Lec7[1]Valores Maximos y Minimos

8/13/2019 Lec7[1]Valores Maximos y Minimos

1/22

MATH 209

Calculus,III

Volker Runde

Maximum andmimimumvalues MATH 209Calculus, III

Volker Runde

University of Alberta

Edmonton, Fall 2011


2/22

MATH 209

Calculus,III

Volker Runde

Maximum andmimimumvalues

Local and absolute maxima and minima

Definition

The function f(x, y) has alocal maximumat (a, b) iff(x, y) f(a, b)for all (x, y) near (a, b). In this case, we call

f(a, b) a local maximum valueoff. Iff(x, y)f(a, b)for all(x, y)Df, we say that f has anabsolute maximumat (a, b).

Definition

The function f(x, y) has alocal minimumat (a, b) iff(x, y) f(a, b)for all (x, y) near (a, b). In this case, we callf(a, b) a local minimum valueoff. Iff(x, y)f(a, b)for all(x, y)Df, we say that f has anabsolute minimumat (a, b).


3/22

MATH 209

Calculus,III

Volker Runde


First derivative test

Theorem (first derivative test)

If f has a local maximum or minimum at(a, b) and if f x(a, b)

and f y(a, b) exist, then

fx(a, b) =fy(a, b) = 0.

IdeaIffhas a local maximum or minimum at (a, b), then thetangent plane to the graph off must be horizontal.


4/22

MATH 209

Calculus,III

Volker Runde


Examples, I

Example

Letf(x, y) =x2 +y2 2x 6y+ 14,

so that

fx(x, y) = 2x 2 and fy(x, y) = 2y 6.

It follows that

fx(a, b) =fy(a, b) = 0 (a, b) = (1, 3).

Hence, a local maximum or minimum off canonlyoccur at(1, 3).


5/22

MATH 209

Calculus,III

Volker Runde


Examples, II

Example (continued)

For any (x, y) R2:

f(x, y) =x2 +y2 2x 6y+ 14

= (x 1)2 + (y 3)2 + 4

4

=f(1, 3)

Hence, fhas an absolute minimum at (1, 3).


6/22

MATH 209

Calculus,III

Volker Runde


Stationary points, I

Definition

A point (a, b) is calledstationaryorcriticalforf iffx(a, b) =fy(a, b) = 0.

Refomulation of the first derivative test

Iffhas a local maximum or minimum at (a, b), then (a, b)

must be a critical point for f.


7/22

MATH 209

Calculus,III

Volker Runde


Stationary points, II

ExampleLet

f(x, y) =x2 +y3.

Then

fx(x, y) = 2x and fy(x, y) = 3y2,

so that (0, 0) is the only stationary point for f.Since f(0, y) =y3 for all y, we get both positive and negativevalues offno matter how close (x, y) gets to (0, 0).

Thus, there is no local maximum or minimum of f at (0, 0).

Definition

Let (a, b) a stationary point for f such that f has neither alocal maximum or minimum at (a, b). Then we call (a, b) a

saddleforf.


8/22

MATH 209

Calculus,III

Volker Runde


Second derivative test, I

Theorem (second derivative test)

Let f have continuous second derivatives, and let(a, b) be astationary point for f . Set

D=fxx(a, b)fyy(a, b) fxy(a, b)2 =fxx(a, b) fxy(a, b)fyx(a, b) fyy(a, b)

.Then:

1 if fxx(a, b)>0 and D>0, then f has a local minimum at

(a, b);

2 if fxx(a, b)0, then f has a local maximum at(a, b);

3 if D


9/22

MATH 209

Calculus,III

Volker Runde


Second derivative test, II

Important

IfD= 0, then anything is possible:1 fmay have a local minimum at (a, b), or

2 fmax have a local maximum at (a, b), or

3 (a, b) may be a saddle for f.

Hence,ifD= 0 nothing can be said.


10/22

MATH 209

Calculus,III

Volker Runde


Examples, III

Example

Letf(

x, y) =

x4

+y4

4xy

+ 1,

so that

f

x = 4x3 4y and

f

y = 4y3 4x.

Thus, (x, y) is a stationary point for f if and only ifx3 =y andy3 =x, which implies x9 =x.


11/22

MATH 209

Calculus,III

Volker Runde


Examples, IV

Example (continued)

As

0 =x9 x

=x(x8 1)

=x(x4 1)(x4 + 1)

=x(x2 1)(x2 + 1)(x4 + 1)

=x(x 1)(x+ 1)(x2 + 1)(x4 + 1),

we have x=1, 0, 1.Hence, the critical points for f are (0, 0), (1, 1), and (1,1).


12/22

MATH 209

Calculus,III

Volker Runde


Examples, V

Example (continued)

Next, use the second derivative test. We have

2f

x2 = 12x2

,

2f

y2 = 12y2

, and

2f

xy =4,

so that

D(x, y) =2f

x2

2f

y2

2f

xy

2

= 144x2y2 16.

