8 Extremum of Function

8/13/2019 8 Extremum of Function

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Mat235 4/13/20

x remumx remum oo

functionsfunctionsObjectives:Objectives:

Compute and classify critical pointsCompute and classify critical points

Hasfazilah Ahmat

Jan 2010

Compute the relativeCompute the relative extremaextrema for afor a

function of two variablesfunction of two variablesSolve appliedSolve applied extremaextrema problemsproblems

Graph of function of twoGraph of function of two

variablesvariables

Hasfazilah Ahmat

Jan 2010


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Mat235 4/13/20

Critical pointsCritical points

Let f(x,y) be a function of two variables. Apoint (a, b) is a critical point of f(x,y) if either

fx (a,b) = 0 and

f a,b = 0 or

Hasfazilah Ahmat

Jan 2010

( ) 2

f x , y 4 xy 2 x y= +fx (a,b) = 0 orfy (a,b) = 0 does not exist

Critical pointsCritical points -- stepssteps

x y

fx = 0 or fy = 0

Solve fx = 0 or fy = 0

Hasfazilah Ahmat

Jan 2010

Determine the critical points


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Mat235 4/13/20

Example 1Example 1Find the critical points for the functionFind the critical points for the function

Solution:Solution:

( ) 2f x , y 4 xy 2 x y= +

LetLet andand

x 4 y 4 x=

yf 4 x 1= +

xf 0= yf 0=

xf 4 y 4 x 0= =

4 x 1 0= + =

Hasfazilah Ahmat

Jan 2010

Solve and simultaneouslySolve and simultaneously

y

xf 0= y 0=

4 y 4 x 0 =

y x=

4 x 1 0+ =1

x4

=

Example 1Example 1Find the critical points for the functionFind the critical points for the function

Solution:Solution:

( ) 2f x , y 4 xy 2 x y= +

Determine the critical pointsDetermine the critical points

y x= 1x

4=

1y

4=

Hasfazilah Ahmat

Jan 2010

Therefore, the critical point isTherefore, the critical point is

1 1,

4 4


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Mat235 4/13/20

Find the critical points for the functionFind the critical points for the function

Solution:Solution:

( ) 2 3x , y 3 4 xy 2 x y= + +

Example 2Example 2

LetLet andand

x 4 y 4 x= + 2

yf 4 x 3 y=

xf 0=

yf 0=

xf 4 y 4 x 0= + =

Hasfazilah Ahmat

Jan 2010

yf 4 x 3 y 0= =


Solution:Solution:

( ) 2 3x , y 3 4 xy 2 x y= + +

Example 2 (cont.)Example 2 (cont.)

Solve and simultaneouslySolve and simultaneouslyxf 0= yf 0=

4 y 4 x 0+ =

y x=

( )224 x 3 y 4 x 3 x 0 = =

( )x 4 3 x 0 =4

x 0 ,3

=

Hasfazilah Ahmat

Jan 2010

Determine the critical pointsDetermine the critical pointsy 0=x 0=

4y

3=

4x

3=

( )0 ,04 4

,3 3


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Mat235 4/13/20


Solution:Solution:

( ) 3 3 2 2f x , y 2 x y 3 x 18 y 81 y 5= + + + +

Example 3Example 3

From (1)From (1)

2

xf 6 x 6 x 0 ( 1 )= + =

2

yf 3 y 36 y 81 0 ( 2 )= + =

( )6 x x 1 0+ =x 0 , 1=

Hasfazilah Ahmat

Jan 2010

From (2)From (2)2

y 12 y 27 0 + =( ) ( )y 9 y 3 0 =

y 9 ,3=

Therefore, thecritical pointsare: (0,3), (0,9),(-1,3) and (-1,9)

