Max

ima

and

Min

ima

1U

nive

rsita

sSask

atch

ewan

en

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

How

toSo

lve

Ap

pli

edM

axM

in“W

ord

”Pro

ble

ms

Step

1:

Dra

wa

sket

ch,i

fap

pro

pri

ate.

Step

2:

Mak

ea

list

of

var

iab

les

.D

on

’tex

pec

tto

get

itri

gh

tim

med

iate

ly,

you

may

hav

eto

com

eb

ack

and

add

more

.

Step

3:

Mak

ea

list

of

sym

bols

.

Step

4:

Mak

ea

list

of

rela

tion

s.

Agai

n,y

ou

may

hav

eto

add

more

asth

ep

rob

lem

dev

elop

s.

Step

5:

Iden

tify

the

dep

end

ent

vari

able

wh

ose

extr

eme

valu

eis

tob

efo

un

d,an

dth

ein

dep

end

ent

vari

able

up

on

wh

ich

itd

epen

ds.

Exp

ress

the

ind

epen

den

tva

riab

leas

afu

nct

ion

of

the

ind

epen

den

tva

riab

le,i

fth

isis

poss

ible

.

Step

6:

Diff

eren

tiat

eth

efu

nct

ion

,or

equ

atio

nre

lati

ng

the

two

vari

able

s.

Step

7:

Solv

eth

em

ath

emat

ical

pro

ble

m.

Step

8:

Tra

nsl

ate

the

pro

ble

mb

ack

into

Engli

sh,a

nd

be

sure

that

itis

are

ason

able

solu

tion

.

2M

axim

aan

dM

inim

aU

nive

rsita

sSask

atch

ewan

en

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

We

wil

lex

amin

eva

riou

sty

pes

of

op

tim

izat

ion

pro

ble

ms

that

aris

e:

Max

imu

mar

eas

encl

ose

din

sid

ea

giv

enre

gio

nIn

its

most

gen

eral

form

,th

isty

pe

of

pro

ble

min

volv

esth

eco

mp

uta

tion

of

the

larg

est

area

regio

nly

ing

insi

de

anoth

ergiv

enre

gio

n.

Usu

ally

the

giv

enre

gio

nh

asa

“sim

ple

”b

ou

nd

ary,

i.e.

,a

fair

lyst

and

ard

geo

met

ric

ob

ject

like

atr

ian

gle

,re

cata

ngle

,or

elli

pse

,or

its

isb

ou

nd

edab

ove

and

bel

ow

by

the

gra

ph

sof

two

rela

tive

lysi

mp

lefu

nct

ion

s.

Pro

ble

m1:

Fin

dth

ear

eaof

the

larg

est

rect

angle

that

can

be

insc

rib

edin

ari

gh

ttr

ian

gle

wit

hle

gs

of

len

gth

s3

cman

d4

cmif

two

sid

esof

the

rect

angle

lie

alon

gth

ele

gs.

Solu

tion

:St

ep1:

Sket

ch:

x

y

(x,y

)

01234

01

23

45

Step

s2&

3:V

aria

ble

s&

Sym

bols

:W

ele

tth

etr

ian

gle

be

loca

ted

soth

atit

sve

rtic

esar

eat(0,0),(4,0),

and(4,3).

We

let

the

insc

rib

edre

ctan

gle

hav

eve

rtic

es(x,y),(4,y),(4,0),

and(x,0),

wh

erey=

3 4x

.

Step

4:

Rel

atio

ns:

Th

ear

eaof

the

rect

angle

isA(x)=(4−x)3 4x=

3x−

3 4x

2.

Step

5:

Iden

tifi

cati

on

:Fi

nd

the

valu

eofx

that

mak

esA

am

axim

um

.

Step

6:

Diff

eren

tiat

e:A′ (x)=

3−

3 2x

,A′′ (x)=−3 2

<0.

Step

7:

Solv

e:A′ (x)=

0w

henx=

2.

Step

8:

Tra

nsl

ate:

Th

em

axim

um

occ

urs

wh

enx=

2,a

nd

equ

als

3.

Max

ima

and

Min

ima

3

Rel

ated

Pro

ble

m:

Fin

dth

ear

eaof

the

larg

est

rect

angle

that

can

be

insc

rib

edin

ari

gh

ttr

ian

gle

wit

hle

gs

of

len

gth

sa>

0(h

ori

zon

tal

leg)

andb>

0(v

erti

cal

leg)

iftw

osi

des

of

the

rect

angle

lie

alon

gth

ele

gs.

Solu

tion

:St

ep1:

Sket

ch:

See

above

.

Step

s2&

3:V

aria

ble

s&

Sym

bols

:W

ele

tth

etr

ian

gle

be

loca

ted

soth

atit

sve

rtic

esar

eat(0,0),(a,0),

and(a,b).

We

let

the

insc

rib

edre

ctan

gle

hav

eve

rtic

es(x,y),(a,y),(a,0),

and(x,0),

wh

erey=b ax

.

Step

4:

Rel

atio

ns:

Th

ear

eaof

the

rect

angle

isA(x)=(a−x)b ax=bx−b ax

2.

Step

5:

Iden

tifi

cati

on

:Fi

nd

the

valu

eofx

that

mak

esA

am

axim

um

.

Step

6:

Diff

eren

tiat

e:A′ (x)=b−

2b ax

,A′′ (x)=−2

b a<

0.

Step

7:

Solv

e:A′ (x)=

0w

henx=a 2

.

Step

8:

Tra

nsl

ate:

Th

em

axim

um

occ

urs

wh

enx=a 2

,an

deq

ual

sab 2

.

4M

axim

aan

dM

inim

aU

nive

rsita

sSask

atch

ewan

en

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m2:

Fin

dth

ear

eaof

the

larg

est

rect

angle

that

can

be

insc

rib

edin

the

elli

psex

2

a2+y

2

b2=

1.

Solu

tion

:St

ep1:

Sket

ch:

(wit

ha=

2an

db=

1)

x2

22+y

2

12=

1:

x

y

x

y

(x,y

)

-101

-2-1

01

2

Step

s2&

3:

Var

iab

les:x

,y,a

ndA

,th

ear

eaof

the

rect

angle

.

Step

4:

Rel

atio

ns:

Let(x,y)

be

any

poin

ton

the

elli

pse

inth

efi

rst

qu

adra

nt.

Th

enx

2

a2+y

2

b2=

1,a

nd

the

rect

angle

wit

h

vert

ices(x,y),(−x,y),(−x,−y),

and(x,−y)

has

areaA(x)=(2x)(

2y)=

4xy=

4xb a

√ a2−x

2=

4b ax(a

2−x

2)1 2

.

Step

5:

Iden

tifi

cati

on

:Fi

nd

the

valu

eofx

that

giv

esth

ela

rges

tva

lue

ofA

.

Step

6:

Diff

eren

tiat

e:

A′ (x)=

4b a

( (1)(a

2−x

2)1 2+x

1 2(a

2−x

2)−

1 2(−

2x)) =

4b a

( (a2−x

2)1 2−x

2(a

2−x

2)−

1 2

) =4b a

(a

2−x

2

(a2−x

2)1 2

−x

2

(a2−x

2)1 2

) =

4b a

( a2−

2x

2√ a

2−x

2

)

Max

ima

and

Min

ima

5U

nive

rsita

sSask

atch

ewan

en

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

7:

Solv

e:

A′ (x)=

0ifx=±√ 2 2

a,

and

we

on

lyco

nsi

der

posi

tive

valu

esofx

.W

eh

aveA′ (x)>

0ifx<√ 2 2a

andA′ (x)<

0if

x>√ 2 2a

sob

yth

eFi

rst

Der

ivat

ive

Tes

tth

em

axim

um

occ

urs

atx=√ 2 2a

.W

eth

enh

ave

A( √

2 2a) =

4b a

( √2 2a) a

2−( √

2 2a) 2 1 2

=

2b( √ 2

)(a

2 2

)1 2

=2b( √ 2

) a √2=

2ab

Step

8:

Tra

nsl

ate:

Th

em

axim

um

area

of

2ab

occ

urs

wh

enth

ew

idth

of

the

rect

angle

is

√ 2 2ti

mes

the

wid

thof

the

elli

pse

.

