Applications Of Derivatives

Chapter VIII: Applications of Derivatives Prof. D. R. Patil
1
CHAPTER VIII:APPICATIONS of
DERIVATIVES
8.1MAXIMA AND MINIMA:
Monotonicity:
The application of the differential calculus to the investigation of functions is based on a simple
relationship between the behaviour of a function and properties of its derivatives and, particularly, of the
first derivative. Increase is associated with positive derivatives and decrease with negative derivatives:
Suppose that a function  
x
f is differentiable at every point x of an interval J. Then
1. f is increasing on J if   J
x
x
f 

 ,
0
'
2. f is decreasing on J if   J
x
x
f 

 ,
0
'
3. f is non-decreasing on J if   J
x
x
f 

 ,
0
'
4. f is non-increasing on J if   J
x
x
f 

 ,
0
'
In geometric terms: Differentiable functions increase on intervals where their graphs have position
slopes and decrease on intervals where their graphs have negative slopes.
Example 4. Show, that the function 0.5
: log
f y x
 is decreasing on  
f
D and the function
x
x
x
y
g 5
2
: 2
3


 is increasing on  
g
D .

2
Example 5. Find intervals of monotonicity for the functions: x
x
y
f 3
: 3
1 
 and 2
3
4
2 4
4
: x
x
x
y
f 


Geometrically, it appears evident that if the derivative of a function takes zero values at some separate
points but retains constant sign at all other points, the function is also strictly monotone (increasing or
decreasing) in the given interval.
For instance, the function x
x
y
f sin
: 
 is increasing and   ,...
2
,
1
,
0
,
0
2
' 



 k
k
f 
Example 6. With the aid of first derivative show, that the equation 0
cos 
 x
x has one and only one
root in the interval
2
,
0

.
Local (Relative) Extrema
Let a function  
x
f be defined in a neighbourhood of a point 0
x . The value  
0
f x is said to be a local
maximum of  
x
f , if there exists such an  
0
x
N , that      
0
0 , x
N
x
x
f
x
f 


 . Then 0
x is called
the point of local maximum.
The value  
0
x
f is said to be a local minimum of  
x
f , if there exists such an  
0
x
N , that
     
0
0 , x
N
x
x
f
x
f 


 . Then 0
x is called the point of local minimum.
If    
0
x
f
x
f  or    
0
x
f
x
f  for   0
0 , x
x
x
N
x 

  , the value  
0
x
f is called the strict local
maximum or minimum, resp. When speaking of a maximum or a minimum at a point, we usually mean
a strict extremum.
If 0
x is point of local extremum of a function f and f is differentiable at 0
x , then   0
' 0 
x
f . It follows
that a function can posses local extrema at points at which the derivative is equal zero (these are called
stationary points) or at points at which the derivative doesn't exist.
The points where 0
'
f or fails to exist are commonly called the critical points of f (for the first
derivative).

3
Example 7. Find all points of local extrema for functions: x
y
f cos
:
1  , 3
2 : x
y
f  , 1
:
3 
 x
y
f
If a function f is differentiable at each   0
0 , x
x
x
N
x 
  and    
 
0
'
0
' 
 x
f
x
f for
 
0
0 ,x
x
x 


 and    
 
0
'
0
' 
 x
f
x
f for  



 0
0,x
x
x then  
0
x
f is strict local maximum
(minimum).
It means that for a local maximum  
0
x
f f is increasing in  
0
x
N 
 and decreasing in  
0
x
N 
 and for a
local minimum  
0
x
f , f is decreasing in  
0
x
N 
 and increasing in  
0
x
N 
 .
The secondderivative test:
If   0
' 0 
x
f and   0
'
' 0 
x
f , then 0
x is a point of local extremum. If   0
'
' 0 
x
f , then  
0
x
f is the
strict local maximum and if   0
'
' 0 
x
f , then  
0
x
f is the strict local minimum.
Example 8. Find all local extrema for functions: 2
3
4
1 4
4
: x
x
x
y
f 

 , 4
2 : x
y
f  ,
3 2
3 : x
y
f 
Example 9. Find two positive numbers whose sum is 20 and whose product is as large as possible.
Example 10. We have to make a can with a given volume 0

V , shaped like a right circular cylinder.
What dimensions will use the least material.

4

5

6

7

8

9
Problem:

10

11
Maximize Volume ofa Box Howto maximize the volumeof abox usingthe firstderivative of the volume.
Problem1: A sheetof metal 12 inchesby10 inchesisto be usedto make a openbox.Squaresof equal sidesx
are cut out of each cornerthenthe sidesare foldedtomake the box.Findthe value of x that makesthe
volume maximum.
Solutionto Problem1:
 We firstuse the formulaof the volume of a rectangularbox.
V = L * W * H
 The box to be made has the followingdimensions:
L = 12 - x
W = 10 - 2x
H = x
 We nowwrite the volume of the box toba made as follows:
V(x) = x (12 - 2x) (10 - 2x) = 4x (6 - x) (5 - x)
= 4x (x 2
-11 x + 30)
 We nowdetermine the domainof functionV(x).Alldimemsionsof the box mustbe positiveorzero,
hence the conditions
x > = 0 and6 - x > = 0 and5 - x > = 0
 Solve the above inequalitiesandfindthe intersection,hence the domainof functionV(x)
0 < = x < = 5

12
Use Derivativesto solve problems:Area Optimization
A problemtomaximize (optimization)the areaof a rectangle witha constantperimeterispresented.An
interactive appletisusedtounderstandthe problem.Thenananalytical method,basedonthe derivativesof a
functionandsome calculustheorems,isdevelopedinordertofindan analytical solutiontothe problem.

13
Problem:You decide toconstructa rectangle of perimeter400 mm and maximumarea.Findthe lengthand
the widthof the rectangle.
We now lookat a solutiontothisproblemusingderivativesandothercalculusconcepts.
Let x ( = distance DC) be the widthof the rectangle andy( = distance DA)itslength,thenthe areaA of the
rectangle maywritten:
A = x*y
The perimetermaybe writtenas
P = 400 = 2x + 2y
Solve equation400 = 2x + 2y for y
y = 200 - x
We nownowsubstitute y= 200 - x intothe areaA = x*yto obtain.
A = x*(200 - x)
AreaA isa functionof x.Asyou change the widthx in the applet,the areaA on the rightpanel change.
Expandthe expressionforthe areaA and write itas a functionof x.
A(x) = -x 2
+ 200x

