Loading in 2 Seconds...

A point ( x 0 , y 0 ) is called a [local minimum / local maximum] , if

Loading in 2 Seconds...

- 48 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' A point ( x 0 , y 0 ) is called a [local minimum / local maximum] , if' - steel-wilkins

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

A local minimum or local maximum for a function of two variables f(x,y) are defined analogously to the way they are defined for a function of one variable:

A point (x0 , y0) is called a [local minimum / local maximum] , if

for all points (x,y) in a disk of positive radius about the point (x0 , y0) .

[ f(x,y) f(x0 , y0) / f(x,y) f(x0 , y0) ]

A point (x0 , y0) is called a [global minimum / global maximum] , if

for all points (x,y) in the domain .

[ f(x,y) f(x0 , y0) / f(x,y) f(x0 , y0) ]

A point at which either a minimum or a maximum occurs is called an

extremum.

In single variable calculus, we begin the search for extrema by setting the first derivative equal to zero to obtain critical points. The critical points of a function f(x,y) are all points at which either

(1) fx = fy = 0 (where the tangent plane is parallel to the xy plane), or (2) f is not differentiable.

If (x0 , y0) is a local extremum of f(x,y), and the first partial derivatives exist at (x0 , y0), then (x0 , y0) must be a critical point of f(x,y).

Theorem 4 on page 208 states this for any function f from Rn to R .

It is possible that (x0 , y0) could be a critical point of f(x,y) and not be a local extremum of f(x,y). Such a point

will be like an inflection point in single variable calculus

or

will be what is called a saddle point. (See the figure on page 210.)

–(x2+y2)

Does the function f(x,y) = 2(x2+y2)e have any extrema?

By inspection, we find that

Find the critical points of the function.

since f(0,0) = 0 and f(x,y) is never negative, the point (0 , 0) is a global minimum.

Does the function f(x,y) = 2(x2+y2)e have any extrema?

By inspection, we find that

since f(0,0) = 0 and f(x,y) is never negative, the point (0 , 0) is a global minimum.

Find the critical points of the function.

f(x,y) = 2(x2+y2) e

fx =

fy =

–(x2+y2)

–(x2+y2)

–(x2+y2)

4x e – 4x(x2+y2) e

–(x2+y2)

–(x2+y2)

4y e – 4y(x2+y2) e

The critical points are

(0,0) and all points on the circle x2 + y2 = 1.

Since the value of the function can never be negative, (0,0) must be a global minimum point. From the graph on page 210, we see that each point on the circle x2 + y2 = 1 is a global maximum point.

Find all critical points of f(x,y) = x2y + xy2

fx = fy =

2xy + y2

x2 + 2xy

2xy + y2 = 0 when either y = 0 or y = –2x

x2 + 2xy = 0 when either x = 0 or y = –x/2

The only critical point is (0,0). Notice that if x = y = t, then f(x,y) = 2t3 which can be either positive or negative no matter how close (t,t) is to (0,0). This implies that (0,0) is not a local extremum.

Suppose f(x,y) is differentiable at a critical point x0 = (x0 , y0). Then, for points x0 + h which are “close” to x0 = (x0 , y0) the Taylor expansion tells us

h1

h2

f(x0 + h) f(x0) +

fx(x0) fy(x0)

fxx(x0) fyx(x0)

fxy(x0) fyy(x0)

h1

h2

1

—

2

+ h1h2

fxx(x0) fyx(x0)

fxy(x0) fyy(x0)

h1

h2

1

—

2

f(x0 + h) –f(x0)

h1h2

1

—

2

fxx(x0) h12 + 2fxy(x0) h1h2 + fyy(x0) h22

=

h22

——

2

h12

—

h22

h1

—

h2

fxx(x0) + 2fxy(x0) + fyy(x0)

=

A

x2

B

x

C

Recall from Precalculus that when the roots of Ax2 + Bx + C are complex numbers, then this function will either always be positive or always be negative (since the graph of the function can never cross the x axis forcing the graph to be either always above the x axis or always below the x axis.)

Consequently, Ax2 + Bx + C will either always be positive or always be negative if B2– 4AC < 0 which is the same as 4AC – B2 > 0 .

A

x2

B

x

C

h22

——

2

h12

—

h22

h1

—

h2

fxx(x0) + 2fxy(x0) + fyy(x0)

f(x0 + h) –f(x0)

will either always be positive or always be negative when

4[fxx(x0)][fyy(x0)] – [2fxy(x0)]2 > 0

[fxx(x0)][fyy(x0)] – [fxy(x0)]2 > 0

We shall label this Hessian matrix as A.

fxx(x0) fyx(x0)

fxy(x0) fyy(x0)

det

> 0

This suggests that if det(A) = fxx(x0) fyy(x0) – [fxy(x0)]2 > 0, then x0 is a , but if det(A) = fxx(x0) fyy(x0) – [fxy(x0)]2 < 0, then x0 is a

local extremum

saddle point.

The determinant det(A) = fxx(x0) fyy(x0) – [fxy(x0)]2 is called the

discriminant of the Hessian and is often represented by D.