We

note

that

wh

ena=b=r,

we

fin

dth

atth

ein

scri

bed

rect

angle

insi

de

aci

rcle

of

rad

iusr

isa

squ

are

of

sid

e√ 2r,

and

area

2r2

.

Rel

ated

Pro

ble

ms:

(a)

Fin

dth

ear

eaof

the

larg

est

rect

angle

that

may

be

dra

wn

insi

de

the

elli

psex

2

a2+y

2

b2=

1,

and

lyin

gab

ove

thex

-axi

s.

(b)

Fin

dth

ear

eaof

the

larg

est

rect

angle

that

may

be

dra

wn

insi

de

the

elli

psex

2

a2+y

2

b2=

1,

and

lyin

gin

sid

eth

efi

rst

qu

adra

nt.

6M

axim

aan

dM

inim

aU

nive

rsita

sSask

atch

ewan

en

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m3:

Fin

dth

ed

imen

sion

sof

the

rect

angle

of

larg

est

area

that

has

its

bas

eon

thex

-axi

san

dit

soth

ertw

ove

rtic

esab

ove

thex

-axi

san

dly

ing

on

the

par

abolay=

8−x

2.

Solu

tion

:St

ep1:

Sket

ch:

x

(x,y

)

02468

-3-2

-10

12

3

Step

s2&

3:V

aria

ble

s&

Sym

bols

:w

idthw

,hei

gh

th

,an

dar

ea,A

.

Step

4:

Rel

atio

ns:

Let

the

up

per

righ

t-h

and

corn

erof

the

rect

angle

hav

eco

ord

inat

e(x,h)=(x,8−x

2).

Th

enth

ear

eaof

the

rect

angle

isA(x)=

wid

thti

mes

hei

gh

t=wh=

2xh=

2x(8−x

2)=

16x−

2x

3.

Step

5:

Iden

tifi

cati

on

:Fi

nd

the

valu

eofx

that

mak

esA

am

axim

um

.

Step

6:

Diff

eren

tiat

e:A′ (x)=

16−

6x

2,A

′′ (x)=−1

2x<

0ifx>

0.

Step

7:

Solv

e:A′ (x)=

16−

6x

2=

0ifx=±√ 1

6 6=±2√ 2 3

.

Step

8:

Tra

nsl

ate:

Sin

ceon

lyp

osi

tive

valu

esar

ere

leva

nt,

the

rect

angle

mu

stb

e4

√ 2 3b

y16 3

.

Rel

ated

Pro

ble

m:

Fin

dth

ed

imen

sion

sof

the

rect

angle

of

larg

est

area

that

has

its

bas

eon

thex

-axi

san

dit

soth

ertw

ove

rtic

esab

ove

thex

-axi

san

dly

ing

on

the

par

abolay=a−x

2,w

her

ea>

0.

Max

ima

and

Min

ima

7U

nive

rsita

sSask

atch

ewan

en

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m4:

Fin

dth

ear

eaof

the

larg

est

isosc

eles

tria

ngle

wit

hve

rtex

atth

eori

gin

and

hori

zon

talb

ase

that

can

be

insc

rib

edin

the

elli

psex

2

a2+y

2

b2=

1.

Solu

tion

:St

ep1:

Sket

ch:

(wit

ha=

2an

db=

1)

x2

22+y

2

12=

1:

x

y

x

y

(x,y

)

-101

-2-1

01

2

Step

s2&

3:

Var

iab

les:x

,y,a

ndA

,th

ear

eaof

the

tria

ngle

.

Step

4:

Rel

atio

ns:

Let(x,y)

be

any

poin

ton

the

elli

pse

inth

efi

rst

qu

adra

nt.

Th

enx

2

a2+y

2

b2=

1,an

dth

etr

ian

gle

wit

h

vert

ices(0,0),(x,y),

and(−x,y)

has

areaA(x)=

1 2(2x)(y)=xy=xb a

√ a2−x

2=b ax(a

2−x

2)1 2

.

Step

5:

Iden

tifi

cati

on

:Fi

nd

the

valu

eofx

that

giv

esth

ela

rges

tva

lue

ofA

.

Step

6:

Diff

eren

tiat

e:

A′ (x)=b a

( (1)(a

2−x

2)1 2+x

1 2(a

2−x

2)−

1 2(−

2x)) =

b a

( (a2−x

2)1 2−x

2(a

2−x

2)−

1 2

) =b a

(a

2−x

2

(a2−x

2)1 2

−x

2

(a2−x

2)1 2

) =b a

( a2−

2x

2√ a

2−x

2

)

8M

axim

aan

dM

inim

aU

nive

rsita

sSask

atch

ewan

en

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

7:

Solv

e:

A′ (x)=

0ifx=±√ 2 2

a,

and

we

on

lyco

nsi

der

posi

tive

valu

esofx

.W

eh

aveA′ (x)>

0ifx<√ 2 2a

andA′ (x)<

0if

x>√ 2 2a

sob

yth

eFi

rst

Der

ivat

ive

Tes

tth

em

axim

um

occ

urs

atx=√ 2 2a

.W

eth

enh

ave

A( √

2 2a) =

b a

( √2 2a) a

2−( √

2 2a) 2 1 2

=

1 2b( √ 2

)(a

2 2

)1 2

=1 2b( √ 2

) a √2=

1 2ab

Step

8:

Tra

nsl

ate:

Th

em

axim

um

area

of

1 2ab

occ

urs

wh

enth

ew

idth

of

the

tria

ngle

is

√ 2 2ti

mes

the

wid

thof

the

elli

pse

.

We

note

that

wh

ena=b=r,

we

fin

dth

atth

ein

scri

bed

tria

ngle

insi

de

aci

rcle

of

rad

iusr

has

area

1 2r2

.

Max

ima

and

Min

ima

9U

nive

rsita

sSask

atch

ewan

en

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m5:

Fin

dth

ear

eaof

the

larg

est

isosc

eles

tria

ngle

wit

hve

rtex

at(0,−b)

and

hori

zon

tal

bas

eth

atca

nb

e

insc

rib

edin

the

elli

psex

2

a2+y

2

b2=

1.

Solu

tion

:St

ep1:

Sket

ch:

(wit

ha=

2an

db=

1)

x2

22+y

2

12=

1:

x

y

(x,y

)

-101

-2-1

01

2

Step

s2&

3:

Var

iab

les:x

,y,a

ndA

,th

ear

eaof

the

tria

ngle

.

Step

4:

Rel

atio

ns:

Let(x,y)

be

any

poin

ton

the

elli

pse

inth

efi

rst

qu

adra

nt.

Th

enx

2

a2+y

2

b2=

1,an

dth

etr

ian

gle

wit

h

vert

ices(0,−b),(x,y),

and(−x,y)h

asar

eaA(x)=

1 2(2x)(y+b)=x(y+b)=x( b a

√ a2−x

2+b) =

b ax(a

2−x

2)1 2+bx

.