14
we mightconsiderthe domainof functionA(x) asbeingall valuesof x inthe closedinterval [0, 200] since x
>= 0 andy = 200 - x >= 0 (if yousolve the secondinequality,youobtainx <= 0).
To findthe value of x thatgivesan area A maximum,we needtofindthe firstderivative dA/dx (A isa
functionof x).
dA/dx = -2x + 200
If A has a maximumvalue,ithappensat x such that dA/dx = 0. At the endpointsof the domainwe have A(0)
= 0 and A(200) = 0.
dA/dx = -2x + 200 = 0
Solve the above equationforx.
x = 100
dA/dx hasone zeroat x = 100.
The secondderivative d 2
A/dx 2
=-2 isnegative.(seecalculustheoremonusingthe firstandsecond
derivative todetermineextremmaof functions).The value of the areaA at x = 100 is equal to10000 mm 2
and itis the largest(maximum).Soif youselectarectangle of widthx = 100 mm and lengthy= 200 - x = 200
- 100 = 100 mm(it isa square!),youobtaina rectangle withmaximumareaequal to10000 mm 2
.
Exercises
1 - Solve the same problemasabove butwiththe perimeterequal to500 mm.
solutiontothe above exercise
widthx = 125 mm and lengthy= 125 mm.
Example1 A windowisbeingbuiltandthe bottomisa rectangle andthe top isa semicircle. If there is12
metersof framingmaterialswhatmustthe dimensionsof the window be toletinthe mostlight?
Solution

15
Oka ,let’saskthis uestio agai isslightl easie tou de sta dte s. We wanta window inthe shape
describedabove tohave amaximumarea(andhence letinthe mostlight) andhave a perimeterof 12 m
(because we have 12 m of framingmaterial). Little biteasiertounderstandinthose terms.
He e’sa sket hof the i do . The heightof the rectangularportionis h andbecause the semicircle isontop
we can thinkof the widthof the rectangularportionat 2r.
The perimeter(ourconstraint) isthe lengthsof the three sidesonthe rectangularportionplushalf the
circumference of acircle of radius r. The area (whatwe wantto maximize) isthe areaof the rectangle plushalf
the area of a circle of radius r. He e a e the e uatio s e’ll e o ki g ithi thise a ple.
I this ase e’ll sol ethe o st ai tfo h and plugthatinto the area equation.

16
The firstandsecondderivativesare,
We can see that the onlycritical pointis,
We a alsosee that the se o dde i ati eisal a s egati e i fa tit’sa o sta t a dso e a see thatthe
maximumareamustoccur at thispoint. So,for the maximumareathe semicircle ontopmusthave a radiusof
1.6803 and the rectangle musthave the dimensions3.3606 x 1.6803 (h x 2r).
Example2 Determine the areaof the largestrectangle thatcan be inscribedinacircle of radius4.
Solution This problemisbestdescribedwithasketch. He e is hat e’ e looki gfo .
We wantthe area of the largestrectangle thatwe can fitinside acircle and have all of its cornerstouchingthe
circle.

17
To dothisp o le it’seasiesttoassu e thatthe i le a d hence the rectangle) iscenteredatthe
origin. Doingthiswe knowthat the equationof the circle will be
and that the rightuppercorner of the rectangle will have the coordinates . This
meansthat the widthof the rectangle will be 2x and the heightof the rectangle will be 2y. The area of the
rectangle will thenbe,
“o, e’ e gotthe fu tio e a tto a i ize the area),butwhatisthe constraint? Well since the
coordinatesof the upperrightcornermust be on the circle we knowthat x andy must satisfythe equationof
the circle. In otherwords,the equationof the circle isthe constraint.
The firstthingto dothenis to solve the constraintforone of the variables.
“i e the poi tthat e’ e looki gatisi the fi st uad a t e k o that y must be positive andsowe can take
the + pa t of this. Pluggingthisintothe areaand computingthe firstderivative gives,
Befo e getti gthe iti al poi tslet’s oti ethat e a li it x tothe range
since we are assumingthat x isinthe firstquadrantand muststay inside the circle. Now the four

18
critical pointswe get(twofromthe numeratorand twofrom the denominator)
are,
We onlywantcritical pointsthatare in the range of possible optimal valuessothat meansthatwe have two
critical pointstodeal with: and . Notice howeverthatthe secondcritical pointisalso
one of the endpointsof ourinterval.
Now,area functioniscontinuousandwe have aninterval of possiblesolutionwithfiniteendpointsso,
“o, e a see that e’ll getthe a i u a eaif and the correspondingvalue of y is,
It lookslike the maximumareawill
be foundif the inscribedrectangle isinfacta square.
Example6 Two poles,one 6 meterstall andone 15 meterstall,are 20 metersapart. A lengthof wire is
attachedto the top of eachpole andit isalso stakedtothe ground somewherebetweenthe twopoles. Where
the wire shouldbe stakedsothat the minimumamountof wire isused?
Solution
As al a slet’ssta toff itha sket hof thissituatio .

19
The total lengthof the wire is andwe needtodetermine the valueof x thatwill minimize
this. The constraintinthisproblemisthatthe polesmustbe 20 metersapart and that x must be inthe range
. The fi stthi gthat e’ll eedtodohe e isto getthe le gthof i e i te sof x,whichis fairly
simple todousingthe PythagoreanTheorem.
Notthe i estfu tio e’ e hadto o k ith ut the e itis. Note however,thatitisa continuousfunction
a d e’ e gota i te al ithfi itee dpoi tsa dsofi di gthe a solute i i u o ’t e ui e u h o e
work than justgettingthe critical pointsof thisfunction. “o,let’sdothat. He e’sthe de i ati e.
Settingthisequal tozerogives,
It’s p o a l ee uite a hile si e ou’ e ee askedtosol e so ethi glike this. Tosol e this e’ll eedto
square bothsidestoget ridof the roots,but thiswill cause problemsaswell soonsee. Let’sfi stjusts ua e
bothsidesandsolve thatequation.

20
Note that if ou a ’t do that fa to i gdo e o , ou a al a sjustuse the uad ati fo ulaa d ou’ll get
the same answers.
Okay twoissuesthatwe needtodiscussbrieflyhere. The fi stsolutio a o e otethatIdid ’t all ita iti al
poi t… does ’t ake a se se e ause itis egati e a doutside of the a ge of possi lesolutio sa dso e
can ignore it.
Secondly,andmaybe more importantly,if youwere toplug intothe derivative youwouldnot
getzero andso is notevena critical point. Howis thispossible? Itisa solutionafterall. We’ll e allthat e
squaredbothsidesof the equationabove anditwasmentionedatthe time thatthiswouldcause
problems. We’ll e’ e hitthose p o le s. Insquaringbothsideswe’ e i ad e te tl i t odu eda e
solutiontothe equation. WhenyoudosomethinglikethisyoushouldALWAYSgobackand verifythatthe
solutionsthatyouare in fact solutionstothe original equation. I this ase e e e lu k a dthe ad
solutio alsohappe edto e outside the i te alof solutio s e e e i te estedi utthat o ’tal a s e
the case.
So, if we go back and doa quickverificationwe caninfact see that the onlycritical pointis
and thisisnicelyinourrange of acceptable solutions.
Now all that we needto dois plugthiscritical pointandthe endpointsof the wire intothe lengthformulaand
identifythe one thatgivesthe minimum
value.
So,we will
getthe minimumlengthof wire if we stake ittothe ground feetfromthe smallerpole.

