ch 16 maximum and minimum problems 2012

8/2/2019 Ch 16 Maximum and Minimum Problems 2012

1/27

Maximum and Minimum Problems 1

Singapore Chinese Girls School

Secondary 4

Additional Mathematics

Second Derivatives and Application

Name: ( ) Date: ________________

Class: Sec 4_____

Worksheet 22: Nature of Stationary Points

Determination of Maximum and Minimum Points

Consider the graph of )(xfy in the diagram below.

Gradient of the curve Gradient of the curve

Along

AB0

dx

dyy increases asx increases AtD 0

dx

dyD is a stationary points

AtB 0dx

dyB is a stationary points

Along

DE0

dx

dyy decreases asx increases

Along

BC0

dx

dyy decreases asx increases AtE 0

dx

dy is a stationary points

At C 0dx

dyCis a stationary points

Along

EF0

dx

dyy increases asx decreases

Along

CD0

dx

dyy increases asx increases

Stationary points

PointsB, C, D andEwhere 0dx

dyare called stationary points.

Minimum points

Point C(orE) is called a minimum pointbecause )(xfy has a minimum value as compared to

the points alongBCand CD (alongDEandEF).

Points CandEare also known as local minimums as they do not represent the minimum value of

the whole curve.


2/27


Maximum points

PointB (orD) is called a maximum point because )(xfy has a maximum value as compared to

the points alongAB andBC(along CD andDE).

PointsB andD are also known as local maximums as they do not represent the maximum value

of the whole curve.

Points of Inflexion

Consider the graphs of )(xfy and )(xgy in the diagram below.

At pointsA andB, 0dx

dy, but they are both neither maximum nor minimum points.

We call these points the stationary points of inflexion.

Summary

Given a curve )(xfy ,

(a) When 0dx

dyat ax , then )(, afa is a stationarypoint. )(, afa is a turning point if it

is either a maximum point or a minimum point.

(b) If dx

dy

changes sign from positive to negative as it passes through ax , then it a maximum

point.

(c) Ifdx

dychanges sign from negative to positive as it passes through ax , then it is a minimum

point.

(d) Ifdx

dydoes not change sign as it passes through ax , then it is a stationary point of inflexion.


3/27


Example 1

Find, by calculus, the stationary point of the curve 642 xxy . Observe the changes in the sign

of the gradient of the curve and determine the nature of the point.

642 xxy 42 x

dx

dy

When 0dx

dy, 042 x

2x

262422 y The stationary point is (2, 2).

x 2 2 2

+

dx

dy 0.02 0 0.02

Slope \ _ /

Asx increases throughx = 2,dx

dychanges sign from negative to positive and hence the point is (2, 2)

is a minimum point.

Example 2

Given that 22 )2( xxy and its gradient function isdxdy )2)(1(4 xxx ,

(a) find the coordinates of the stationary points on the curve.

(b) observe the change in sign ofdx

dyasx increases through each of the stationary points. Hence,

deduce the nature of the points.

(a) When 0dx

dy, 0)2)(1(4 xxx

2,1,0x

When 0x , 0y When 1x , 1y

When 2x , 0y

The stationary points are (0, 0), (1, 1) and (2, 0).

x 0 0 0+ 1 1 1

+ 2 2 2

+

dx

dy < 0 0 > 0 > 0 0 < 0 < 0 0 > 0

Slope \ __ / / \ \ __ /

(0, 0) and (2, 0) are minimum points.

(1, 1) is a maximum point.


4/27


Example 3 (Pg348 Ex16.1Q5)

The diagram shows part of the curve of4

522

x

xy where 2x and 0x .

(a) Finddx

dy,

(b) Find the x coordinates of the stationary points.

(a)dx

dy

22

2

)4(

)52(2)4(2

x

xxx

22

22

)4(

10482

x

xxx

22

2

)4(

8102

x

xx

(b) When 0dx

dy, 08102 2 xx

0)45(22 xx

0)4)(1(2 xx

4,1 xx

Example 4 (Pg348 Ex16.1Q1)

Given that ,36 23 xxy find

(a) an expression fordx

dy,

(b) the x coordinates of the stationary points.

Show that the gradient of the curve between the stationary points is always negative. (a)dx

dy

xx 1232

(b) When 0dx

dy, 0123

2 xx

0)4(3 xx

4,0

xx

xx 1232 )444(3 2 xx

12)2(32 x

For 40 x , 222 x

4)2(0 2 x

012)2(3122 x

0

dx

dy

Hence for 40 x ,dx

dyis always negative.