Step

5:

Iden

tifi

cati

on

:Fi

nd

the

valu

eofx

that

giv

esth

ela

rges

tva

lue

ofA

.

Step

6:

Diff

eren

tiat

e:

A′ (x)=b a

( a2−

2x

2+a√ a

2−x

2√ a

2−x

2

)

Step

7:

Solv

e:

A′ (x)=

0ifa

2−2x

2+a

√ a2−x

2=

0,o

r( a2

−2x

2) 2 =

( −a√ a

2−x

2) 2 ,o

rei

therx=

0orx

2=

3a

2

4,s

ow

eta

kex=√ 3 2a

asth

eon

lyvi

able

solu

tion

.T

he

Firs

tD

eriv

avti

vete

stte

lls

us

that

we

hav

ea

max

imu

m.

10

Max

ima

and

Min

ima

Step

8:

Tra

nsl

ate:

Th

em

axim

um

area

of

3√ 34ab

occ

urs

wh

enth

ew

idth

of

the

tria

ngle

is√ 3a

.

Max

ima

and

Min

ima

11

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m6:

Fin

dth

ear

eaof

the

larg

est

isosc

eles

tria

ngle

wit

hve

rtex

at(0,−b)

and

hori

zon

tal

bas

eth

ath

asit

sve

rtex

atth

eori

gin

and

its

oth

ertw

ove

rtic

esab

ove

thex

-axi

san

dly

ing

on

the

par

abolay=

8−x

2.

Solu

tion

:St

ep1:

Sket

ch:

x

y

(x,y

)

02468

-3-2

-10

12

3

Step

s2&

3:V

aria

ble

s&

Sym

bols

:w

idthw

,hei

gh

th

,an

dar

ea,A

.

Step

4:

Rel

atio

ns:

Let

the

up

per

righ

t-h

and

corn

erof

the

tria

ngle

hav

eco

ord

inat

es(x,h)=(x,8−x

2).

Th

enth

ear

eaof

the

tria

ngle

isA(x)=

1 2w

idth

tim

esh

eigh

t=

1 2wh=

1 22xh=x(8−x

2)=

8x−x

3.

Step

5:

Iden

tifi

cati

on

:Fi

nd

the

valu

eofx

that

mak

esA

am

axim

um

.

Step

6:

Diff

eren

tiat

e:A′ (x)=

8−

3x

2,A

′′ (x)=−6x<

0ifx>

0.

Step

7:

Solv

e:A′ (x)=

8−

3x

2=

0ifx=±√ 8 3

=±2√ 2 3

.

Step

8:

Tra

nsl

ate:

Sin

ceon

lyp

osi

tive

valu

esar

ere

leva

nt,

the

tria

ngle

mu

stb

e2

√ 2 3w

ide

by

16 3

hig

h.

How

cou

ldw

eh

ave

ded

uce

dth

isre

sult

from

Pro

ble

m3

wit

hn

ofu

rth

erca

lcu

lati

on

?

12

Max

ima

and

Min

ima

Pro

ble

m7:

Fin

dth

ear

eaof

the

larg

est

tria

ngle

wit

hve

rtic

esat(−

2,4)

and

(0,0

)w

hose

thir

dve

rtex

lies

inth

efi

rst

qu

adra

nt

and

on

the

par

abolay=

8−x

2.

Solu

tion

:St

ep1:

Sket

ch:

x

y

(x,y

)

02468

-3-2

-10

12

3St

eps

2&

3:V

aria

ble

s&

Sym

bols

:w

idthw

,hei

gh

th

,an

dar

ea,A

.

Step

4:

Rel

atio

ns:

Th

ed

ista

nce

bet

wee

nth

etw

ofi

xed

vert

ices

isw=√ 2

2+

42=√ 2

0=

2√ 5

.W

eta

ke

this

asa

bas

eof

the

tria

ngle

,an

dm

ust

nex

tfi

nd

the

alti

tud

eon

this

bas

e.T

he

lin

eth

rou

gh

the

two

fixe

dve

rtic

esh

aseq

uat

iony=−2x

,or,

ingen

eral

form

,2x+(1)y+

0=

0.

Let

the

up

per

righ

t-h

and

corn

erof

the

rect

angle

hav

eco

ord

inat

es(x,h)=(x,8−x

2).

Its

dis

tan

cefr

om

the

bas

eish=∣ ∣ 2(

x)+(1)(

8−x

2)+

0∣ ∣

√ 22+

12

=∣ ∣ 2x

+8−x

2∣ ∣

√ 5=

Th

enth

ear

eaof

the

tria

ngle

isA(x)=

1 2w

idth

tim

esh

eigh

t=

1 2wh=

1 22√ 5

∣ ∣ 2x+

8−x

2∣ ∣

√ 5=∣ ∣ 2x

+8−x

2∣ ∣ .

Sin

cew

eh

ave

assu

med

that

the

mova

ble

vert

exli

esin

the

firs

tq

uad

ran

t,w

eca

nre

move

the

abso

lute

valu

esi

gn

s:A(x)=

2x+

8−x

2

Step

5:

Iden

tifi

cati

on

:Fi

nd

the

valu

eofx

that

mak

esA

am

axim

um

.

Step

6:

Diff

eren

tiat

e:A′ (x)=

2−

2x

,A′′ (x)=−2

<0.

Step

7:

Solv

e:A′ (x)=

0ifx=

1.

Max

ima

and

Min

ima

13

Step

8:

Tra

nsl

ate:

Th

etr

ian

gle

has

its

thir

dve

rtex

at(1

,7).

14

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m4.7

:25.

Ari

gh

tci

rcu

lar

cyli

nd

eris

insc

rib

edin

asp

her

eof

rad

iusr.

Fin

dth

ela

rges

tp

oss

ible

volu

me

of

such

acy

lin

der

.

Solu

tion

:St

ep1:

Sket

ch:

Max

ima

and

Min

ima

15

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

s2&

3:V

aria

ble

s&

Sym

bols

:Le

tth

era

diu

s,h

eigh

tan

dvo

lum

eof

the

cyli

nd

erb

eR

,han

dV

.

Step

4:

Rel

atio

ns:

Th

ed

ista

nce

from

the

cen

tre

of

the

sph

ere

toth

eb

ase

of

the

cyli

nd

eris√ r

2−R

2.h

wil

lth

enb

e

twic

eth

isam

ou

nt,

orh=

2√ r

2−R

2.

Th

usV=πR

2h

can

be

exp

ress

edin

term

sofh

orR

:

V(h)=π( r2

−( h 2

) 2)h=πr2h−

π 4h

3or

V(R)=πR

2( 2√ r

2−R

2) =

2πR

2(r

2−R

2)1 2

.

Step

5:

Iden

tifi

cati

on

:Fi

nd

the

valu

esofR

andh

that

mak

eV

am

axim

um

.

Th

ere

are

seve

ral

way

sto

app

roac

hth

ep

rob

lem

:

16

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Th

eEas

yW

aySt

ep6:

Diff

eren

tiat

e:

V′ (h)=πr2−

3π 4h

2=

0ifh=±

2 √ 3r.

Step

7:

Solv

e:W

eh

aveV′ (

0)=πr2>

0,a

ndV′ (

2r)=−2πr2<

0,s

o(b

yth

eFi

rst

Der

ivat

ive

Tes

t)w

eh

ave

am

axim

um

ifh=

2 √ 3r.

