Các bài toán về giá trị lớn nhất và giá trị nhỏ nhất

Ket luan bai toan... Bien thien cua y la difcfng net duft... Noi dung phan nay chi neu cac ufng dung sau : 1.. Chufng minh cac bat dang thufc 2.. Giai cac phi/ang trinh c6 dang "khong ma

C101B NGUYEN THAI HOE

CAC BAI TOAN

T R I L O N NHA^

NGUYEN THAI HOE

GlA TRILON NHAT

VA GlA T R I N H O N H A T

Nhitng k

xac din

h

- GTL

N : Gia tr

i nh

o nha

t

- : Gia

m

-N ha xu

eo

n g

b o ta

c pham

PHAN MOT

GIAI C A C BAI TOAN G T L N , GTNN

BANG PHUONG PHAP C A O C A P (dung dao hdm)

Cifc d a i , circ t i e u v a G T L N , G T N N cua h a m so

S u p h a n b i e t do du'Oc m o t a m o t each h i n h hoc cho bori h i n h ve sau d a y

max f(x) = f(b)

min f(x) =

fcTi^u

= ^x^

)

Luu y :

- Ta

tieu

- Ta

i die

m c

6 hoan

h do

X 4

thi

f'(x4

)

= 0 nhifng

khong la diem

n thu

f ha

i tron

g din

h nghi

a : ton

xo) =

M hoS

c f(xo) =

m

Vi nh

ii ta biet :

sinx < 15 la bat dSn

5 v

i trong

TCr din

a c

6

II PHtfOfNG

P HA

P CHUN

G D

E T IM G TL

m s

o y = fix)

Buac 1.

Tim MXD ciia ha

h l

a X

Buac 2.

Tim trong

f'(x) khon

g to

n t

ai

Goi ta

h gi

a t

ri cu

a fix) t

ai ca

c bie

n cu

a fix) (ne

u c6) ;

,

r

- M a x f(x) = so 16'n nhat trong

- m i n fix) = so nho nhat trong •

- T i n h gia t r i cua fix) t a i moi diem thuoc mien xac d i n h ma f'(x) = 0

hoac f'(x) khong xac d i n h

- So sanh cac gia t r i bien va moi f ( x i ) , i = l , 2 , , k

Buac 4

Ket luan bai toan

cac gia t r i bien va c^c f ( x i ) ( i = l , 2 , , k ) ' cac gia t r i bien va

cac f ( X i ) , ( i = l , 2 , , k ) "

Lilu y :

Neu bai toan chi doi h o i t i m G T L N va G T N N ciia h a m so t h i khong nen

t i m gia t r i ctfc t r i de suy r a G T L N va G T N N cua h a m so

min y = y(0) =

Ta C

O y(t

) = 2t t

^ 3

Tim GTLN

va GTN

N cu

a y(t) tron

g [0, 1] :

Ta

CO

y'(t) =

2 4t2

Trong doa

n [

0 ; 1] , y'(t) =

0 ch

i kh

i t =

72

Tinh y(0), y V2

v2,

va yd) :

i mo

i xma

t = sinx = —

x = 0

2

Tim GTLN

n X

trong [1

21nx ln

^x X ^

Ldl GIAI

In x(

2

In x)

=> y' =

Tinh y d ) , y(e^), y(e^)

Giai ba

t phircfn

g trin

h y' >

Ta

dtro-c y' >

0 <

^ ijil

- >

(1 x)^ >

x^

< => 1

- X > X

'1'

[2)

-4 /

miny = y(0) =

Tim GTLN

m a x y = y(72) = 2V2

{Li/u y : nhieu hoc sinh ket luan m i n y = -2N/2 la sai v i khong c6

xo e [ - 2 ; 2] de y(xo) = -2V2)

De t i m miny, ta nhan xet r k n g :

K h i X = - 2 t h i ca 2 so hang x va V 4 - x ^ dong thcfi dat G T N N , vay

Chon (pe[0; 7i

] k

hi d

o x- 2c os 9e [-

2;

2]

sin cp > 0 sin

(p = si

n c

p

va

y = 2(sin cp + co s

9) =

2^2 sin

do O<

x<

7i :o -<

x + -

-2

<y

<272

(xem h in

h ve )

(diem ngon

M cu

a cun

g x + — chay tren cung

xy

- 2V

2

mi ny = -2

i to an

7

Tim GTL

N v

a GTN

Ldl GIAI

Ta c6 y' = -

5 si

n x + 5 sin 5x = 5(si

n 5

x sin x)

-y' =

0 o sin5x =

sinx

< =>

5x

= X +

Xo = - + /

0 ch

i k

hi x = 0, x = ±

De tin

h gi

a t

ri cu

a y, t

a d

e y rkn

g cosx, cos5x.l

a ham

so chin

y

6

_

7 1 57

= 37

= 6

^ =

373 2

va y(0) =

y =

y ± — =

373 ; min

y = y(0) =

4

I 6j

Ba

i to an

8

Tim GTL

N v

a GTN

N cu

a ham

so y = 2

sin^

x + cos'^ 2

x

LOI GIAI

Ta bien doi

+ cos'