5/27



Secondary 4



Name: ( ) Date: ________________

Class: Sec 4_____

Assignment 23: Nature of Stationary Points

1. Given that xxy 7)5( , find


dy,

(b) thex-coordinates of the stationary point. (Pg348 Ex16.1Q3)

Solution

(a)dx

dy

xxx

72

1)5(7

x

xx

72

5)7(2

x

x

72

93

(b) When 0dx

dy, 093 x

3x


6/27


2. Find the coordinates of the stationary point on the curve defined byx

xy

1 . Determine the

nature of this point. (Pg 395 Rev Ex 15 Q6(a))

dx

dy

2

2

1

1)1(21

x

xxx

2

112

x

xx

x

12

)1(22

xx

xx

12

2

2

xx

x

When ,0dx

dy 02 x

2x

2

1y

The stationary point is

2

1,2 .

x 2 2 2

+

dx

dy > 0 0 < 0

Slope / \

Hence

2

1,2 is a maximum point.


7/27


3. A curve has an equation of the form2

x

baxy ,where a and b are constants.

Given that the curve has a stationary point at (3, 5), find the value ofa and ofb.

(Pg348 Ex16.1Q6)

32xba

dxdy

At (3, 5), 0dx

dy, 0

27

2

ba

27

2ba (1)

5y ,9

35b

a (2)

Sub (1) into (2)927

235

bb

53

b

15b

9

10a


8/27


4. Given that2

122

x

xy , find


dy,

(b) thex-coordinates of two stationary points.

Show thaty increases asx increases between the stationary points. (Pg348 Ex16.1Q2)

Solution

(a)dx

dy

22

2

)2(

)12(2)2(2

x

xxx

22

22

)2(

2442

x

xxx

22

2

)2(

422

x

xx

(b) When 0dx

dy, 0422

2 xx

0)2)(1(2 xx

2,1 xx

dx

dy

22

2

)2(

422

x

xx

22)2(

)2)(1(2

x

xx

For 21 x , 310 x 023 x

0)2( 22 x

0dx

dy

Hence for 21 x , y increases as x increases between the stationary points.


9/27



Secondary 4



Name: ( ) Date: ________________

Class: Sec 4_____

Worksheet 24: Second Derivative

Determining Maximum and Minimum Points Using the Second Derivative ofyThe diagrams below shows the curve )(xfy and the graph of

dx

dyagainstx.

From the graph ofdx

dyagainstx, it is observed that

(a)dx

dy decreases asx increases throughA,

(b) the rate of change ofdx

dyatA, i.e.

2

2

dx

yd

dx

dy

dx

d

< 0 atA.

Thus a turning point is a maximumwhen 0dx

dyand 0

2

2

dx

yd.

(c)dx

dyincreases asx increases throughB,

(d) the rate of change ofdx

dyatB, i.e.

2

2

dx

yd

dx

dy

dx

d

> 0 atB.

Thus a turning point is a minimum when 0dx

dyand 0

2

2

dx

yd.


10/27


Summary

Given a curve )(xfy ,

(a) When 0dx

dyand 0

2

2

dx

ydat ax , then ))(,( afa is a turning point.

(b) If 2

2

dxyd > 0, then ))(,( afa is a minimum point.

(c) If2

2

dx

yd< 0, then ))(,( afa is a maximum point.

Example 1

Given that )2()1(2 xxy , find

(a)dx

dyand

2

2

dx

yd,

(b) the stationary values ofy and determine the nature of these values.

(a)dx

dy 2)1()2)(1(2 xxx

1242222 xxxx

33 2 x

2

2

dx

yd )33( 2 x

dx

d

x6

(b) Whendx

dy= 0, 1,1x

When 1x , 4y

2

2

dx

yd< 0

y is a maximum value.

When 1x , 0y

2

2

dx

yd> 0

y is a minimum value.


11/27


Example 2

Find the coordinates of the turning point of the curve22

18

xxy and determine whether this

point is a maximum or minimum point. (Pg354 Ex 16.1Q3)

dx

dy

22

18 xx

dx

d

38 x

2

2

dx

yd )8( 3 x

dx

d

43

x

Whendx

dy= 0, 08

3 x

2

1x

y 2

2

12

1

2

18

6

When2

1x ,

2

2

dx

yd=

4

2

13

= 48 > 0

6,

2

1is a minimum point.