Warm up exerciseWarm up exercise

( ) 3 3x , y 3 y x y 3 x= +

Hasfazilah Ahmat

Jan 2010


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Mat235 4/13/20

ExtremumExtremum

Minimum point Maximum point

Hasfazilah Ahmat

Jan 2010

Saddle point

Second partial derivative testSecond partial derivative testSuppose the second partial derivativesSuppose the second partial derivatives of f areof f are

continuoucontinuous on critical point (s on critical point (a,ba,b) and suppose) and supposethatthat ffxx((a,ba,b) = 0 and) = 0 and ffyy((a,ba,b)=0. Let)=0. Let

2

xx yy xya , a , a , a ,= =

Hasfazilah Ahmat

Jan 2010

D > 0 And

fxx > 0

D > 0 And

fxx< 0

D < 0 D = 0 Inconc

lusive


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Mat235 4/13/20

Identify critical point, (a,b)

Second partial derivative testSecond partial derivative testDiscriminantDiscriminant :: ( ) ( ) ( ) ( )

2

xx yy xyD a ,b f a ,b f a ,b f a ,b = =

Compute fxx, fyy and fxy

At critical point, evaluate Fxx (a,b), Fyy (a,b) and Fxy(a,b)

2

D =

Hasfazilah Ahmat

Jan 2010

,

Check D and fxx (a,b) (if necessary)

Classify as minimum, maximum, saddle point or noconclusion

xx yy xy

The critical points for the functionThe critical points for the function

are (0, 0) and (4/3,are (0, 0) and (4/3, --4/3). Classify the critical points4/3). Classify the critical points

( ) 2 3x , y 3 4 xy 2 x y= + +

Example 4Example 4

Solution:Solution:

Identify theIdentify the critical pointscritical points

ComputeCompute ffxxxx,, ffyyyy andand ffxyxy

( )0 ,0 4 4,3 3

=

xf 4 y 4 x= + 2

y 4 x 3 y=

Hasfazilah Ahmat

Jan 2010

xx yy xy

Critical point

(0,0) 4 0 4

(4/3, -4/3) 4 8 4

xxf 4= yyf 6 y= xyf 4=


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Mat235 4/13/20

The critical points for the functionThe critical points for the function

are (0, 0) and (4/3,are (0, 0) and (4/3, --4/3). Classify the critical points4/3). Classify the critical points

( ) 2 3x , y 3 4 xy 2 x y= + +


ComputeCompute discriminantdiscriminant

Critical pointD

(0,0) 4 0 4 -16

(4/3, -4/3) 4 8 4 16

xxf 4= yyf 6 y= xyf 4=

Hasfazilah Ahmat

Jan 2010

Check D andCheck D and xxxx (if necessary) and classify nature(if necessary) and classify nature

Critical point D Classification

(0,0) -16 < 0 Not necessary to check Saddle point

(4/3, -4/3) 16 4 > 0 Relative minimum

xxf 4=

Locate and classifyLocate and classify the critical point forthe critical point for

Solution:Solution:

( ) 2 2x , y 3 x 2 xy y 8 y= +

Example 5Example 5

xf 6 x 2 y 0= =

yf 2 x 2 y 8 0= + =

( )y 3 x ..... 1=

( )2 x 2 3 x 8 0 + =x 2= Into (1)

( )y 3 2 6= = Critical point

Hasfazilah Ahmat

Jan 2010

xxf 6=

yyf 2= xy 2=

Critical point

(2,6) 6 2 -2

xxf 6= yyf 2= xyf 2=

(2,6)


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Mat235 4/13/20

Locate and classify the critical point forLocate and classify the critical point for

ComputeCompute discriminantdiscriminant


( ) 2 2x , y 3 x 2 xy y 8 y= +

Check D andCheck D and ffxxxx (if necessary) and classify nature(if necessary) and classify nature

Critical pointD

(2,6) 6 2 -2 8

xxf 6= yyf 2= xyf 2=

Hasfazilah Ahmat

Jan 2010

Critical point D Classification

(2,6) 8 > 0 Fxx > 0 Minimum point

xxf 6=

Locate allLocate all relativerelative extremumextremum and saddle points ofand saddle points of

Solution:Solution:

ExampleExample 66

( ) 3 3x , y x y 3 xy= +

2

xf 3 x 3 y 0= + =

2

yf 3 y 3 x 0= + =

( )2y x ..... 1=

( )2

23 x 3 x 0 + =

4x x 0 + =

3x x 1 0 + =

Hasfazilah Ahmat

Jan 2010

Into (1)x 0 ,1=

Critical point (0, 0) and (1, -1)

y 0=x=0

y 1= X=1


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Mat235 4/13/20

Locate all relativeLocate all relative extremumextremum and saddle points ofand saddle points of


( ) 3 3x , y x y 3 xy= +

xx 6 x= 6 y=

xf 3=

Critical pointD

(0,0) 0 0 3 -9

(1,-1) 6 6 3 27

xxf 6 x= yyf 6 y= xyf 3=

2

xx yy xyD f f f=

Hasfazilah Ahmat

Jan 2010

ass ca on

(0,0) -9 Saddle point

(1,-1) 27 Fxx > 0 Minimum point

xx

( ) 3 2 3f x , y 2 x 6 xy 3 y 150 x= +

Find the stationary point of the function andFind the stationary point of the function anddeterminedetermine the naturethe nature

ExampleExample 77

Solution:Solution:

2 2

xf 6 x 6 y 150 0= + =

212 x 9 0= =

2 26 x 6 y 150+ =

( )2 2y 25 x .... 1=

Hasfazilah Ahmat

Jan 2010

y

Into (1)

( )3 y 4 x 3 y 0 =4 x

y 0 ,3

=

225 x 0 =y=0 x 5=


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Mat235 4/13/20

( ) 3 2 3f x , y 2 x 6 xy 3 y 150 x= +

Find the stationary point of the function andFind the stationary point of the function and

determinedetermine the naturethe nature

ExampleExample 7 (cont.)7 (cont.)

Solution:Solution:

4 xy

3=

2216 x 25 x

9=

3 3=

225 x25 0

9 =

5 x 5 x5 5 0

+ =

Critical point :(5, 0), (-5, 0),(3 4) (-3 -4)

Hasfazilah Ahmat

Jan 2010

x 3,=

3 3

4 xy3=

( )4 3y 43= =

x 3,= ( )4 3

y 43

= =

( ) 3 2 3f x , y 2 x 6 xy 3 y 150 x= +

Find the stationary point of the function andFind the stationary point of the function anddeterminedetermine the naturethe nature


Solution:Solution:

point :(5, 0), (-5, 0),(3, 4), (-3, -4)

Criticalpoint D Conclusion

(5, 0) 60 60 0 3600 Minimum

xxf 12 x= yyf 12 x 18 y= xyf 12 y=

xxf 12 x=

yyf 12 x 18 y=

xyf 12 y=

2

xx yy xyD f f f=

Hasfazilah Ahmat

Jan 2010

(-5, 0) -60 -60 0 3600 Maximum

(3, 4) 36 -36 48 -3600 Saddle

(-3, -4) -36 36 -48 -3600 Saddle


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Mat235 4/13/20

Find all the criticalFind all the critical points of the functionpoints of the function

Determine which give relative minima, maxima orDetermine which give relative minima, maxima orsaddle pointssaddle points

( ) 4 4x , y 4 xy x y=

ExampleExample 88

Solution:Solution:

3

xf 4 y 4 x 0= =

3f 4 x 4 y 0= =

( )3y x .... 1=

Hasfazilah Ahmat

Jan 2010x 0 ,1 , 1=

( )3

34 x 4 x 0 =

9x x 0 =

( )8x 1 x 0 =x 0 , y 0

x 1, y 1

x 1, y 1

= =

= =

= =

Find all the critical points of the functionFind all the critical points of the function


ExampleExample 88 (cont.)(cont.)