Step

8:

Tra

nsl

ate:

Th

em

axim

um

isV( 2 √ 3

r) =4π

3√ 3r3

Max

ima

and

Min

ima

17

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Th

eH

ard

Way

Step

6:

Diff

eren

tiat

e:V′ (R)=

2π( 2R√ r2

−R

2+R

2(1 2(r

2−R

2)−

1 2(−

2R)) =

2πR( 2√ r2

−R

2−R

2(r

2−R

2)−

1 2

) =2πR( 2

r2−R

2√ r

2−R

2−

R2

√ r2−R

2

) =

2πR

2r2−

3R

2√ r

2−R

2=

0ifR=√ 2 3

r.

18

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

7:

Solv

e:B

yth

eFi

rst

Der

ivat

ive

Tes

t,th

isgiv

esth

em

axim

um

.

We

hav

eh=

2

√ 1 3r,

and

V

√ 2 3r =

π

√ 2 3r 2

2√ 1 3

r =4π

3√ 3r3

.

Max

ima

and

Min

ima

19

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Th

eT

rigon

om

etri

cW

ay:

Letθ

be

the

angle

bet

wee

nth

eax

isan

dth

eli

nes

from

the

cen

tre

of

the

sph

ere

and

the

poin

tsof

con

tact

of

the

cyli

nd

erw

ith

the

sph

ere.

Th

enh=

2r

cosθ

andR=r

sinθ

,so

V(θ)=πR

2h=π(r

sinθ)2(2r

cosθ)=

2π

sin

2θ

cosθr3

Step

6:

Diff

eren

tiat

e:

V′ (θ)=

2πr3( 2

sinθ

cos2θ−

sin

3θ) =

2πr3

sinθ( 2

cos2θ−

sin

2θ) =

2πr3

sinθ( 3

cos2θ−

1)

Step

7:

Solv

e:V′ (θ)=

0if

sinθ=

0or

cosθ=±

1 √ 3.

Cle

arly

the

max

imu

mocc

urs

for

cosθ=

1 √ 3,a

nd

equ

als

4π

3√ 3r3

.

20

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m4.7

:27.

Ari

gh

tci

rcu

lar

cyli

nd

eris

insc

rib

edin

asp

her

eof

rad

iusr.

Fin

dth

ela

rges

tp

oss

ible

surf

ace

area

of

such

acy

lin

der

.

Solu

tion

:St

ep1:

Sket

ch:

Don

eju

stab

ove

.

Step

s2&

3:V

aria

ble

s&

Sym

bols

:A

sab

ove

.

Max

ima

and

Min

ima

21

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

4:

Rel

atio

ns:

Th

esu

rfac

ear

eaca

nb

eex

pre

ssed

inte

rms

ofR

:

A(R)=

2πR

2+

2πRh=

2πR

2+

2πR(2√ r2

−R

2)=

2πR

2+

4πR(r

2−R

2)1 2

,

or

inte

rms

ofh

:

A(h)=

2πR

2+

2πRh=

2π( r2

−( h 2

) 2)+

2π

√ r2−( h 2

) 2 h=

2πr2−π 2h

2+πh(4r2−h

2)1 2

.

Step

5:

Iden

tifi

cati

on

:Fi

nd

the

valu

esofR

andh

that

giv

esth

ela

rges

tva

lue

ofS.

22

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

As

inPro

ble

m4.7

:25,t

her

eis

more

than

on

ew

ayof

doin

gth

ep

rob

lem

:

Th

e“E

asy

”W

ay:

A(h)=

2πr2−π 2h

2+πh(4r2−h

2)1 2

.

Step

6:

Diff

eren

tiat

e:

A′ (h)=−π

h+π(1)(

4r2−h

2)1 2+πh

1 2(4r2−h

2)−

1 2(−

2h)=

−πh+π(4r2−h

2)1 2−πh

2(4r2−h

2)−

1 2=

π( −h

+4r2−h

2

(4r2−h

2)1 2

−h

2

(4r2−h

2)1 2

) =

π( −h

+4r2−

2h

2

(4r2−h

2)1 2

) =π−h√ 4r2−h

2+

4r2−

2h

2√ 4r2−h

2

Max

ima

and

Min

ima

23

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

7:

Solv

e:A′ (h)=

0if

h√ 4r2−h

2=

4r2−

2h

2or

h2(4r2−h

2)=

16r4−

16r2h

2+

4h

4or

4r2h

2−h

4=

16r4−

16r2h

2+

4h

4or

5h

4−

20r2h

2+

16r4=

0.

wh

ich

isa

qu

adra

tic

equ

atio

ninh

2:

5(h

2)2−

20r2h

2+

16r4=

0,

and

has

solu

tion

24

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

h2=−(−2

0r2)±√ (−

20r2)2−

4(5)(

16r4)

2(5)

=20±√ 4

00−

320

10

r2=

20±

4√ 5

10

r2=

2

( 1±√ 5 5

) r2

soth

em

axim

um

mig

ht

occ

ur

eith

erw

hen

h=√ √ √ √ 2

( 1−√ 5 5

) rorh=√ √ √ √ 2

( 1+√ 5 5

) r,if

thes

eva

lues

sati

sfy

h√ 4r2−h

2=

4r2−

2h

2.

Max

ima

and

Min

ima

25

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

anW

etr

yh=√ √ √ √ 2

( 1+√ 5 5

) r:

√ √ √ √ 2( 1+√ 5 5

) r√ √ √ √ 4r2−

2

( 1+√ 5 5

) r2?=

4r2−

2

( 2

( 1+√ 5 5

) r2) ,o

r

2

√ 1+√ 5 5

√ √ √ √ 2−( 1+√ 5 5

)?=−4√ 5 5

wh

ich

isn

ot

tru

e,so

we

reje

ctth

isva

lue

ofh

.

26

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

anW

etr

yh=√ √ √ √ 2

( 1−√ 5 5

) r:

√ √ √ √ 2( 1−√ 5 5

) r√ √ √ √ 4r2−

2

( 1−√ 5 5

) r2?=

4r2−

2

( 2

( 1−√ 5 5

) r2) ,o

r

2

√ 1−√ 5 5

√ √ √ √ 2−( 1−√ 5 5

)?=

4

√ 5 5,o

r2

√ 1−√ 5 5

√ 1+√ 5 5

?=

4

√ 5 5,o

r2

√ 1−

1 5?=

4

√ 5 5w

hic

his

tru

e.

Max

ima

and

Min

ima

27

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

anSi

nceV′ (

0)=

2πr>

0an

dli

mh→

2rV′ (h)=−∞

,th

efi

rst

der

ivat

ive

tell

su

sth

atth

em

axim

um

occ

urs

ath=√ √ √ √ 2

( 1−√ 5 5

) r.

Iteq

ual

sπr2( 1+√ 5) .

and

then

ther

e’s

also

:

28

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Th

eH

ard

Way

:

Step

6:

Diff

eren

tiat

e:

A′ (R)=

4πR+

4π[ (1

)(r2−R

2)1 2+R

1 2(r

2−R

2)−

1 2(−

2R)] =

4πR+

4π[ r

2−R

2√ r

2−R

2−

R2

√ r2−R

2

] =4π[ R

+r2−

2R

2√ r

2−R

2

] =

4πR√ r

2−R

2+r2−

2R

2√ r

2−R

2

Max

ima

and

Min

ima

29

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

A′′ (R)=

4π

√ r2−R

2(R√ r

2−R

2+r2−

2R

2)′−(R√ r

2−R

2+r2−

2R

2)1 2(r

2−R

2)−

1 2(−

2R)

(√r2−R

2)2

=

(r2−R

2)( √

r2−R

2+R

1 2(r

2−R

2)−

1 2(−

2R)−

4R) +

R( R√

r2−R

2+r2−

2R

2)

√ r2−R

2(r

2−R

2)

=

(r2−R

2)( r

2−2R

2√ r

2−R

2−

4R) +

R( R√

r2−R

2+r2−

2R

2)

√ r2−R

2(r

2−R

2)

=

30

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

(r2−

2R

2)√r2−R

2−

4R(r

2−R

2)+R

2√ r

2−R

2+R(r

2−

2R

2)

√ r2−R

2(r

2−R

2)

=(r

2−R

2)√r2−R

2−R(3r2−R

2)

√ r2−R

2(r

2−R

2)

Max

ima

and

Min

ima

31

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

7:

Solv

e:

A′ (R)=

0if

R√ r

2−R

2+r2−

2R

2=

0or

R√ r

2−R

2=

2R

2−r2

or,

squ

arin

g,

R2(r

2−R

2)=(2R

2−r2)2

.