12/27


Example 3

Show that the curve3

2

x

xy where 3x has neither a maximum nor a minimum point.

dx

dy

2)3(

)2()3(

x

xx

2)3(

5

x

Since 3x , 2)3( x > 0 , 0)3(

52

x

.

Hence,dx

dy 0.

3

2

x

x

y does not have any turning points.

Example 4 (Pg354 Ex 16.1Q4)

Show that the curvex

xy

1

12where 1x has neither a maximum nor a minimum point.

dx

dy

2)1(

)12()1(2

x

xx

2)1(

1

x

Since 2)1( x > 0 , 0)1(

12

x.

Hence,dx

dy 0.

x

xy

1

12does not have any turning points.


13/27


Example 5

Given that curve 4)1( xy , find

(a) an expression forx

y

d

d,

(b) the coordinates of the stationary point and determine its nature.

(a)x

y

d

d 3)1(4 x

(b) Whenx

y

d

d= 0, 0)1(4 3 x

1x 0y

2

2

d

d

x

y 2)1(12 x

When 1x ,2

2

d

d

x

y= 0 (inconclusive).

x 1 1 1+

x

y

d

d < 0 0 > 0

Slope \ /

Hence 0),1( is a minimum point.


14/27


Example 6

Given that curve 2)1(3 xy , find

(a) an expression forx

y

d

d,

(b) the coordinates of the stationary point and determine its nature.(a)

x

y

d

d 2)1(3 x

(b) Whendx

dy= 0, 0)1(3

2 x

1x 2y

2

2

d

d

x

y)1(6 x

When 1x ,2

2

ddxy = 0 (inconclusive).

x 1 1 1

+

x

y

d

d > 0 0 > 0

Slope / /

Hence 2),1( is a point of inflexion.


15/27



Secondary 4



Name: ( ) Date: ________________

Class: Sec 4_____

Assignment 25: Second Derivative

1. Find the coordinates of the stationary point(s) of the following curves and determine the nature

of each point.

(a) 2)6( xxy

(b)x

xy162

(a)x

y

d

d 2)6( xx

)6(2)6( 2 xxx

)26)(6( xxx

)63)(6( xx

)2)(6(3 xx

When ,0

d

d

x

y 0)2)(6(3 xx

2,6x

32,0y

The stationary points are (6, 0) and (2, 32).

2

2

d

d

x

y)6(3)2(3 xx

246 x

When ,6x 2

2

d

d

x

y012

(6, 0) is a minimum point.

When ,2x 2

2

d

d

x

y06

(2, 32) is a maximum point.


16/27


(b)x

y

d

d

2

162

xx

2

3)8(2

x

x

When ,0dd xy 0)8(2 3 x

83 x 2x 12y

The stationary points are (2, 12).

2

2

d

d

x

y

3

322

x

When ,2x 2

2

d

d

x

y06

(2, 12) is a minimum point.

2. The graph of baxxy 232 has a stationary point )19,3( . Find the value ofa and ofb.

Determine whether this stationary point is a maximum or a minimum.

axxx

y

26d

d 2

At )19,3( , ,0d

d

x

y 0)3(2)3(6

2 a

0654 a 9a

At )19,3( , ,19y b 23 )3(9)3(219

b 815419 8b

1812d

d2

2

xx

y

When ,3x 018)3(12d

d2

2

x

y

)19,3( is a maximum point.


17/27


3. Prove that the curve kxxy 43 has only one turning point. Find the turning point where

12k and determine the nature of this point.

kxx

y 312

d

d

When ,0d

d

x

y 012

3 kx

12

3 kx

3

12

kx

Hence, kxxy 43 has only one turning point at 312

kx .

2

2

2

36d

dx

x

y

When 12k , 1x 9y

036d

d2

2

x

y

)9,1( is a minimum point.


18/27



Secondary 4



Name: ( ) Date: ________________

Class: Sec 4_____

Worksheet 26: Maxima and Minima Problems

1. A rectangle has sides x cm and y cm. If the area of the rectangle is 16 cm2, show that itsperimeter, P cm, is given by

xxP

322 . Hence, calculate the value ofx which gives P a

stationary value and show that this value ofP is a minimum.