21
Example1 An apartmentcomplex has250 apartmentstorent. If theyrentx
apartmentsthentheirmonthlyprofit,indollars,isgivenby,
How manyapartmentsshouldtheyrentinorderto maximize their profit?
Solution
All that e’ e eall ei gaskedtodohe e isto a i ize the p ofitsu je ttothe o st ai tthat x
mustbe inthe range .
Fi st, e’ll eedthe de i ati e a dthe iti al poi t s thatfall i the a ge
.
Since the profitfunctioniscontinuousandwe have aninterval withfiniteboundswe canfindthe
maximumvalue bysimplyplugginginthe onlycritical pointthatwe have (whichnicelyenoughinthe
range of acceptable answers) andthe endpointsof the range.
So, itlookslike theywill generate the mostprofitif theyonlyrentout200 of the apartmentsinsteadof
all 250 of them.
Note that iththese p o le s oushould ’tjustassu ethat e ti gall the apa t e ts illge e atethe ost
profit. Do notforgetthat there are all sortsof maintenance costsandthatthe more tenantsrentingapartments
the more the maintenance costswill be. Withthisanalysiswe cansee that,forthis complex atleast,something
probablyneedstobe done to get the maximumprofitmore towardsfull capacity. Thiskindof analysiscanhelp
themdetermine justwhattheyneedtodotomove towardsgoal that whetheritbe raisingrentor finda way to
reduce maintenance costs.

22
Note as well thatbecause mostapartmentcomplexeshave atleastafewunitemptyaftera tenantmovesout
a d the like thatit’spossi le thatthe oulda tuall like the a i u p ofittofall slightl u de full apa it
to take thisintoaccount. Again,anotherreasontonot justassume that maximumprofitwillalwaysbe atthe
upperlimitof the range.
Let’stake a ui klookat a othe p o le alo gthese li es.
Example2 A productionfacilityiscapable of producing60,000 widgetsina day andthe total dailycostof
producingx widgetsina dayis givenby,
How manywidgetsperdayshouldtheyproduce inorderto minimizeproductioncosts?
Solution
Here we needto minimizethe costsubjecttothe constraintthat x must be inthe range
. Note that inthiscase the cost fu tio is ot o ti uousatthe lefte dpoi ta dso e o ’t e a le to just
plugcritical pointsandendpointsintothe costfunctiontofindthe minimumvalue.
Let’sget the fi st ouple of de i ati esof the ostfu tio .
The critical pointsof the cost functionare,
No , lea l the egati e alue does ’t ake a se se i thissetti ga dso e ha e a si gle iti al poi ti
the range of possible solutions:50,000.

23
Now,as longas the secondderivativeispositive andso,inthe range of possible
solutionsthe functionisalwaysconcave upandsoproducing50,000 widgetswill yieldthe absoluteminimum
productioncost.
1 A ladder which is 5 m long slips down a wall at a rate of 2 m/s. How fast, in m/s, is the base of the
ladder moving away from the wall at the instant when its height above the ground is 3 metres?
( 4 marks)
a) 1 b) 1.5 c) 2 d) 2.5 e) 3
2 The position of a particle is given by the formula x  t
3
 4t
2
10 . At t= 2, which of the following
statements is correct? Circle each CORRECT STATEMENT. (3 marks)
I) Its velocity is increasing II) Its speed is increasing III) It is moving towards 0
3. The x coordinate of the pointon the curve y  2 x whichisnearest(4,0) isx=
a) 1 b) 2 c) 3 d) 4 e) 5 (5 marks)
4. A cone has radius5 cm andheight15 cm. It isemptyand isbeingfilledwithata constantrate of 12
cm
3
/s. Findthe rate of change of the radius,incm/s,whenthe radiusof the water is2 cm.
( 5marks)
a) 0.5 b) 1 c) 1.5 d) 2.5 e) 3
5. The motiondescribedbythe formula s  2 t 
1
4
t hasa maximumpositionof s 
a) 8 b) 4 c) 0 d) 3 e) 10 ( 4 marks)
6. The same motiondescribedbythe formula s  2 t 
1
4
t inthe interval [0,100] . The time at whichthe
average velocityoverthe entireintervalequalsthe instantaneousvelocityisatt=

24
( 4 marks)
a) 15 b) 25 c) 35 d) 50 e) 100
Section B-
7. The graphs shownbelowdescribethe motionof twoships. The firstship(A)istravellingdue westfroma
harbour. The secondrepresentsashiptravellingdue northfromthe same harbour.The shipsleave the
harbourat the same time.Use the graphs to estimate howquicklythe shipsare separatingafter1hour. Ship
A inkm, time onx axisisinhours

25
ShipB in km,time onx axisisin hours
(7 marks)
a) 64 km/h b) 42 km/h c) 34 km/h d) 10 km/h e) –20km/h
8 a) A 3 metre longtrough hasa cross sectioninthe shape of an isoscelestriangle withadepthof 80 cm and a
widthof 60 cm. If the trough isinitiallyfull andwaterallowedtodrainfromitat a rate of 450 cm
3
/minute,
findthe rate at whichthe waterlevel ischangingatthe instantwhenthe wateris1 cm deep.
b) Findthe rate at whichthe waterlevel ischangingafter2minutes.
9. a) Findthe maximumareaof a rectangle ABCDwithA at (0,0),B onthe x-axis,Ca pointinthe firstquadrant
on the curve y 
90
x
2
 9
, and D on the y-axis. ( 5 marks)
b) Suppose we considerthe same GENERALrectangle ABCD describedinparta). Write downa formulafor
the perimeterof ABCD. ( 1 mark)

26
c) Whenthe widthof the rectangle is3 units(ie whenx=3),itisknownthat the perimeterisdecreasingat2
units/second.Findthe rate at whichthe widthof the rectangle ischangingat thisinstant.
(3 marks)
8.2 MEAN VALUE THEOREM
Definition Let )
(x
f
y  be a functiondefinedonaninterval I.f issaid to have an absolutemaximumatc if


 x
x
f
c
f ),
(
)
( I and )
(c
f is calledthe absolutemaximumvalue.
Similarly, f issaidtohave an absoluteminimum at d if 

 x
x
f
d
f ),
(
)
( I and f(d) iscalled
the absoluteminimumvalue.
Theorem Fermat'sTheorem
Let )
(x
f
y  be definedanddifferentiableonanopeninterval (a,b).If )
(x
f attainsits
absolutemaximum or absoluteminimum (bothare called absoluteextremum) at c
x  , where
)
,
( b
a
c , then 0
)
(
' 
c
f .
Proof For any ),
,
( b
a
x  there existsareal numberhsuch that )
,
(
)
( b
a
h
x 
 and )
,
(
)
( b
a
h
x 
 .
Now,suppose )
(x
f attainsitsabsolute maximumat c
x  .Then we have (a,b)
h)
(c 
 and
(a,b)
h)
(c 
 , and so )
(
)
( c
f
h
c
f 
 and )
(
)
( c
f
h
c
f 
 . Now,the leftandrighthand
derivativesare givenby
0
lim
0