Th

usR

2r2−R

4=

4R

4−

4R

2r2+r4

or

5R

4−

5R

2r2+r4=

0or

5(R

2)2−

5r2(R

2)+r4=

0or

R2=−(−5r2)±√ (−

5r2)2−

4(5)r

4

2(5)

=5±√ 5

10

r2.

32

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

anT

och

eck

ou

rw

ork

,we

sub

stit

uteR=√ 5

+√ 5

10

ran

dR

2=

5+√ 5

10

r2in

toth

eeq

uat

ion

R√ r

2−R

2=

2R

2−r2

:

Firs

t,

√ 5+√ 5

10

r√ √ √ √ √ r2− √ 5

+√ 5

10

r 2

?=

2

√ 5+√ 5

10

r 2

−r2

or

√ 5+√ 5

10

r√ r2−

5+√ 5

10

r2?=

25+√ 5

10

r2−r2

or

√ 5+√ 5

10

√ 1−

5+√ 5

10

?=

25+√ 5

10

−1

or

√ 5+√ 5

10

√ 5−√ 5

10

?=

5+√ 5

5−

1or

√ (5+√ 5)(

5−√ 5)

100

?=√ 5 5

or

√20

100

?=√ 5 5

,wh

ich

istr

ue.

Max

ima

and

Min

ima

33

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

anSe

con

d,√ 5

−√ 5

10

r√ √ √ √ √ r2− √ 5

−√ 5

10

r 2

?=

2

√ 5−√ 5

10

r 2

−r2

or

√ 5−√ 5

10

r√ r2−

5−√ 5

10

r2?=

25−√ 5

10

r2−r2

or

√ 5−√ 5

10

√ 1−

5−√ 5

10

?=

25−√ 5

10

−1

or

√ 5−√ 5

10

√ 5+√ 5

10

?=

5−√ 5

5−

1or

√ (5−√ 5)(

5+√ 5)

100

?=−√ 5 5

or

√20

100

?=−√ 5 5

,wh

ich

isim

poss

ible

,so

we

reje

ctth

isva

lue

ofR

.

34

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

anN

ext

we

com

pu

teA

√ 5+√ 5

10

r =

2π

√ 5+√ 5

10

r 2

+4π

√ 5+√ 5

10

r √ √ √ √ √ r2− √ 5

+√ 5

10

r 2

=

2πr2 5

+√ 5

10

+2

√ 5+√ 5

10

√ 1−

5+√ 5

10

=

2πr2 5

+√ 5

10

+2

√ 5+√ 5

10

√ 5−√ 5

10

=

2πr2 5

+√ 5

10

+2

√20

100

=2πr2[ 5

+√ 5

10

+2

2√ 510

] =

2πr2

5+

5√ 5

10

=πr2( 1+√ 5) .

Max

ima

and

Min

ima

35

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

4.7

:28

AN

orm

anw

ind

ow

has

the

shap

eof

are

ctan

gle

surm

ou

nte

db

ya

sem

icir

cle.

Ifth

ep

erim

eter

of

the

win

dow

is30

feet

,fin

dth

ed

imen

sion

sof

the

win

dow

soth

atth

egre

ates

tp

oss

ible

amou

nt

of

ligh

tis

adm

itte

d.

Solu

tion

:St

ep1:

Sket

ch:

36

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

s2&

3:V

aria

ble

s&

Sym

bols

:Le

tx

be

the

wid

thof

the

win

dow

,hth

eh

eigh

tof

the

rect

angu

lar

par

t,an

dA

the

area

of

the

win

dow

.

Step

4:

Rel

atio

ns:

30=x+

2h+πx 2

,A=xh+

1 2π( x 2

) 2 .

Step

5:

Iden

tifi

cati

on

:W

em

ust

fin

dei

ther

the

valu

eofx

orh

that

mak

esA

am

axim

um

.W

eu

seth

efi

rst

rela

tion

toex

pre

ssh

inte

rms

ofx

:

30−x−πx 2=

2h

,h=

30−x−πx 2

2=

15−x+πx 2

2=

15−x

1+

π 2

2

and

then

we

exp

ressA

asa

fun

ctio

nofx

:

A(x)=A=x( 1

5−x

1+π 2

2

) +1 2π( x 2

) 2 =x(1 2(π 4+

1)x−

15)=

0ifx=

0orx=

120

π+

4.

Th

us

the

max

imu

mar

eaocc

urs

for

x=

60

π+

4.

Note

that

ther

ew

asn

on

eed

tod

iffer

enti

ate.

Max

ima

and

Min

ima

37

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m4.7

:30.

Ap

ost

eris

toh

ave

anar

eaof

180

in2

wit

h1-i

nch

mar

gin

sat

the

bott

om

and

sid

esan

da

2-i

nch

mar

gin

atth

eto

p.

Wh

atd

imen

sion

sw

ill

giv

eth

ela

rges

tp

rin

ted

area

?

Solu

tion

:St

ep1:

Sket

ch:

Step

s2&

3:V

aria

ble

s&

Sym

bols

:w

idth

,w,h

eigh

t,h

,an

dp

rin

ted

area

,A.

Step

4:

Rel

atio

ns:

Rel

atio

ns:wh=

180,s

oh=

180

w.

A(w)=(w

−2)(h−

3)=(w

−2)(

180

w−

3)=

186−

3w−

360

w.

Step

5:

Iden

tifi

cati

on

:W

em

ust

fin

dth

eva

lue

ofw

that

mak

esA

am

axim

um

.

Step

6:

Diff

eren

tiat

e:A′ (w)=−3

+360

w2

.

38

Max

ima

and

Min

ima

Step

7:

Solv

e:T

he

max

imu

mocc

urs

for

w=

2√ 3

0an

dh=

3√ 3

0.

Max

ima

and

Min

ima

39

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m4.7

:32

Ap

iece

of

wir

e10

mlo

ng

iscu

tin

totw

op

iece

s.O

ne

pie

ceis

ben

tin

toa

squ

are

and

the

oth

eris

ben

tin

toa

circ

le.

How

shou

ldth

ew

ire

be

cut

soth

atth

eto

tal

area

encl

ose

dis

(a)

am

axim

um

?(b

)a

min

imu

m?

Solu

tion

:St

ep1:

Sket

ch:

40

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

s2&

3:V

aria

ble

s&

Sym

bols

:Le

tx

be

the

len

gth

of

the

wir

eth

atis

ben

tin

toa

squ

are,

soth

atth

ere

is10−x

left

ove

rfo

rth

eci

rcu

mfe

ren

cec

of

the

circ

le.

We

letr

be

the

rad

ius

of

the

circ

lean

dA

be

the

area

con

tain

edb

yb

oth

the

squ

are

and

circ

le.