* 2

cos2x^

^l-^^

(l-cos2x)^

+cos

^2

Chon phep bien doi

t = cos2x

t e [-1 ; 1]

Voi bien t , h a m y = ^ ( 1 - t ) ^ +

8

Ta t i m G T L N va G T N N cua y(t) trong [ - 1 ; 1]

(2t)^ - (1 - 1 ) ^ Taco y'(t) = ^ ( l - t ) 3 ( - l ) + 4 t 3 = ^

maxy = 3, dat t a i moi x ma cos2x = 1 x = krc

m i n y = — , dat t a i moi x ma cos 2x = — = cos 2a => x = ± a + /:t

Ap dung bat dSng thufc Cosi doi v d i t i c h 5 thiTa so khong a m , trong do c6

— sin*^ X va 3 thiifa so' — 1

2

2 f

l 2 ' ' 1

108

3125

Dau = xay r

a kh

i : — sin

^ x — cos

^ x)

^ (cos

^ x)

^ = (sin^ x)

^ x => u

e [

0 ; 1,]

Vdri bie

n u, ha

m y c6 dan

g

y(u) =

u 2

u)^

vdriu G

[0 ; 1]

Ta tim GTLN

va GTN

N cu

a y(u) tron

g [

0 ; 1]

Taco y'(u) =

u2 3(l-u)2

.(-l) + (l-u)^.2u

= u(

l - uf

[2(1 u) - 3u;

-= u(l-u

)2

(2-5u)

y' =

0 kh

i u = 0, u = 1, u = -

Bai toan 10

Ldl GlAl

Ta CO y' = sin^ x.q cos^"^ x(- sin x) + cos*' x.p sin^"^ x cos x

= p cos^"^^ X sinP"-^ X - q sin^"^^ x cos^~^ x

= sinP~^ x cosP"^ x.[q cos^ x - p sin^ x]

P q (p + q)2 (p + q)2

>0

min y =^(0) = y = 0

i to an

11

Tim GTL

N v

a GTN

N cu

a ha

m s

o y = sin^

^ x + cos^

° x

Ldl GlAl

Chon

u sin

x =

> u

e [

0 ; 1]

Kh

i d

o :

y ^u^

^ + (l-u)

^°

Tim GTL

N v

a GTNN cua ha

m y(u) tron

g [

0 ; 1]

Ta

CO

y' = lOu^

- 10(

1 u

l ^ _ b)[a2

" + a^"-^b

+ •

• + ab^""^

- (

1 u)] [u

-^ + u

"(l

- u) + •••

+ u(l - u)"^ + (1 -

i bSn

g 0 nen

A >

0

=> y' =

0 k

hi u = - 2

Ta

CO

y(0) =

1, yd) =

(me

Taco y(4) =

21 + -ln

2 2

yd) =

0

/1 \

Vay

4 2

2 2

maxy = y(4) =

21 +

-ln

2 2 miny =

y(l)

= 0

Ba

i toa

n 13

Tim GTLN

n

X

- I

n 5 = In

x

5 5 n - = I X In 0 o ' = = :> y <=>

X

= - e [1; 5

Ta

CO

yd) =

- ln5,

0

min y = y '5^

5

e

Bai toan 14

Tim GTLN cua ham so

y = 2x^ - 3x^ - 12x + 1 trong doan [-3 ; 3]

Ldl GIAI

Goi z = 2x^ - 3x2 - I2x + 1, khi do y =

Ta xet bien thien cua ham so z tren [-3 ; 3] roi suy ra bien thien cua

ham so y de tii do c6 ket luan cua bai toan

=> z' = O o x = - l hoac x = + 2

De hieu bai toan mot each tiTdng minh, ta nen dung bang bien thien sau :

Luu y tren bang bien thien :

Bien thien cua z Ja diidng net lien

Bien thien cua y la difcfng net duft

TCr

M, I,

N bie

n th ie

n cii

a y v

a z trCing

nh au

Doc tr en b an

g bie

n th ie

n ta t hu

duac

ma

x y = y (- 3) =

44

Ch

u y r kn

g, ba

i to an thiTc ra c

hi ca

n gia

i n gS

m z ' va g ia

i z ' =

0 t im cac ng hi

em

- Ti nh

z ( -3 ), z(3 ) va g ia t

ri cii

a z(x ) ta

i ca

c ng hi em m

a a

day

la

3) =

- 44 , z{

-l ) =

8, z(2 ) =

- 19 , z(3 ) = -

8

- D

o y = |

z | ch

o ne

n GT LN ciia ha

m y c

hi la

so Io

n n ha

t t ro ng cac s

= 8 , |z(2)

| = 1

9, |z(3)

| =

8

suy ra m ax

y = |z(-3)|

= 4

4

Nh if ng

de hie

u

ro ti nh c ha

t bie

n th ie

n cii

a ha

m y t

n to

t ho'n ,

Ba

i to an

15

Go

i X

j , X2 la cac ng hi em ciia phiTdn

g t ri nh :

x^

+ p

x + = 0

1) (

p ^ 0)

Ha

y ti

m p

de :

u = x

f + x^

da

t gia t

ri nh

d nh

at va t im G TN

N cu

a u

Ldl GIAI

Qu

y tr in

h gia

i b

ai to an bao go

u ki en ciia p

de phiTcrn

g t ri nh 1) c

6 ng hi em (Buoc

Budc 2.