 


h
f(c)
h)
f(c
(c)
f'
h
, ( since 0


 f(c)
h)
f(c )
and 0
lim
0



 


h
f(c)
h)
f(c
(c)
f'
h
. ( since
0


 f(c)
h)
f(c )

27
Since )
(x
f isdifferentiable at c
x  ,the leftandright handderivatives mustbe equal,
i.e. )
(
'
)
(
' c
f
c
f 
  .This ispossible onlyif 0
)
(
' 
c
f .
The proof for )
(x
f attainingitsabsolute minimumat c
x  issimilarandisleftasan exercise.
Remark1. 0
)
(
' 
p
f NOT IMPLIES absolute max.ormin.at p
x  .
e.g. 3
)
( x
x
f  at 0

x , notmax.and min.
2. Fermat'sTheoremcan't applyto functioninclosedinterval.absolute max.orminmaybe
attainedat the end-points.Asaresult,one of the leftandright handderivativesatc may
not exist.
e.g. 1
)
2
(
)
( 2


 x
x
f definedon[ 0, 5] attainsits absolute max.at 5

x but its
righthand derivativedoesnotexist.
3. Fermat's Theoremcan't applyto functionwhichare notdifferentiable.
e.g. x
x
f 
)
( . Not differentiableat 0

x butmin.at 0

x .
Theorem: Rolle's Theorem
If a function )
(x
f satisfiesall the followingthree conditions:
(1) )
(x
f is continuousonthe closedinterval ]
,
[ b
a ,
(2) )
(x
f is differentiableinthe openinterval )
,
( b
a ,
(3) )
(
)
( b
f
a
f  ;
thenthere existsatleasta point )
,
( b
a

 suchthat 0
)
(
' 

f .

28
Proof Since )
(x
f iscontinuouson ]
,
[ b
a  )
(x
f isbounded
(i) M
m  , where m(min), M(Max) are constant.
 ]
,
[
,
)
( b
a
x
M
x
f
m 



 ]
,
[
,
)
( b
a
x
M
x
f 


 )
,
(
,
0
)
(
' b
a
x
x
f 


(ii) M
m  , the max.and min.cannot bothoccur at the endpointsa,b.
 )
,
( b
a
p 
 such that M
p
f 
)
(
i.e. )
(
)
( x
f
p
f  x
 sufficientlyclosedto p.
By Fermat'sTheorem, )
(
' p
f existandequal to 0.
Example Define 1
)
2
(
)
( 2


 x
x
f on[0,4].Note that 5
)
4
(
)
0
( 
 f
f .
We have 2)
2(
)
(
' 
 x
x
f and so 0
(2)
' 
f . Since )
4
,
0
(
2 ,Rolle's Theoremisverified.
The geometricsignificance of Rolle'stheoremisillustratedinthe followingdiagram.

29
If the line joining the end points ))
(
,
( a
f
a and ))
(
,
( b
f
b is horizontal (i.e. parallel to the x-axis) then
there must be at least a point  (or more than one point) lying between a and b such that the
tangentat thispointishorizontal.
Theore :Lagra ge’s Mea Value Theore
If a function )
(x
f is
(1) Continuousonthe closedinterval ]
,
[ b
a and
(2) Differentiableinthe openinterval )
,
( b
a ,
thenthere existsatleasta point )
,
( b
a

 suchthat
)
(
'
)
(
)
(

f
a
b
a
f
b
f



.
Proof Considerthe function g definedby
)
(
)
(
)
(
)
(
)
( a
f
a
x
a
b
a
f
b
f
x
g 



  )
(x
g isdifferentiable andcontinuouson )
,
( b
a .
Let )
(
)
(
)
( x
f
x
g
x
h 

 )
(x
h is alsodifferentiableandcontinuouson )
,
( b
a .
We have .
0
)
(
,
0
)
( 
 b
h
a
h
 By Rolle'sTheorem, )
,
( b
a

 such that 0
)
(
' 

h
 0
)
(
'
)
(
' 
 
 f
g

a
b
a
f
b
f
f



)
(
)
(
)
(
' 

30
Remark: 1. The Mean Value Theoremstill holdsfor b
a  .
a
b
a
f
b
f
p
f



)
(
)
(
)
(
' .
2. Anotherformof Mean Value Theorem )
)(
(
'
)
(
)
( a
b
p
f
a
f
b
f 


3. The value of p can be expressedas )
( a
b
a
p 

  , 1
0 
 .
 )
))(
(
(
'
)
(
)
( a
b
a
b
a
f
a
f
b
f 



 
Example Use the Mean Value Theorem.show b
a 
(a) b
a
b
a 

sin
sin
(b) b
a
x
bx
ax


 cos
cos
, 0

x
(c) p
x
px

sin
, 0
,
0 
 x
p .
Example By usingMean Value Theorem, show that
)
( a
y
e
e
e a
a
y



for all real values y and a .
Example Let R
b
a 
, such that b
a  and )
(x
f be a differentiablefunctionon R suchthat 0
)
( 
a
f ,
0
)
( 
b
f and )
(
' x
f is strictlydecreasing.Show that 0
)
(
' 
b
f .
Example Let )
(x
f be a continuousfunctiondefinedon[ 3, 6 ]. If )
(x
f isdifferentiable on( 3, 6 ) and
3
9
)
(
' 

x
f . Showthat 36
)
3
(
)
6
(
18 

 f
f .

31
Example Let 0
1
1
)
( a
x
a
x
a
x
P n
n
n
n 


 
  be a polynomialwithreal coefficients.
If ,
0
1
0
1






a
n
a
n
a n
n
 byusingMean Value Theorem,show thatthe equation 0
)
( 
x
P has
at leastone real root between0and 1.
Example Let f be a real-valuedfunctiondefined on )
,
0
(  .If )
(
' t
f isan increasingfunction,
showthat )
)(
1
(
'
)
(
)
(
)
)(
(
'
)
( n
t
n
f
n
f
t
f
n
t
n
f
n
f 





 ))
1
,
(
( 
 n
n
t
Example Let f be a real-valuedfunctionsuchthat
2
)
(
)
(
)
( y
x
y
f
x
f 

 , )
,
( R
y
x 

Showthat f is a differentiablefunction.
Hence deduce that k
x
f 
)
( for all R
x , where k isa real constant.
Example Let )
(x
f be a functionsuchthat )
(
' x
f isstrictlyincreasingfor 0

x .
(a) UsingMean Value Theorem,orotherwise,showthat
)
1
(
'
)
(
)
1
(
)
(
' 



 k
f
k
f
k
f
k
f
(b) Hence,deduce that
)
(
'
)
3
(
'
)
2
(
'
)
1
(
)
(
)
1
(
'
)
2
(
'
)
1
(
' n
f
f
f
f
n
f
n
f
f
f 
 








