now, recall that we only want
critical points in the region that we’re given
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The only value of x that will satisfy this is the first one so we can ignore the
last two for this problem
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Plugging x = 0 into the equation for y gives us,

02
y=
=0
4
The single critical point, in the region (and again, that’s important), is ( 0, 0 )
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f ( 0, 0 ) = 4

Eventually we will compare this to values of the function found in the next step and take the
largest and smallest as the absolute extrema of the function in the rectangle
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We need to find the absolute extrema of the
function along the boundary of the rectangle
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Let’s first take a look at the right side
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Let’s take advantage of this by defining a
new function as follows,

g ( y ) = f (1, y ) = 12 + 4 y 2 − 2 (12 ) y + 4 = 5 + 4 y 2 − 2 y

Now, finding the absolute extrema of f ( x, y ) along the right side will be equivalent to finding
the absolute extrema of g ( y ) in the range −1 ≤ y ≤ 1
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We find the critical points of g ( y ) in the range −1 ≤ y ≤ 1 and then evaluate g ( y )
at the critical points and the end points of the range of y’s
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g′ ( y ) = 8 y − 2

1
y=
4

⇒

This is in the range and so we will need the following function evaluations
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75
⎝4⎠ 4

g (1) = 7

Notice that, using the definition of g ( y ) these are also function values for f ( x, y )
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75
⎝4⎠
⎝ 4⎠ 4
We can now do the left side of the rectangle which is defined by,

x = −1, − 1 ≤ y ≤ 1

Again, we’ll define a new function as follows,

g ( y ) = f ( −1, y ) = ( −1) + 4 y − 2 ( −1) y + 4 = 5 + 4 y 2 − 2 y
2

2

2

Notice however that, for this boundary, this is the same function as we looked at for the right
side
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We know that the critical point is y = 14 and we know
that the function value at the critical point and the end points are,

g ( −1) = 11

g (1) = 7

⎛ 1 ⎞ 19
g ⎜ ⎟ = = 4
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In this case these will correspond to the following function values for
f ( x, y )
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75
4⎠ 4
⎝4⎠
⎝
We can now look at the upper side defined by,

y = 1, − 1 ≤ x ≤ 1

We’ll again define a new function except this time it will be a function of x
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First find the critical
point(s)
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The first two function values have already been
computed when we looked at the right and left side
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Finally, we need to take care of the lower side
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Here they are,

f ( 0, 0 ) = 4
⎛ 1⎞
f ⎜1, ⎟ = 4
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75
4⎠
⎝

f (1, −1) = 11

f (1,1) = 7

f ( −1,1) = 7

f ( −1, −1) = 11

f ( 0,1) = 8

f ( 0, −1) = 8

The absolute minimum is at ( 0, 0 ) since gives the smallest function value and the absolute
maximum occurs at (1, −1) and ( −1, −1) since these two points give the largest function value
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As this example has shown these can be very long problems
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Example 2 Find the absolute minimum and absolute maximum of f ( x, y ) = 2x 2 − y 2 + 6 y
on the disk of radius 4, x 2 + y 2 ≤ 16
Solution
First note that a disk of radius 4 is given by the inequality in the problem statement
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Of course, this also means that the boundary of the disk is a circle of radius 4
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This will require the
following two first order partial derivatives
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4x = 0
−2 y + 6 = 0

This is actually a fairly simple system to solve however
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So the only critical point for this function is ( 0,3) and this is
inside the disk of radius 4
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This one will be somewhat different from the previous
example
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Instead we have,

x + y = 16
2

2

We can solve this for x and plug this into the x in f ( x, y ) to get a function of y as follows
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We’ll first need the critical points of this function
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We
can do this by plugging the value of y into our equation for the circle and solving for y
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87

:

The function values for g ( y ) then correspond to the following function values for f ( x, y )
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So, comparing these values to the value of the function at the critical point of f ( x, y ) that we
found earlier we can see that the absolute minimum occurs at ( 0, −4 ) while the absolute

(

)

maximum occurs twice at − 15,1 and

(

)

15,1
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In both of these examples one of the absolute extrema actually occurred at more than one place
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Also note that, as we’ve seen, absolute extrema will often occur on the boundaries of these
regions, although they don’t have to occur at the boundaries
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