Step

4:

Rel

atio

ns:

Th

esq

uar

eh

assi

de

of

len

gthx 4

and

areax

2

16

,

and

the

rad

iusr

of

the

circ

leisr=

c 2π=

10−x

2π

,wit

har

eaπ( 1

0−x

2π

) 2 .T

hu

sth

eto

tal

area

encl

ose

dis

A(x)=x

2

16+π( 1

0−x

2π

) 2=x

2

16+

100−

20x+x

2

4π

=

πx

2

16π+

400−

80x+

4x

2

16π

=400−

80x+(π+

4)x

2

16π

,0≤x≤

10

Max

ima

and

Min

ima

41

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

5:

Iden

tifi

cati

on

:W

em

ust

fin

dth

eva

lue

ofx

that

mak

esA

(a)

am

axim

um

,an

d(b

)a

min

imu

m.

Step

6:

Diff

eren

tiat

e:A′ (x)=−8

0+

2(π+

4)x

16π

=0

ifx=

40

π+

4

Step

7:

Solv

e:W

eh

aveA(0)=

400

16π=

25 π�

7.9

6,

A(

40

π+

4

) =400−

80( 4

0π+4) +

(π+

4)(

40

(π+4) 2

16π

=

400−

32

00

π+4+

16

00

π+4

16π

=400π+4

π+4−

32

00

π+4+

16

00

π+4

16π

=4

00π

π+4

16π=

25

π+

4�

3.5

A(1

0)=

100

16=

6.2

5.

42

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

8:

Tra

nsl

ate:

Th

em

inim

um

valu

eis

25

π+

4an

dth

em

axim

um

valu

eis

25 π

.

Pro

ble

m4.7

:34.

Afe

nce

8fe

etta

llru

ns

par

alle

lto

ata

llb

uil

din

gat

ad

ista

nce

of

4fe

etfr

om

the

bu

ild

ing.

Wh

atis

the

len

gth

of

the

short

est

lad

der

that

wil

lre

ach

from

the

gro

un

dove

rth

efe

nce

toth

ew

all

of

the

bu

ild

ing?

Solu

tion

:St

ep1:

Sket

ch:

Step

s2&

3:V

aria

ble

s&

Sym

bols

:W

ele

tx

be

the

dis

tan

ceof

the

foot

of

the

lad

der

from

the

fen

cean

dy

the

hei

gh

tof

the

top

of

the

lad

der

,an

d�

the

len

gth

of

the

lad

der

.

Step

4:

Rel

atio

ns:

By

sim

ilar

tria

ngle

s,y

x+

4=

8 x,s

oy=

8( 1+

4 x

) .

�=√ (x

+4)2+y

2,

Step

5:

Iden

tifi

cati

on

:W

efi

nd

the

min

imu

mof

the

squ

are

of

the

len

gth

of

the

lad

der

,let

tin

gf(x)=(x+

4)2+

64(1+

4 x)2=x

2+

8x+

80+

8x−1+

16x−2

.

Max

ima

and

Min

ima

43

Step

6:

Diff

eren

tiat

e:f′ (x)=

2x+

8−

8x−2−

32x−3=

2 x3(x+

4)(x

3−

4),

and

f′′ (x)=

2+

16x−3+

96x−4<

0ifx>

0.

Step

7:

Solv

e:f′ (x)=

0ifx=−4

orx=

3√ 4.

Step

8:

Tra

nsl

ate:

Th

em

inim

um

occ

urs

for

x=

3√ 4=

22 3

.T

he

min

imu

mle

ngth

is

√ (2

2 3

) 2 +8( 2

2 3

) +80+

8( 2

2 3

) −1+

16( 2

2 3

) −2=

√ 24 3+

23·2

2 3+

80+

23·2

−2 3+

24·2

−4 3=

√ 24 3+

21

1 3+

80+

27 3+

28 3=

√ 2

√ 3·2

1 3+

6·2

2 3+

40

44

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m4.7

:35.

Aco

nic

ald

rin

kin

gcu

pis

mad

efr

om

aci

rcu

lar

pie

ceof

pap

erof

rad

iusR

by

cutt

ing

ou

ta

sect

or

and

join

ing

the

edgesCA

andCB

.Fi

nd

the

max

imu

mca

pac

ity

of

such

acu

p.

Solu

tion

:St

ep1:

Sket

ch:

Max

ima

and

Min

ima

45

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

s2&

3:V

aria

ble

s&

Sym

bols

:T

he

volu

meV

,th

eh

eigh

th

,an

dth

era

diu

sr.

Step

4:

Rel

atio

ns:V=

π 3r2h

,an

dr2+h

2=R

2.

Th

us

we

hav

etw

ow

ays

of

wri

tin

gV

asa

fun

ctio

nof

just

on

eva

riab

le:

V(h)=π 3(R

2−h

2)h=π 3R

2h−π 3h

3,o

rV(r)=π 3r2√ R

2−r2

Step

5:

Iden

tifi

cati

on

:Fi

nd

eith

erth

eva

lue

ofh

or

the

valu

eofr

that

mak

esV

am

axim

um

.

46

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Th

eEas

yW

ay

V(h)=π 3R

2h−π 3h

3

V′ (h)=π 3R

2−πh

2=

0ifh=R √ 3

.

V′ (

0)=π 3R

2>

0,

V′ (R)=π 3R

2−πR

2=−2π 3R

2<

0,

soth

efi

rst

der

ivat

ive

test

giv

esa

max

imu

m,

ath=R √ 3

.

We

hav

eV( R √

3

) =2π

9√ 3R

3

Max

ima

and

Min

ima

47

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Th

eH

ard

Way

-for

Mas

och

ists

V(r)=π 3r2(R

2−r2)1 2

V′ (r)=π 3

( 2r(R

2−r2)1 2+r2

1 2(R

2−r2)−

1 2(−

2r)) =

rπ 3

( 2(R

2−r2)1 2−r2(R

2−r2)−

1 2

) =rπ 3

( 2(R

2−r2)

(R2−r2)1 2

−r2

(R2−r2)1 2

) =

rπ 3

2R

2−

3r2

√ R2−r2

=0

ifr=

0or±√ 2 3

R.

For

smal

lp

osi

tiver,V′ (r)>

0,a

nd

forr

close

toR

,V′ (r)<

0,s

oth

efi

rst

der

ivat

ive

test

giv

esa

max

imu

m

atr=√ 2 3

R.

We

hav

eV

√ 2 3R

=2π

9√ 3R

3.

Th

eR

eall

yH

ard

Way

-for

the

Tru

lyD

emen

ted

!

Th

eci

rcu

mfe

ren

ceof

the

giv

enp

iece

of

pap

eris

2πR

,an

dth

ele

ngth

of

arc

of

the

sect

or

rem

ove

disRθ

,so

the

circ

um

-fe

ren

ceof

the

circ

ula

rto

pof

the

cup

is(2π−θ)R

,an

dth

eref

ore

its

rad

ius

is:

r=(2π−θ)R

2π

=(1−θ 2π)R

.

Th

ush=√ R

2−r2=√ R

2−[( 1

−θ 2π

) R] 2 =

√ 1−( 1−θ 2π

) 2 R=√ 2

θ 2π−θ

2

4π

2R=

√ 4θπ−θ

2R 2π=√ θ(4π−θ)R 2π

48

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

and

thu

s

V(θ)=π 3r2h=π 3

[( 1−θ 2π

) R] 2√

θ(4π−θ)R 2π=

( 1−θ 2π

) 2√θ(4π−θ)R

3 6

Con

tin

uin

gp

ast

this

poin

tin

dic

ates

an

eed

for

cou

nse

llin

g!