Suppose det(A) = fxx(x0) fyy(x0) – [fxy(x0)]2 > 0. Then if fxx(x0) and fyy(x0) are both positive, x0 is a , but if fxx(x0) and fyy(x0) are both negative, x0 is a

local minimum

local maximum.

If det(A) = fxx(x0) fyy(x0) – [fxy(x0)]2 = 0, then we cannot tell whether x0 is an extremum or not.

Theorem 6 on page 216 is a statement of the Second Derivative test for functions of two variables.

Find all critical points of f(x,y) = x2 + 3xy + y2 , and determine whether each is a local minimum, a local maximum, or a saddle point.

fx = fy =

2x + 3y

3x + 2y

The only critical point is (0,0).

fxx = fyy = fxy =

D =

2

2

3

(2)(2) – (3)2 = – 5

Since D < 0, then (0,0) is a saddle point.

Find all critical points of f(x,y) = 3x2 – 5xy + 3y2 , and determine whether each is a local minimum, a local maximum, or a saddle point.

fx = fy =

6x – 5y

–5x + 6y

The only critical point is (0,0).

fxx = fyy = fxy =

D =

6

6

–5

(6)(6) – (–5)2 = 11

Since D > 0, and fxx(0,0) > 0, then (0,0) is a local minimum.

Find all critical points of f(x,y) = x sin y , and determine whether each is a local minimum, a local maximum, or a saddle point.

fx = fy =

sin y

x cos y

sin y = 0 when y = n for any integer n

x cos y = 0 when either x = 0 or y = (n + 1/2) for any integer n

The critical points are (0 , n) for any integer n.

fxx = fyy = fxy =

D =

0

– x sin y

cos y

– cos2y

Since D < 0 for all critical points, then all critical points are saddle points.

Find all critical points of z = (x2 – y2) e , and determine whether each is a local minimum, a local maximum, or a saddle point.

fx =

fy =

(–x2–y2) / 2

x[2 – x2 + y2] e

(–x2–y2) / 2

y[– 2 – x2 + y2] e

The critical points are

(0,0), (2,0), (–2,0), (0,2), (0,–2)

fxx =

fyy =

fxy =

(–x2–y2) / 2

[2 – 5x2 + x2(x2 – y2) + y2] e

(–x2–y2) / 2

[– 2 + 5y2 + y2(x2 – y2) – x2] e

(–x2–y2) / 2

xy(x2 – y2) e

fxx(0,0) = fyy(0,0) = fxy(0,0) =

D = (0,0) is

fxx(2,0) = fxx(–2,0) = D =

fyy(2,0) = fyy(–2,0) = (2,0) and (–2,0) are

fxy(2,0) = fxy(–2,0) =

fxx(0,2) = fxx(0,–2) = D =

fyy(0,2) = fyy(0,–2) = (0,2) and (0,–2) are

fxy(0,2) = fxy(0,–2) =

2

– 2

0

– 4

a saddle point

16e–2

– 4e–1

– 4e–1

each a local maximum

0

16e–2

4e–1

4e–1

each a local minimum

0

Find the minimum distance from the origin to a point on the plane

x + 3y – z = 6 .

The distance from the origin to any point (x,y,z) is

Minimizing this distance is the same as minimizing

If (x,y,z) is a point on the plane, then

Consequently, we want to minimize the function f(x,y) =

fx = fy =

The only critical point is

(x–0)2 + (y–0)2 + (z–0)2 .

x2 + y2 + z2 .

z = x +3y – 6 .

x2 + y2 + (x +3y – 6)2 .

2x + 2(x +3y – 6)

2y + 6(x +3y – 6)

(6/11 , 18/11) .

Since f(x,y) only becomes large as x and y become large, then f(x,y) must attain a minimum somewhere and at (6/11 , 18/11) is the only possibility.

(6/11 , 18/11) .

Since f(x,y) only becomes large as x and y become large, then f(x,y) must attain a minimum somewhere and at (6/11 , 18/11) is the only possibility.

The minimum distance from the origin to a point on the plane is the distance from the origin to the point

This distance is

(6/11 , 18/11 , –6/11) .

6/11 .

Find the minimum distance from the origin to a point on the surface

g(x,y) = 1/ (xy) .

The distance from the origin to any point (x,y,z) is

Minimizing this distance is the same as minimizing

If (x,y,z) is a point on the surface, then

Consequently, we want to minimize the function f(x,y) =

fx = fy =

The critical points are

(x–0)2 + (y–0)2 + (z–0)2 .

x2 + y2 + z2 .

z = 1/ (xy) .

x2 + y2 + 1/(x2y2) .

2x – 2/(x3y2)

2y – 2/(x2y3)

(1,1), (–1,1), (1,–1), (–1,–1) .

Since f(x,y) only becomes large as (x,y) moves away from any one of these four points in the same quadrant, at least one of these four points must be a minimum point.

(1,1), (–1,1), (1,–1), (–1,–1) .

Since f(x,y) only becomes large as (x,y) moves away from any one of these four points in the same quadrant, at least one of these four points must be a minimum point.

The distance from the origin to each of the points

(1,1,1), (–1,1,–1), (1,–1,–1), (–1,–1,1)

on the surface is , which must be the minimum distance.

3

Download Presentation

Connecting to Server..