2cm16rectangleofArea

16xy

xy

16

cm22rectangleofPerimeter yx

P

xx

1622

xx 322

2

322

d

d

xx

P

32

2 64

d

d

xx

P

,0d

dWhen

x

P 0

322

2

x

0162 x 4x

4,0Since xx

,4When x 02

2

dx

Pd

minimum.aisHence,P


19/27


2. A piece of wire of length 104 cm, is bent to form a trapezium as shown in the diagram. Expressy in terms ofx and show that the area,A cm, enclosed by the wire is given by 220208 xxA .Find the value ofx and ofy for whichA is a maximum. (Pg356 Ex 16.3Q5)

cm104trapeziumofPerimeter

104216 yx

xy 161042

xy 852

2cm4)26(2

1trapeziumofArea xyx

A )26(2 yxx

)852(412 2 xxx 22 3220812 xxx

220208 xx

xx

A40208

d

d

40dd

2

2

xA

,0d

dWhen

x

A 040208 x

2.5x 4.10y

,2.5When x 02

2

dx

Ad

maximum.aisHence,A

y cm

cm

5x cm 5x cm

y cm

cm

5x cm 5x cm4x cm

3x cm


20/27


3. A rectangle block has a total surface area of 1.08 m2. The dimensions of the block are x m, 2xm and h m. Show that

x

xh

6

408.12

and hence express the volume of the block in terms ofx.

Find the value ofx that makes this volume a maximum.

2m.081areasurfaceTotal 08.1)22(2 hxxxhx

08.146 2 xxh

x

xh

6

408.12

32 cm2blockrrectangulaofVolume hx

V

x

xx

6

408.12

22

3

3

436.0 xx

2436.0

d

dx

x

V

8d

d2

2

x

V

,0

d

dWhen

x

V 0436.0

2 x

09.02 x 3.0x

3.0,0Since xx

,3.0When x 02

2

dx

Vd

.3.0whenvaluemaximumahasHence, xV


21/27


4. A piece of wire 120 cm long, is bent to form the shape shown in the diagram. This shapeencloses a plane region, of area 2cmA , consisting of a semi-circle of radius cm4r , a rectangle

of length cmx , and an isosceles triangle having two equal sides of length cm5r .

(i) Express x in terms of rand hence show that22 288480 rrrA .

Given that r can vary,

(ii) calculate, to 1 decimal place, the value or r for which

A has a maximum value.

(i) cm120Perimeter 120102)4( rxr

rrx 1041202 rrx 5260

22 cm382

18)4(

2

1Area rrxrr

A 22 12)5260(8)4(2

1rrrrr

2222 1240164808 rrrrr 22 288480 rrr

(ii)r

A

d

d rr 5616480

2

2

d

d

r

A5616

,0When dr

dA 05616480 rr

r5616

480

5.4

,52.4When r 02

2

dx

Ad

.5.4whenvaluemaximumahasHence, rA

4rcm

x cm

5rcm 5rcm


22/27


5. The diagram shows a greenhouse standing on a horizontal rectangular base. The verticalsemicircular ends and the curved roof are made from polythene sheeting.

The radius of each semicircle is r m and the length of

the greenhouse is ml . Given that 120 m2

of polythene

sheeting is used for the greenhouse, express l in terms

of r and show that the volume, V m3

of the

greenhouse is given by2

603r

rV

.

Given that r can vary, find, to 2 decimal places, the

value of r for which V has a stationary value. Find

this value of V and determine whether it is a

maximum or a minimum.(Nov 02)

22 mareaSurface rlr

1202

rlr

r

rl

2120

22 cm2

1greenhousetheofVolume lr

V

r

rr

22 120

2

1

2

603r

r

r

V

d

d

2

360

2r

2

2

d

d

r

Vr 3

,0d

dWhen

x

S 0

2

360

2

r

040 2 r

r

40

57.3 143V

,57.3When r 02

2

dr

Vd

.57.3whenvaluemaximumahasHence, rV

r m

lm


23/27


6. The diagram below shows the curve 142

xy . A variable point P, moves on the curve such

that PQ is always parallel to the y-axis and Q is on thex-axis. Let the x-coordinate ofP be p,

p > 0, and the area of triangle OPQ beA square units, where O is the origin.

(i) Obtain an expression forA in terms ofp.(ii) Calculate the smallest possible area of triangle OPQ.

(i) Coordinates ofP =

1

4,

2p

p

Area of triangle OPQ, A =

1

4

2

12

pp unit

2

=

1

4

2 2p

punit

2

=2

2 pp

unit2

(ii)p

A

d

d=

2

122

p

2

2

d

d

p

A=

3

4

p> 0 A is a minimum value.