Therefore:

1 as D(0, 0) =16, fhas a saddle at (0, 0);


13/22

MATH 209

Calculus,III

Volker Runde


Examples, VI

Example (continued)

2 as D(1, 1) =D(1,1) = 128>0 and

2f

x2(1, 1) =

2f

x2(1,1) = 12>0,

fhas a local minimum at both (1, 1) and (1,1) with

corresponding local minimum valuesf(1, 1) =f(1,1) =1.


14/22

MATH 209

Calculus,III

Volker Runde


Examples, VII

ExampleFind those points on the surface Sgiven by z2 =xy+ 1 thatare closest to the origin.Let P(x, y, z) be a point on S. Then:

distance from (0, 0, 0) to (x, y, z) =|x, y, z|

=

x2 +y2 +z2

=

x2 +y2 +xy+ 1.

Now:x2 +y2 +xy+ 1 is minimal

f(x, y) :=x2 +y2 +xy+ 1 is minimal.


15/22

MATH 209

Calculus,III

Volker Runde


Examples, VIII

Example (continued)As

fx= 2x+y and fy = 2y+x,

a point (x, y) is stationary for f if and only ifx= 12 y and

x=2y, i.e., if and only ifx= 0 =y.Since

fxx= 2>0, fyy = 2, and fxy = 1,

we haveD= 4 1 = 3>0,

so that fhas a local minimum at (0, 0).As z=1 for x=y= 0, the points (0, 0, 1) and (0, 0,1) are

the points on Sclosest to (0, 0, 0).


16/22

MATH 209

Calculus,III

Volker Runde


Examples, IX

Example

Make a rectangular box out of 12 m2 of cardboard without lidsuch that its volume is maximal.Let

x= length,

y= width

z= height,

so that

V= volume =xyz,

S= surface area = 2xz+ 2yz+xy= 12.


17/22

MATH 209

Calculus,III

Volker Runde


Examples, X

Example (continued)

As S= 12, we obtain

z= 12 xy

2(x+y),

so that

V =xy

12 xy

2(x+y) =

12xy x2y2

2(x+y) .


18/22

MATH 209

Calculus,III

Volker Runde


Examples, XI

Example (continued)

It follows that

V

x =

1

2

(12y 2xy2)(x+y) (12xy x2y2)

(x+y)2

=1

2

12xy 2x2y2 + 12y2 2xy3 12xy+x2y2

(x+y)2

=1

2

12y2 2xy3 x2y2

(x+y)2

= y2

2

12 2xy x2

(x+y)2

and, similarly, Vy

= x2

2 122xyy2

(x+y)2 .


19/22

MATH 209

Calculus,III

Volker Runde


Examples, XII

Example (continued)

To maximize V, we need x, y>0 such that Vx

= Vy

= 0.As

12 2xy

x2

= 0 = 12 2xy

y2

if and only ifx2 =y2 if and only ifx=y, solve

0 = 12 2x2 x2 = 12 3x2

forx. It follows that x=y= 2 (and thus z= 1). As themaximum ofVmust occur, we can skip the second derivativetest.Hence, V is maximal for x=y= 2 and z= 1 with V = 4.


20/22

MATH 209

Calculus,III

Volker Runde


Examples, XIII

Example

Find the absolute minimum and maximum values of

f(x, y) =x2 2xy+ 2y

on the rectangle

D :={(x, y) : 0x3, 0y2}.

with sides

L1 :={(x, 0) : 0 x3},

L2 :={(3, y) : 0 y2},

L3 :={(x, 2) : 0 x3},

L4

:={(0, y) : 0 y2}.


21/22

MATH 209

Calculus,III

Volker Runde


Examples, XIV

Example (continued)

Asfx= 2x 2y and fy =2x+ 2

there is only one critical point, namely (1, 1); we havef(1, 1) = 1.On L1: y= 0, f(x, 0) =x

2, 0 x3.Hence, the minimum value of f on L1 is f(0, 0) = 0, and themaximum value is f(3, 0) = 9.On L2: x= 3, f(3, y) = 9 4y, 0y2.Hence, the minimum value of f on L2 is f(3, 2) = 1, and themaximum value is f(3, 0) = 9.


22/22

MATH 209

Calculus,III

Volker Runde


Examples, XV

Example (continued)

On L3: y= 2, f(x, 2) =x2 4x+ 4 = (x 2)2, 0x3.

Hence, the minimum value of f on L3 is f(2, 2) = 0, and the

maximum value is f(0, 2) = 4.On L4: x= 0, f(0, y) = 2y, 0 y2.Hence, the minimum value of f on L4 is f(0, 0) = 0, and themaximum value is f(0, 2) = 4.Hence:

the absolute minimum value off on D isf(0, 0) =f(2, 2) = 0;

the absolute maxiimum value off on D is f(3, 0) = 9.

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Lec7[1]Valores Maximos y Minimos