( ) 4 4x , y 4 xy x y=

Solution:Solution:

point :(0, 0), (1, 1),(-1, -1)

Critical2=

2= 4=

2

xx 12 x= 2

yy 12 y=

xyf 4=

2

xx yy xyD f f f=

Hasfazilah Ahmat

Jan 2010

po n

(0, 0) 0 0 4 -16 saddle

(1, 1) -12 -12 4 128 Maximum

(-1, -1) -12 -12 4 128 maximum

xx yy xy


13/15

Mat235 4/13/20

Find all the criticalFind all the critical points of the functionpoints of the function


( ) 4 4x , y 4 xy x y=

ExampleExample 99

Solution:Solution:

3

xf 4 y 4 x 0= =

3f 4 x 4 y 0= =

( )3y x .... 1=

Hasfazilah Ahmat

Jan 2010x 0 ,1 , 1=

( )3

34 x 4 x 0 =

9x x 0 =

( )8x 1 x 0 =x 0 , y 0

x 1, y 1

x 1, y 1

= =

= =

= =

A rectangular boxA rectangular box opeopen at the top is to have an at the top is to have acolumecolume of 32 cmof 32 cm33. What must be the dimensions so. What must be the dimensions sothat the total surface is minimum?that the total surface is minimum?

ExampleExample 99

Solution:Solution:

V lwh 32= =

( )S l ,w ,h lw 2lh 2wh= + + lw

h32h

lw=

( ) 32 32

S l , w lw 2l 2wlw lw

= + +

w 264

S l 0= =

Hasfazilah Ahmat

Jan 2010

( ) 64 64

S l , w lww l

= + +

l 2

64S w 0

l= = ( )2

64w ..... 1

l=

w

( )264

l ..... 2w

=


14/15

Mat235 4/13/20

A rectangular boxA rectangular box opeopen at the top is to have an at the top is to have a

columecolume of 32 cmof 32 cm33. What must be the dimensions so. What must be the dimensions sothat the total surface is minimum?that the total surface is minimum?


Solution:Solution:

V lwh 32= = 32

hlw

=

( )264

w ..... 1l

=64

l=

4ll 0

64 =

3

(1) Into (2)

Hasfazilah Ahmat

Jan 2010

2l ..... 2

w=

So, l = 4

( )264 l4l

l64

=

l 1 064

=

l 0 ,4=

2

64w 4

4= =

( ) ( )32

h 24 4

= =

A rectangular boxA rectangular box opeopen at the top is to have an at the top is to have acolumecolume of 32 cmof 32 cm33. What must be the dimensions so. What must be the dimensions sothat the total surface is minimum?that the total surface is minimum?


Solution:Solution:

ll 3

128S

l= ww 2

128S

w= lwS 1=

2

ll ww lwD S S S=

Hasfazilah Ahmat

Jan 2010

( )23 3D 1l w= ( )2

3 3D 1 34 4= =

llD 3 0 , f 0= > > The computed value produce

minimum total surface


15/15

Mat235 4/13/20

FindFind the dimensions of a sixthe dimensions of a six--faced box that has thefaced box that has the

shape of a rectangular prism with the largestshape of a rectangular prism with the largestpossible volume that you can make with 12 squarepossible volume that you can make with 12 square

ExampleExample 1010

. .

Solution:Solution:A 2 xy 2 yz 2zx 12= + + =

V xyz=

6 xyz

x y

=

+

( ) 6 xy

V x , y xyx y

=

+

( )2 2x 2

y x 2 xy 6V 0

x y

+ = =

+

Hasfazilah Ahmat

Jan 2010

( )

( )

2 2

y 2

x y 2 xy 6V 0

x y

+ = =

+2

x 2 xy 6 0+ =

2y 2 xy 6 0+ =

2 2x y 0 =

FindFind the dimensions of a sixthe dimensions of a six--faced box that has thefaced box that has theshape of a rectangular prism with the largestshape of a rectangular prism with the largestpossible volume that you can make with 12 squarepossible volume that you can make with 12 square


. .

Solution:Solution:2x 2 xy 6 0+ =

2y 2 xy 6 0+ =

2 2x y 0 =

x y , y=

Since it is a dimension, x -y

Hasfazilah Ahmat

Jan 2010

x y= 2 2x 2 x 6 0+ =

x 2= y 2= 6 2

z 22 2

= =

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8 Extremum of Function