Max

ima

and

Min

ima

49

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m4.7

:40

Aw

om

anat

ap

oin

tA

on

the

shore

of

aci

rcu

lar

lake

wit

hra

diu

s2

mil

esw

ants

tob

eat

ap

oin

tC

dia

met

rica

lly

op

posi

teA

on

the

oth

ersi

de

of

the

lake

inth

esh

ort

est

poss

ible

tim

e.Sh

eca

nw

alk

atth

era

teof

4m

iles

/hou

r,an

dsh

eca

nro

wa

boat

at2

mil

es/h

ou

r.A

tw

hat

angleθ

toth

ed

iam

eter

shou

ldsh

ero

w?

Solu

tion

:St

ep1:

Sket

ch:

Step

s2&

3:V

aria

ble

s&

Sym

bols

:Le

tB

be

the

poin

tsh

ero

ws

to,a

nd

letθ

be

the

angle

BA

C.

Letdw

anddr

be

the

dis

tan

ces

wal

ked

and

row

ed,a

nd

letT w

andT r

be

the

tim

esp

ent

wal

kin

gan

dro

win

g,r

esp

ecti

vely

.

Step

4:

Rel

atio

ns:

By

the

Law

of

Cosi

nes

,

d2 r=|AB|2=|AO|2+|OB|2−

2|AO||O

B|c

os(π−

2θ)=

22+

22−

2(2)(

2)c

os

2θ=

8−

8co

s2θ=

8(1−

cos

2θ)=

8(1−(1−

2co

s2θ))=

16

cos2θ

,sodr=|AB|=

4co

sθ

.

Th

ed

ista

nce

wal

ked

isdw=

2(2θ)=

4θ

.T

he

tim

esp

ent

row

ing

ist r=

4co

sθ

2=

2co

sθ

and

the

tim

esp

ent

wal

kin

gis

dw=

4θ 4=θ

,so

the

tota

lti

me

elap

sed

wil

lb

e

T(θ)=t r+t w=

2co

sθ+θ

,wh

ere

0≤θ≤π 2

.

50

Max

ima

and

Min

ima

Step

5:

Iden

tify

:Fi

nd

the

angleθ

bet

wee

n0

andπ 2

that

mak

esT(θ)

am

inim

um

.

Step

6:

Diff

eren

tiat

e:T′ (θ)=−2

sinθ+

1=

0

Step

7:

Solv

e:T′ (θ)=

0if

sinθ=

1 2,

orθ=π 6

.B

ut

the

Firs

tD

eriv

ativ

eT

est

tell

su

sth

atth

isis

am

axim

um

,so

the

min

imu

mm

ust

occ

ur

forθ=

0orθ=π 2

.W

eh

aveT(0)=

2,a

nd

T( π 2

) =π 2�

1.5

7,w

hic

his

the

min

imu

m.

Step

8:

Tra

nsl

ate:

She

shou

ldw

alk

all

the

way

arou

nd

the

lake.

Max

ima

and

Min

ima

51

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m4.7

:42

Fin

dan

equ

atio

nof

the

lin

eth

rou

gh

the

poin

t(3,5)

that

cuts

off

the

leas

tar

eafr

om

the

firs

tq

uad

ran

t.

Solu

tion

:St

ep1:

Sket

ch:

Step

s2&

3:V

aria

ble

s&

Sym

bols

:In

dic

ated

inth

eab

ove

dia

gra

m.

Step

4:

Rel

atio

ns:

Th

eli

ne

thro

ugh

(3,5

)an

d(x,0)

inte

rsec

tsth

ey

-axi

sat

5+

15

x−

3=

5(x−

3)

x−

3+

15

x−

3=

5x

x−

3,

soth

ear

eaof

the

tria

ngle

wit

hve

rtic

es(0

,0),(x,0),

and(0,

5x

x−

3)

isA(x)=

1 2x

5x

x−

3=

5 2x

2

x−

3.

Step

5:

Iden

tifi

cati

on

:Fi

nd

the

valu

eofx

that

mak

esA(x)

am

inim

um

.

Step

6:

Diff

eren

tiat

e:A′ (x)=

5 2(x−

3)(

2x)−x

2(1)

(x−

3)2

=5 2x

2−

6x

(x−

3)2

.

Step

7:

Solv

e:so

the

min

imu

mocc

urs

forx=

6b

yth

efi

rst

der

ivat

ive

test

.T

hey

-in

terc

ept

of

the

lin

eth

rou

gh

(6,0

)

52

Max

ima

and

Min

ima

and

(3,5

)is

10,s

oan

equ

atio

nof

the

req

uir

edli

ne

isx 6+y 10=

1.

Max

ima

and

Min

ima

53

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m4.7

:44

Th

efr

ame

for

akit

eis

tob

em

ade

from

six

pie

ces

of

wood

.T

he

fou

rex

teri

or

pie

ces

hav

eb

een

cut

wit

hth

ele

ngth

sa

andb

ind

icat

edin

the

figu

re.

To

max

imiz

eth

ear

eaof

the

kit

e,h

ow

lon

gsh

ou

ldth

ed

iagon

alp

iece

sb

e?

Solu

tion

:St

ep1:

Sket

ch:

54

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

s2&

3:V

aria

ble

s&

Sym

bols

:In

dic

ated

inth

eab

ove

dia

gra

m.

Step

4:

Rel

atio

ns:

Th

ear

eaisA=x(y

+z)

,an

dw

eh

avex

2+y

2=a

2an

dx

2+z2

=b

2,

soy=√ a

2−x

2,

and

z=√ b

2−x

2,s

ow

eh

ave:

A(x)=x( √ a

2−x

2+√ b

2−x

2) =

x( ( a

2−x

2)1 2+( b2

−x

2)1 2

) .

Max

ima

and

Min

ima

55

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

5:

Iden

tifi

cati

on

:Fi

nd

the

valu

eofx

that

mak

esA

am

axim

um

.

Step

6:

Diff

eren

tiat

e:A′ (x)=

(1)( ( a

2−x

2)1 2+( b2

−x

2)1 2

) +x( 1 2

( a2−x

2) −1 2

(−2x)+

1 2

( b2−x

2) −1 2

(−2x)) =

a2−x

2√ a

2−x

2+b

2−x

2√ b

2−x

2−x

2(

1√ a

2−x

2+

1√ b

2−x

2

) =

a2−

2x

2√ a

2−x

2+b

2−

2x

2√ b

2−x

2

56

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

7:

Solv

e:A′ (x)=

0if

a2−

2x

2√ a

2−x

2=−b

2−

2x

2√ b

2−x

2or

(a2−

2x

2)2

a2−x

2=(b

2−

2x

2)2

b2−x

2or

(a2−

2x

2)2(b

2−x

2)=(b

2−

2x

2)2(a

2−x

2)

or (a

4−

4a

2x

2+

4x

4)(b

2−x

2)=(b

4−

4b

2x

2+

4x

4)(a

2−x

2)

or a4b

2−a

4x

2−

4a

2b

2x

2+

4a

2x

4+

4b

2x

4−

4x

6=b

4a

2−b

4x

2−

4a

2b

2x

2+

4b

2x

4+

4a

2x

4−

4x

6

or

4a

2x

4−

4b

2x

4+

4b

2x

4−

4a

2x

4−a

4x

2+b

4x

2−

4a

2b

2x

2+

4a

2b

2x

2+a

4b

2−b

4a

2=

0

or (b

2−a

2)(b

2+a

2)x

2−(b

2−a

2)a

2b

2=

0

assu

min

gb

2�=a

2,t

his

sim

pli

fies

to:(b

2+a

2)x

2−a

2b

2=

0w

hic

hh

asas

solu

tionx=

ab

√ a2+b

2

Ifa=b

,th

enA(x)=

2x√ a

2−x

2,a

ndA′ (x)=

2a

2−

2x

2√ a

2−x

2=

0ifx=

√ 2 2a

,wh

ich

agre

esw

ith

the

above

.