Theorem:Cauchy Mean Value Theorem
Let )
(x
f and )
(x
g such that

32
(i) )
(x
f and )
(x
g are continuouson[ a, b ].
(ii) )
(x
f and )
(x
g are differentiable on( a,b ).
Thenthere isat leastone points )
,
( b
a
p  suchthat
)
(
'
)]
(
)
(
[
)
(
'
)]
(
)
(
[ p
f
a
g
b
g
p
g
a
f
b
f 

 .
Proof Let )
(
)]
(
)
(
[
)
(
)]
(
)
(
[
)
( x
f
a
g
b
g
x
g
a
f
b
f
x
h 


 , b
x
a 
 .
 (i) )
(x
h is continuouson[ a, b ].
(ii) )
(x
h is differentiableon( a, b ).
By Mean Value Theorem, )
,
( b
a
p 
 suchthat
a
b
a
h
b
h
p
h



)
(
)
(
)
(
' , hence the resultisobtained. (
Why? )
Remark: Suppose that )
(x
f and )
(x
g are differentiableon( a, b ) and that 0
)
(
'
)
(
' 
 x
g
x
f ,
)
,
( b
a
x 
 then
)
(
'
)
(
'
)
(
)
(
)
(
)
(
p
g
p
f
a
g
b
g
a
f
b
f



.
Thisis useful toestablishaninequalitybyusinggeneralizedmeanvalue theorem.
Example (a) Let f and g be real-valuedfunctionscontinuouson ]
,
[ b
a anddifferentiable in
)
,
( b
a .
(i) By consideringthe function )]
(
)
(
)[
(
)]
(
)
(
)[
(
)
( a
f
b
f
x
g
a
g
b
g
x
f
x
h 

 on ]
,
[ b
a , or otherwise,
show that there is )
,
( b
a
c  such that )]
(
)
(
)[
(
'
)]
(
)
(
)[
(
' a
f
b
f
c
g
a
g
b
g
c
f 


(ii) Suppose 0
)
(
' 
x
g forall )
,
( b
a
x  . Showthat 0
)
(
)
( 
 a
g
x
g for any )
,
( b
a
x  .
If,in addition,
)
(
'
)
(
'
x
g
x
f
isincreasingon )
,
( b
a ,showthat
)
(
)
(
)
(
)
(
)
(
a
g
x
g
a
f
x
f
x
P


 isalso

33
Increasingon )
,
( b
a .
(b) Let




















0
1
4
,
0
1
cos
sin
1
cos
)
(
x
if
x
if
x
x
x
e
x
Q
x

Showthat Q is continuousat 0

x and increasingon 





4
,
0

.
Hence or otherwise,deduce thatfor 






4
,
0

x ,  
x
0
x
dt
)
t
(
Q .
8.3 CURVE SKETCHING
Sample Problem#1:
f(x) = x3 - 6x2 + 9x + 1
1. Lookfor any asymptotes: Polynomial functionsdonothave
asymptotes:
a) vertical: No vertical asymptotesbecause f(x)
continuousforall x
b) horizontal: No horizontal asymptotesbecause
f(x) isunboundedas x  
2. Intercepts:

34
a) y-intercepts: f(0) = 1 y-intercept:(0,1)
b) x-intercepts: difficulttofind - skip
3. Increasing/decreasing:
a) take the firstderivative: f'(x) = 3x2 - 12x + 9
b) set itequal to zero: 3x2 - 12x + 9 = 0
c) solve forx: 3(x2 - 4x + 3) = 0
3(x - 1)(x - 3) = 0
x = 1, x = 3
d) where isf'(x) undefined? nowhere
e) signanalysis:
Plotthe numbersfoundabove ona number
line. Choose testvaluesforeachinterval
createdand evaluate the firstderivative
f'(0) = 3(0)2 -12(0) + 9 = 9 positive f(x) increasingon(  ,1).
f'(2) = 3(2)2 -12(2) + 9 = -3 negative  f(x) decreasingon(1,3).
f'(4) = 3(4)2 -12(4) + 9 = 9 positive  f(x) increasingon(3, ).

35
f '(x)
1 3
+ _
+
4. Critical points:
a) forwhichvaluesof x (foundabove
in 3) isf(x) defined? x = 1 and x = 3
Note: The valuesof x foundinsteps3a - 3c will alwaysbe inthe domainof f(x) (and therefore defined).
However,valuesof x foundinstep3dmay or may not be defined.
b) findcorrespondingvaluesof y: f(1) = (1)3 - 6(1)2 + 9(1) + 1 = 5
f(3) = (3)3 - 6(3)2 + 9(3) + 1 = 1
c) critical points: (1,5) and (3,1)
5. Testcritical pointsformax/min:
SECONDDERIVATIVETEST
a) take the secondderivative: f''(x) = 6x - 12
b) substitute x-coordof crit.pt(s): f''(1) = 6(1) - 12
= -6 (negative  max)
f''(3) = 6(3) - 12
= 6 (positive  min)

36
c) label yourpoint(s): max: (1,5)
min: (3,1)
or :
FIRST DERIVATIVETEST
a) f(x) isincreasingbefore x =1 and
decreasingafterx = 1. (1,5) is a maximum
b) f(x) isdecreasingbefore x =3 and
increasingafterx = 3. (3,1) is a minimum
6. Concave up/concave down:
a) setf''(x) equal tozero 6x - 12 = 0
b) solve forx x = 2
c) where isf''(x) undefined nowhere
d) sign analysis:
createdand evaluate the secondderivative
f''(1) = 6(1) - 12 = -6 negative f(x) concave downon ( ,2)


37
f''(3) = 6(3) - 12 = 6 positive  f(x) concave upon (2, )

f ''(x)
+
2
_
7. Findany inflectionpoints:
a) forwhichvaluesof x (foundin6)
is f(x) defined? x = 2
Note: The valuesfoundinsteps6a - 6b will alwaysbe inthe domainof f(x) (and therefore defined). However,
valuesof x foundinstep6c may or may not be defined.
b) findcorrespondingvalue of y: f(2) = (2)3 -6(2)2 + 9(2) + 1 = 3
c) f(x) changesfromconcave upto
concave downat x = 2, so (2,3) is
an inflectionpoint. inflectionpoint:(2,3)
8. Note ina chart yourpointsobtained: x y
______
1 5 (maximumpoint.)