Whenp

A

d

d= 0,

2

122

pp = 2

MinimumA =2

2

2

2 unit

2

= 2 unit2


24/27



Secondary 4



Name: ( ) Date: ________________

Class: Sec 4_____

Assignment 27: Maxima and Minima Problems

1. The diagram shows a container made by fixing a hollow hemisphere of radius r cm to ahollow right circular cylinder of the same radius. Water is filled to a height of h cm.

If the volume of water in the container is 144 cm3, show that

3

1442

r

rh .

Hence, find the value of r and of h such that the surface area of the container that is in

contact with the water is a minimum.

Volume of water = 144 cm3

3144

3144

1443

2

1443

2

2

32

323

23

rr

h

rhr

rhrr

rhrr

Let surface area beA cm2.

A = rhrr 22 2

=

r

r

rrr

3

14422

2

2

=3

2288 2r

r

34288

dd

2r

rrA

34576

32

2

rdr

Ad

When 0d

d

r

A, 0

3

42882

r

r

6r 6h

When 6r , 0d

d2

2

r

A

Surface area in contact with the water is a minimum when 6 hr cm.

rcm

h cm

rcm

(hr)cm


25/27


2. The diagram shows part of a curve .2

92x

y PQRS is a rectangle with )0,(pP is inscribed

in the region bounded by the curve and the x-axis. Find the dimensions of the rectangle if its

area is to be a maximum.

)0,(pP

2

9,

2

ppQ .

29

2p

PQ units

pSP 2 units

Area of PQRS ,A

292

2p

p unit2

318 pp unit2

p

A

d

d= 2318 p

2

2

d

d

p

A= a6

Whenp

A

d

d= 0, 0318

2 p

6p (p > 0)

62

92

p

When 6p ,2

2

dp

Ad= 66 < 0

The dimensions of the rectangle is 62 units by 6 units.


26/27


3. The diagram shows the cross-section of a hollow cone of height 30 cm and base radius 12 cmand a solid cylinder of radius rcm and height h cm. Both stand on a horizontal surface with the

cylinder inside the cone. The upper circular edge of the cylinder is in contact with the cone.

(i) Express h in terms ofrand hence show that the volume of the cylinder, ,cm3V is given by

322

530 rrV .

Given that r can vary,

(ii) find the volume of the largest cylinder which can stand inside the cone and show that, inthis case, the cylinder occupies

9

4of the volume of the cone.

(i)1230

30 rh

25

30 rh

rh 5260

2

560 rh

(ii) V hr2

2

5602 rr

322

530 rr

r

V

d

d

2

15

60

2r

r

2

2

d

d

r

Vr 1560

When 0d

d

r

V, 0

2

1560

2

r

r

082 rr

0)8( rr

8r ( 0r )

When ,8r V 640

2

2

d

d

r

V01560 r

Volume of the largest cylinder

640 cm3

conetheofVolume

cylinderlargesttheofVolume

)30)(12(3

14062

9

4


27/27

4. A diagram shows a toy which is made up of a right circular cone fixed to a closed cylinderwith height h cm. The slant height of the cone makes an angle of 60 with the horizontal axis

and the radius of the cone is rcm. Given that the volume of the toy is 50 cm3, show that

3

3502

r

rh .

If the total surface area of the toy isA, show that

rrA

100

3

3232 .

Hence, find the value ofrfor whichA has a stationary value.

Determine ifA is a maximum or a minimum.

Height of cone = 60tanr cm

= 3r cm

Volume of toy = hrrr

22

)3(3

1

cm

3

50 = hrr 23

3

3

32

3

350 rhr

2

3

3

350

r

r

h

3

350

2

r

rh

60cosl

r rl 2

Total surface area of the toy isA

= rlrhr 22 cm2

= )2(3

3502

2

2rr

r

rrr

cm2

=3

321003

22 r

rr cm2

=r

r

100

3

3232 cm

2

r

A

d

d

= 2100

3

32

32 rr

Whenr

A

d

d= 0,

2

100

3

3232

rr

= 0

2

100

3

3232

rr

3r

3

3232

100

r=3

3

3232

100

r = 3.0035 cm

r 3.00 cm

2

2

d

d

r

A=

3

200

3

3232

r

= 34.8 > 0

A is a minimum when r= 3.00 cm

60

rcm

h cm

ch 16 maximum and minimum problems 2012

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