Max

ima

and

Min

ima

57

We

hav

ey=√ a

2−(

ab

√ a2+b

2)2=√ a

2a

2+b

2

a2+b

2−(a

2b

2

a2+b

2)=√

a4

a2+b

2=

a2

√ a2+b

2

andz=√ b

2−(

ab

√ a2+b

2)2=√ b

2a

2+b

2

a2+b

2−(a

2b

2

a2+b

2)=√

b4

a2+b

2=

b2

√ a2+b

2

Step

8:

Tra

nsl

ate:

Th

eop

tim

alle

ngth

of

the

hori

zon

tal

stru

tis

2x=

2ab

√ a2+b

2an

dth

eop

tim

alle

ngth

of

the

vert

ical

stru

tisx+y=

a2

√ a2+b

2+

b2

√ a2+b

2=a

2+b

2√ a

2+b

2=

√ a2+b

2

58

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m4.7

:48

Tw

ove

rtic

alp

ole

sPQ

and

STar

ese

cure

db

ya

rop

ePR

Sgoin

gfr

om

the

top

of

the

firs

tp

ole

toa

poin

tR

on

the

gro

un

db

etw

een

the

pole

san

dth

ento

the

top

of

the

seco

nd

pole

asin

the

figu

reb

elow

.Sh

ow

that

the

short

est

len

gth

of

such

aro

pe

occ

urs

wh

enθ 1=θ 2

Solu

tion

:St

ep1:

Sket

ch:

Max

ima

and

Min

ima

59

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Step

s2&

3:V

aria

ble

s&

Sym

bols

:Le

ta=|PQ|,b=|ST|,d=|QT|,

andx=|QR|,

soth

at|ST|=

d−x

.Le

tθ 1=∠PRQ

,an

dle

tθ 2=∠SRT

.

Step

4:

Rel

atio

ns:|PR|=

√ a2+x

2an

d|RS|=√ b

2+(d−x)2

,

Step

5:

Iden

tifi

cati

on

:W

ew

ish

tom

inim

izef(x)=√ a

2+x

2+√ b

2+(d−x)2=(a

2+x

2)1 2+(b

2+(d−x)2)1 2

.

Step

6:

Diff

eren

tiat

e:f′ (x)=

1 2(a

2+x

2)−

1 2(2x)+

1 2(b

2+(d−x)2)−

1 2(2(d−x)(−1))=

x(a

2+x

2)−

1 2−(d−x)(b

2+(d−x)2)−

1 2=

x√ a

2+x

2−

d−x

√ b2+(d−x)2=|QR|

|PR|−

|ST|

|RS|=

cosθ 1−

cosθ 2=

0ifθ 1=θ 2

.

Step

7:

Solv

e:N

ote

thatf′ (

0)=−

d√ b

2+(d)2<

0an

df′ (d)=

d√ a

2+d

2>

0,s

oth

efi

rst

der

ivat

ive

test

tell

su

sth

atw

e

hav

ea

min

imu

mw

henf′ (x)=

0.

60

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m4.7

:50

Ast

eel

pip

eis

bei

ng

carr

ied

dow

na

hal

lway

9fe

etw

ide.

At

the

end

of

the

hal

lth

ere

isa

righ

t-an

gle

dtu

rnin

toa

nar

row

erh

allw

ay6

feet

wid

e.W

hat

isth

ele

ngth

of

the

lon

ges

tp

ipe

that

can

be

carr

ied

hori

zon

tall

yar

ou

nd

the

corn

er?

Solu

tion

:St

ep1:

Sket

ch:

Step

s2&

3:V

aria

ble

s&

Sym

bols

:In

dic

ated

inth

eab

ove

dia

gra

m.

Step

4:

Rel

atio

ns:

From

the

dia

gra

m,

we

hav

esi

nθ=

9 L 1,

and

cosθ=

6 L 2,

soth

atw

eh

aveL 1=

9si

nθ=

9cs

cθ

and

L 2=

6co

sθ=

6se

cθ

.

Step

5:

Iden

tify

:Fi

nd

the

min

imu

mva

lue

off(θ)=L 1+L 2=

9cs

cθ+

6se

cθ

.

Step

6:

Diff

eren

tiat

e:f′ (θ)=

9(−

cscθ

cotθ)+

6(s

ecθ

tanθ)=

0if

Step

7:

Solv

e:f′ (θ)=

0if

9cs

cθ

cotθ=

6se

cθ

tanθ

or

91

sinθ

cosθ

sinθ=

61

cosθ

sinθ

cosθ

or

Max

ima

and

Min

ima

61

3 2=

sin

3θ

cos3θ

or

3 2=

tan

3θ

or

tanθ=

3√ 3 2.

For

this

valu

eofθ

we

hav

ecs

cθ=√ √ √ √ 1

+( 2 3

)2 3

and

secθ=√ √ √ √ 1

+( 3 2

)2 3

,

sof(θ)=

9

√ √ √ √ 1+( 2 3

)2 3

+6

√ √ √ √ 1+( 3 2

)2 3

�21.0

7

62

Max

ima

and

Min

ima

Uni

vers

itas

Saskat

chew

anen

sis

DE

O

ET

PA

T-

RIÆ

2002 D

ou

g M

acLe

an

Pro

ble

m4.7

:52

Ara

ingu

tter

isto

be

con

stru

cted

from

am

etal

shee

tof

wid

th30

cmb

yb

end

ing

up

on

e-th

ird

of

the

shee

ton

each

sid

eth

rou

gh

anan

gleθ

.H

ow

shou

ldθ

be

chose

nso

that

the

gu

tter

wil

lca

rry

the

max

imu

mam

ou

nt

of

wat

er?

Solu

tion

:St

ep1:

Sket

ch:

Step

s2&

3:V

aria

ble

s&

Sym

bols

:In

dic

ated

inth

eab

ove

dia

gra

m.

Step

4:

Rel

atio

ns:

From

the

dia

gra

m,w

eh

ave

0≤θ≤π 2

,d 10=

cosθ

,sod=

10

cosθ

,an

dh 10=

sinθ

,soh=

10

sinθ

.

Th

ear

eaisA(θ)=

1 2dh+

10h+

1 2dh=dh+

10h=(1

0co

sθ

10)(

sinθ)+

10(1

0si

nθ)=

100(s

inθ

cosθ+

sinθ)=

50

sin

2θ+

100

sinθ

.

Step

5:

Iden

tify

:Fi

nd

the

valu

eofθ

that

max

imiz

esA(θ).

Step

6:

Diff

eren

tiat

e:A′ (θ)=

50(2

cos

2θ)+

100

cosθ=

100(2

cos2θ−

1+

cosθ)=

100(2

cosθ−

1)(

cosθ+

1)

Max

ima

and

Min

ima

63

Step

7:

Solv

e:A′ (θ)=

0if

cosθ=−1

(b

utθ=π

isou

tsid

eth

ed

om

ain

ofA

),or

2co

sθ=

1or

θ=π 3

.

Note

thatA′ (θ)=

200>

0,

andA′( π

2

) =−1

00<

0,

so,

by

the

Firs

tD

eriv

ativ

eT

est,

we

do

ind

eed

hav

eth

em

axim

um

valu

e.

Step

8:

Tra

nsl

ate:

Th

ean

gle

shou

ldb

e60◦ .