38
3 1 (minimumpoint.)
2 3 (inflectionpoint.)
0 1 (y - intercept)
9. Plotall pointsonthe coordinate plane,andsketchinthe restof the graph. Be sure to
include all maximumpoints,minimumpoints,andinflectionpoints:
_
_
_
_
_
_
5
4
2
1
-1
| | | | | |
| | |
-3 -2 -1 1 2 3 4 5 6
f(x) = x - 6x + 9x + 1
3 2
Sample Problem#2:
f(x) = 3x5 - 5x3

39
1. Lookfor any asymptotes:
a) vertical: No vertical asymptotesbecause f(x)
iscontinuousforall x.
b) horizontal: No horizontal asymptotesbecause
f(x) isunboundedas x  
2. Intercepts:
a) y-intercepts: f(0) = 0 y-int:(0,0)
b) x-intercepts: 3x5 - 5x3 = 0
x3(3x2 - 5) = 0
x = 0, x = +
- 5/3
intercepts: (0,0)
( 5/3 ,0)
(- 5/3 ,0)
a) take the firstderivative: f'(x)=15x4 - 15x2
b) set itequal to zero: 15x4 - 15x2 = 0
c) solve forx 15x2(x2 -1) = 0
x = 0, x = 1, x = -1
d) where is f'(x) undefined? nowhere

40
e) signanalysis:
createdand evaluate the firstderivative
f'(-2)=15(-2)4 -15(-2)2 = 180 positive  f(x) increasingon  
, 1
 
f'(-1/2)=15(-1/2)4 - 15(-1/2)2 = -45/16 negative  f(x) decreasingon  
1,0

f'(1/2) = 15(1/2)4 - 15(1/2)2 = -45/16 negative  f(x) decreasingon  
0,1
f'(2) = 15(2)4 - 15(2)2 = 180 positive  f(x) increasingon  
1,
f '(x)
+
0
_
+ _
-1 1
4. Critical points:
in 3) isf(x) defined? x = 0, x = -1, and x = 1
b) findcorrespondingvaluesof y: f(0) = 3(0) - 5(0) = 0
f(-1) = 3(-1)5 - 5(-1)3 = 2

41
f(1) = 3(1)5 - 5(1)3 = -2
c) critical points: (0,0), (-1,2), (1,-2)
5. Testcritical points formax/min:
a) take the secondderivative: f''(x) = 60x3 - 30x
b) substitute x-coord(crit.pts.): f''(-1) = 60(-1)3 - 30(-1)
= -30 (negative  max)
f''(0) = 60(0) - 30(0)
= 0 (zero  testfails,
mustuse the first
derivative test)
f''(1) = 60(1)3 - 30(1)
= 30 (positive  min)
c) label yourpoints: (-1,2) : max
(1,-2) : min
(0,0) : unknown at this time
or:
a) f(x) isincreasingbefore x =-1 and

42
decreasingafterx = -1 (-1,2) isa maximum.
b) f(x) isdecreasingbefore x =0 and
decreasingafterx = 0 (0,0) is neithera max nor a min.
c) f(x) isdecreasingbefore x =1 and
increasingafterx = 1 (1,-2) isa minimum.
a) setf''(x) equal tozero 60x3 - 30x = 0
b) solve forx 30x(2x2 - 1) = 0
x = 0, x = 2 /2, x= - 2 /2
c) where isf''(x) undefined nowhere
d) sign analysis:
createdand evaluate the secondderivative
f''(-2) = 60(-2)3 - 30(-2) = -420 negative 
f(x) concave down on  
, 2 / 2
 
f''(-1/2) = 60(-1/2)3 - 30(-1/2) = 15/2 positive 
f(x) concave up on  
2 / 2,0


43
f''(1/2) = 60(1/2)3 - 30(1/2) = -15/2 negative 
f(x) concave down on  
0, 2 / 2
f''(2) = 60(2)3 - 30(2) = 420 positive 
f(x) concave up on  
2 / 2,
f ''(x)
+
0
_ + _
2 /2
- 2 /2
a) forwhichvaluesof x (foundabove)
is f(x) defined? x = 0, x = 2 /2, x= - 2 /2
b) findcorrespondingvaluesof y: f(0) = 3(0) - 5(0) = 0
f( 2 /2) = 3( 2 /2)5 - 5( 2 /2)3 = -7 2 /8
f(- 2 /2) = 3(- 2 /2)5 - 5(- 2 /2)3 = 7 2 /8
c) f(x) changesfromconcave downto
concave up at x = - 2 /2 so
(- 2 /2,7 2 /8) isan inflectionpoint.inflectionpoint.:(- 2 /2, 7 2 /8)
f(x) changesfromconcave upto

44
concave downat x = 0 so(0,0) is an
inflectionpoint. inflectionpoint.:(0,0)
f(x) changesfromconcave downto
concave up at x = 2 /2 so
( 2 /2,-7 2 /8) is an inflectionpoint.inflectionpoint.:( 2 /2,-7 2 /8)
8. Note ina chart yourpointsobtained: x y
______
0 0 (y-intercept,inflectionpoint)
-1 2 (maximumpoint)
1 -2 (minimum.point)
2 /2 -7 2 /8 (inflectionpoint)
- 2 /2 7 2 /8 (inflectionpoint)
5/3 0 (x-intercept)
- 5/3 0 (x-intercept)
9. Plotall pointson the coordinate plane,andsketchinthe restof the graph. Be sure
to include all maximumpoints,minimumpoints,andinflectionpoints:

45
3
2
1
-2
-3
f(x) = 3x - 5x
5 3
1 2 3
_
_
_
_
_
_
| | |
| | |
-3 -2 -1
f(x) = 3x - 5x
5 3
1 2 3
3
2
1
-2
-3
Sample Problem#3:
f(x) =
x2 + 1
x2 - 9
1. Lookfor any asymptotes:
a) vertical: forwhichvaluesof x
is f(x) undefined?(i.e.:whenis
the denominatorzero?) x2 - 9 = 0
x2 = 9
x = 3, x= -3
3
lim ( )
x
f x


  AND
3
lim ( )
x
f x


  therefore x= -3 is a vertical asymptote.

46
3
lim ( )
x
f x


=  AND
3
lim ( )
x
f x


=  therefore x= 3 isa vertical asymptote.
b) horizontal:
2
2
1
lim
9
x
x
x



=
2
2 2
2
2 2
1
lim
9
x
x
x x
x
x x



=
2
2
1
1
lim
9
1
x
x
x



=
1 0
lim
1 0
x


=
1
lim
1
x
= 1
therefore y= 1 isa horizontal asymptote
2. Intercepts:
a) y-intercepts: f(0) = -1/9 y-int: (0,-1/9)
b) x-intercepts: no x-int.(the numeratorisalways
positive)
a) take the firstderivative: f'(x) =
(2x)(x2 - 9) - (x2 + 1)(2x)
(x2 - 9)2
=
2x3 - 18x -2x3 - 2x
(x2 - 9)2
b) set itequal to zero: =
-20x
(x2 - 9)2 = 0

47
c) solve forx (whendoesthe -20x = 0
numerator= 0?) x = 0
d) where isf'(x) undefined? x = 3, x = -3
e) signanalysis:
f'(-5) =
-20(-5)
((-5)2 - 9)2 =
25
64
positive  f(x) isincreasingon  
, 3
 
f'(-1) =
-20(-1)
((-1)2 - 9)2 =
5
16
positive  f(x) isincreasingon(-3,0)
f'(1) =
-20(1)
((1)2 - 9)2 =
-5
16
negative  f(x) isdecreasingon(0,3)
f'(5) =
-20(5)
((5)2 - 9)2 =
-25
64
negative  f(x) isdecreasingon  
3,
f '(x)
+
0
_
+ _
-3 3
4. Critical points:
in 3) isf(x) defined? x = 0

48
b) findcorrespondingvaluesof y: f(0) = 1/9
c) critical points: (0,1/9)
5. Testcritical pointsformax/min:
a) take the secondderivative: f''(x) =
-20(x2 - 9)2 - (-20x)(2(x2 - 9)(2x))
(x2 - 9)4
f''(x) =
60x2 + 180
(x2 - 9)3
b) substitute x-coord(crit.pt.) f''(0) =
60(0)2 + 180
((0)2 - 9)3 =
20
-81
(negative  max)
c) label yourpoints: (0,1/9) : maximum
or:
a) f(x) isincreasingbefore x =0 and

49
decreasingafterx = 0 (0,1/9) : maximum
a) setf''(x) equal to0
60x2 + 180
(x2 - 9)3 = 0
b) solve forx (whendoesthe
numerator= 0?) 60(x2 + 3) = 0
x2 + 3 is never0
c) where isf''(x) undefined x = 3 and x = -3
d) sign analysis:
f''(-4) =
60(-4)2 + 180
((-4)2 - 9)3 =
1140
343
positive  f(x) concave upon  
, 3
 
f''(0) =
60(0)2 + 180
((0)2 - 9)3 =
20
-81
negative  f(x) concave downon (-3,3)
f''(4) =
60(4)2 + 180
((4)2 - 9)3 =
1140
343
positive  f(x) concave upon (3,  )
f ''(x)
-3 3
+ _
+

50
a) forwhichvaluesof x (foundin6)
is f(x) defined? f(x) isundefinedatx = -3 andat x = 3
therefore we canhave noinflectionpoints
8. Note ina chart yourpointsobtained x y
_______
0 -1/9 (maximumpoint.)
9. Plotall pointsandasymptotesonthe coordinate plane andsketchinthe restof the graph usingthe
informationfoundabove.
1 2 4 5
| | | | |
_
_
_
_
_
_
_
4
3
2
-1
-2
-3
| | | | |
-5 -4 -2 -1
x + 1
2
x - 9
2
f(x) =
1 2 4 5

51
I. Maximum/Minimum; Concavity
A. Consider f(x)=x2
-4x. Use algebra (complete the square) to find the following.
Vertex__________
Is the vertex a maximum or a minimum?_______________
Is f(x) is concave up (CU) or concave down (CD)?_______________
Store the function, its first derivative and its second derivative in your graphing calculator. Graph them one
at a time to answer the following questions.
Y1=f(x)=x2
-4x Y2=f=(x)=__________Y3=f@(x)=__________
What is the root of Y2?_____ Are the y values of f@(x) positive or negative?___
Using the example A, predict the zero of y= and the sign of y@ after finding the vertex and concavity
algebraically.
Algebraic work for B: Algebraic work for C:

52
Algebraic work for D:
Function Vertex Max/Min Concavity Zero of y= Sign of y@
B. y=-x2
+8x
C. y=x2
+6x-4
D. y=-2x2
+4x+1
Conjecture 1: Write a statement that relates the derivative and the max/min of a function.

53
Conjecture 2: Write a statement that relates the second derivative and the concavity of a function.

54
Conjecture 3: Write a statement that relates the sign of the second derivative and whether the zero of the first
derivative is a max/min of the function.
Find the vertex and concavity of the following functions using what you have learned about the first and second
derivatives.
Function dy
dx
zero of
dy
dx
Vertex of
y
2
2
d y
dx
Sign of
2
2
d y
dx
Vertex of y
is max/min
E. y=x2
-3x
F. y=-2x2
+8x+1
II. Increasing/Decreasing Functions
A. Consider y=x2
-5x-1. Use the graph of the function to determine the following; you may use the max/min
function on the calculate menu to speed the process. Graph and label y on grid A.
For which interval(s) is y increasing (use x-values to write intervals)?_______________
For which interval(s) is y decreasing (use x-values to write intervals)?_______________

55
Find Dx[y]:____________________ Graph and label it on grid A.
What is the sign of Dx[y] on the interval(s) for which y is increasing?________________
What is the sign of Dx[y] on the interval(s) for which y is decreasing?_______________
Note--I am asking for the sign of the y-values on a particular intervals of x-values: choose an x-value in the
interval and replace that value for x in the derivative to determine the sign of y-values of the derivative for that
interval of x-values or use the graph of the derivative to determine the sign.
Using example A in this section, predict the sign of f=(x) after finding the interval(s) for which f(x) is increasing/decreasing graphically.
Then, graph and label each function and its derivative on the corresponding grid.
Function Interval(s) for
which f(x) is
increasing
Sign of
f=(x)
Interval(s)
for which
f(x) is
decreasing
Sign of
f=(x)
f=(x)
B. f(x)=-x2
+2x
C. f(x)=x2
+6x+5
D. f(x)=-x2
+10x

56
Conjecture 4: Write a statement that relates the sign of the first derivative and whether the function increases or
decreases.
Create a number line using the critical values of y= (where y==0 or dne) and what you have learned about the
sign of the first derivative to determine the intervals for which the following functions are increasing/decreasing.
Function y= Critical
values of y=
y= number
line to
determine
sign of y=
Interval(s)
for which y
is increasing
Interval(s)
for which y
is
decreasing
E. y=x2
-6x+5
F. y=-x2
+x+1
III. Summary of Curve Sketching
An inflection point is a point on the graph where the function changes concavity. To find possible inflection points,
find where y=>=0 or dne. Using this information and the conjectures you have made in this activity, analyze the
following functions completely.
1. y=5+3x2
-x3
y= Critical y= number line to Interval(s) Interval(s)

57
values of
y=
determine sign of
y=
for which y
is increasing
for which y
is
decreasing
y=> Critical
values of
y@
y@ number line to
determine sign of
y@
Interval(s)
for which y
is CU
Interval(s)
for which y
CD
Use the critical values of y= and their sign in y=> to determine the relative max/min(s) of y:
Use the critical values of y=> to determine any inflection points of y:
IV. Application
A rectangular plot is to be enclosed by a fence and then divided into two plots by another fence joining the mid -
points of the two opposite sides. If the total fenced area is to be
24 yd2
, what dimensions will minimize the length of the fence used?

58
(i) Draw a diagram of the rectangular plot. Label the two opposite sides x and the 3 parallel fence pieces y.
(ii) Write an equation for the area of the plot using x, y and the given area. Solve the equation for y in terms of
x.
(iii) Write an equation for the total amount of fencing in terms of x and y. Call this equation P for the total
perimeter of the two plots
(iv) We want to minimize P. Using the equation in (ii), rewrite P in terms of x only. In order to minimize the
perimeter, differentiate P(x) with respect to x, find its critical values, and evaluate the critical value(s) in P@(x) to
determine whether or not a maximum or minimum exists at the value.
(v) Since the problem asks for the dimensions, find the y value in (ii). Be sure to include units in your answer.
(vi) Choose x values less than and greater than your answer; evaluate the corresponding y values using (ii). Find
the total amount of fencing required with your answer and the two dimensions you have just chosen using (iii).
Does your answer minimize the length of fence used compared with the other two dimensions?
Exercise:
In questions 1 to 4, use the second derivative to identify the location and nature of the stationary points, and
hence sketch the graph.
1. 2
1
9 3
4
y x x
  

59
2. 3 2
3 24
y x x x
  
3. 3 5
15
y x x
 
4. 6 4
6
y x x
 
In questions 5 to 8, write down the values of x for which the function is undefined (i.e. the values where the
function cannot be evaluated). Then use the second derivative to locate and identify the stationary points, and
sketch the graph.
5.
1
9
y x
x
 
6.
2
y x
x
 
7. 2
2
1
y x
x
 
8.
3
2
6
y x x
 
9. For the function 4
y x x
 
(i) find
dy
dx
and
2
2
d y
dx
(ii) find the co-ordinates of the turning point and determine its nature.
10. The equation of a curve is y = (6 )
x x
 .
Find the x co-ordinate of the only turning point and show that the turning point is a maximum.

60
11. For the curve
5 3
2 2
10
y x x
  ,
(i) find the values of x for which y = 0,
(ii) show that there is a minimum on the curve when x = 6 and calculate the y value of this
minimum, giving the answer correct to 1 decimal place,
(iii) and hence sketch the curve.
12. Charlie wants to add an extension with a floor area of 18m2
to the back of his house. He wants to use
the minimum possible number of bricks, so he wants to know the smallest perimeter he can use. The
dimensions, in metres, are x and y as shown.
HOUSE
x
y
(i) Write down an expression for the area A in terms of x and y.
(ii) Write down an expression, in terms of x and y, for the total length, T, of the outside walls.
(iii) Hence show that
T = 2x +
18
x
(iv) Find
dT
dx
and
2
2
d T
dx
(v) Find the dimensions of the extension that give a minimum value of T, and confirm that it is a
minimum.
13. fish tank with a square base and no top is to be made from a thin sheet of toughened glass. The
dimensions are as shown.
(i) Write down an expression for the volume V in terms of x and y.

61
(ii) Write down an expression for the total surface area A in terms of x and y.
The tank needs a capacity of 0.5m3
and the manufacturer wishes to use the minimum possible amount
of glass.
(iii) Deduce an expression for A in terms of x only.
(iv) Find
dA
dx
and
2
2
d A
dx
(v) Find the values of x and y that use the smallest amount of glass, and confirm that these give the
minimum value.
14. An open cylindrical rubbish bin, with volume is
8
 3
,
m is to be made from a large sheet of metal. To
keep costs as low as possible, the bin should have the minimum possible surface area, A m2
.
(i) Using the fact that the volume is 3
8
m

, write down an expression for the height, h, in terms of
the radius, r (both in metres).
(ii) Use this expression to show that A can be written as:
A = 2
4
r
r

 
(iii) Find
dA
dr
and
2
2
d A
dr
, and hence find the dimensions which make the area a minimum.
(iv) Find the corresponding minimum area.
15. A closed rectangular box is made of thin card. The height is h cm, the width is x cm, its length is three
times its width, and the volume of the box is 972 cm3
.
(i) Write down an expression for h in terms of x.
(ii) Show that the surface area, A can be written as 2 2592
6
A x
x
  .
(iii) Find
dA
dx
and use it to find a stationary point.

62
(iv) Find
2
2
d A
dx
and use it to verify that the stationary point gives the minimum value of A.
(v) Hence find the minimum surface area and the corresponding dimensions of the box.
16. A ship is to make a voyage of 100 km at a constant speed of v km h-1
. The running cost of the ship is
2 2000
£ 0.8v
v
 

 
 
per hour, and this is to be kept to a minimum.
(i) Write down the time taken to go 100 km at v km h-1
.
(ii) Hence write down the total cost, £C, of travelling 100 km at v km h-1
.
(iii) By considering
dC
dv
find the speed which keeps the cost of the journey to a minimum. (Give
your answer to the nearest km h-1
).
(iv) Find the minimum cost of the voyage.
17. The diagram shows a right-angled triangle with an area of 8 cm2
.
y
(i) Write down an expression for y in terms of x.
(ii) Write down an expression for the sum, S, of the squares of these two numbers in terms of x.
(iii) Find the least value of the sum of their squares by considering
dS
dx
and
2
2
d S
dx
.
(iv) Hence write down the shortest possible length for the hypotenuse.
x

63
1. Maximum at ( 3
/2, 7) 2. Maximum at ( 2, 28)
Minimum at ( 4, 80)
3. Minimum at ( 3, 162) 4. Minimum at ( 2, 32)
Point of inflection at ( 0, 0) Maximum at ( 0, 0)
Maximum at ( 3, 162) Minimum at ( 2, 32)
5. The function cannot be evaluated when x = 0.
Maximum at ( 1
/3, 6)
Minimum at ( 1
/3, 6 )
6. The function cannot be evaluated when x  0.
Minimum at ( 1, 3)
7. The function cannot be evaluated when x = 0.
Minimum at ( 1, 2)
Minimum at ( 1, 2)
8. The function cannot be evaluated when x < 0.
Maximum at (16, 32)
9. (i)
2
1
dy
dx x
 
3
2
2
2
d y
x
dx


(ii) The turning point at ( 4, 4) is a minimum.
10. 2
11. (i) x = 0, 10
(ii) y = 58.8 (to 1 decimal place)
12. (i) A = xy = 18

64
(ii) T = 2x + y
(iv) 2
18
2
dT
dx x
 
2
2 3
36
d T
dx x

(v) x = 3 and y = 6
13. (i) 2
V x y

(ii) 2
4
A x xy
 
(iv) 2 2
A x
x
 
(v) 2
2
2
dA
x
dx x
 
2
2 3
4
2
d A
dx x
 
(vi) x = 1 and y = ½
14. (i) 2
1
8
h
r

(iii) 2
2
4
dA
r
dr r


 
2
2 3
2
2
d A
dr r


  r = ½ m, h = ½ m
(iv) Minimum area = 2
3
4
m

(= 2.36 m2
to 3 s.f.)
15. (i) 2
324
h
x

(iii) 2
2592
12
dA
x
dx x
  stationary point when x = 6 and h = 9.
(iv)
2
2 3
5184
12
d A
dx x
 
(v) Minimum area = 648 cm2
. Dimensions: 6 cm  18 cm  9 cm.
16. (i) Time taken =
100
v

65
(ii) C = 80v + 2
200000
v
(iii) v = 17 kmh –1
(iv) £2,052 (to nearest £).
17. (i) y =
16
x
(ii) S = x2
+ 2
256
x
(iii) Minimum value of S = 32
(iv) Shortest possible length of hypotenuse = 4 2 cm  5.66cm (to 3 s.f.)

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