T im

u t he

o p

Budc 3.

T im G TN

N cii

a u v

di mo

i p da t im

dJcfc

t ro ng bi/dc 1

+ Die

u ki en

de ph ifo ng t ri nh 1) c

6 ng hi em

la

A >

0 * )

oi u

ve da ng t hu an Icfi d

e du ng diToc di nh l

i V ie

t ch

o phifofn

g

tr in

h 1)

+ Tim GTNN ciia u(p) vo'i moi p thoa man dieu kien *)

(co the giai khong dung, nhiT sau :

Do p'^ + > 2 => min u = 2 - 4 = -2 la sai vi voi dieu kien *) thi

P

p n - L > 2 )

P

Ta chon : t = p'^ ^ t > 4

Khi do ta diipc u(t) = t t - - 4

Ta tim min u(t) voi t > 4

=:> u(t) la ham so dong thien vdti t > 2 va khi do thi miny(t) = —

4

Ta ket luan diToc :

minu =^ - , dat khi t = p'' = 4 => P = ± N / 2

4

Bai toa

n 16

3a^x^

Sa^x + a^ - a^ + -

f + x| da

t GTNN, GTL

6 nghie

m :

A >

0 *)

^ +

12a'* -9a^

>0

Sa^Ca^ 4a2 + 3) <

0 (d

o 33

^ >

0)

(a^ -1) (a

^ 3) <

i +

X2)

Vdri

moi a thoa ma

n die

u kie

n *) t

-4

Khi d

o u = a

n 2 mien 1

; y

/sl

va T-Vs

; -1

i a

=> Tre

n 2 mien tho

a ma

n die

u kie

n *), ha

m u(a) l

o bd

i ban

g sa

u :

CO

u'(t) = (t +

2)(2) (2

- 2t)

(t +

2Y

(t + 2)2

1 da

t kh

i t = 0 => x = 0 hoac y = 0

min u = u

dinh ve sU c6 gia trj Id

Dieu dang luu

N

Dieu do c6 nghTa

Khi do, d

u :

Bai toa

n 18

Cho ha

m so ' y = 7

x

-1 + 7

9

- x

Hay ti

m G TL

N v

a G TN

3 ; 6) ; c) [

3 ; 6) ; d) (

3 ; 6]

Ldl GIAI

Ta co y' =

_ l=

+

L=

l) =

1

V9-x.Vx-

-1 <^

x

<5

Bien thien cu

a ha

m y cho bd

i ban

g sa

u

X 3 5 6

y

y

TCr bang bien t h i e n ta suy ra ket qua ciia bai toan trong cac triiong hop :

b) K h i mien xac dinh la (3 ; 6)

m a x y = y(5) = 4

m i n y : khong c6 v i t a i x = 3, y khong xac d i n h

c) K h i mien xac dinh la : [3 ; 6)

T i m G T L N va G T N N ciia ham so trong cac mien :

Chu y : N/2 + \/6 < \/5 + \/3 va y lien tuc trong cac mien da cho

a) K h i mien xac dinh la [3 ; 6] :

Ldl GIAI

Ta CO y ' = 3x^ + 6x - 9 = 3(x^ + 2x - 3) ^ y ' = 0 o x = l

x = - 3

va ca ham y va y' deu lien tuc trong mien da cho

Bien t h i e n ciia ham y cho bdi bang :

T Cr bang, t

maxy = y(-3) =

y(3) =

16

min

y = y(-5) = yd) =-1

maxy = y(-3) =

y(3)-16

miny = y(l) =

-16 ; (x = -5 khong thuo

c mie

n xa

c dinh)

c) Mie

n xa

c din

h = (-5 ; -3)

maxy y(-3) =

miny = y(l) =

-16 (

x = -5 khong thuo

c mie

n xa

c dinh)

Ba

i toa

n 20

Cho ha

m s

o y = x + - Ti

c mie

a)

[-2 ; 2] ; b) (

0 ; 2]

Ldl GIAI

Taco y' =l-

do li

m y

= o o

-o

o

Bien thien

Tif bang va do thi, ta c6 ket qua

a) MXD = [-2 ; 2], ham y khong Hen tuc trong [-2 ; 2]

max y : khong c6 vi Hm y = +oo

min y : khong c6 vi Um y = -<x>

x->0'

Ta chi thu dugfc :

y^,^=y(l) = 2 ; y c p = y ( - l ) = -2

b) MXD = (0 ; +2] : ham y Hen tuc trong (0 ; +oo)

(bang bien thien cua y la nufa bang ben phai x = 0 Do thi y chi la nhanh

ben phai true Oy cua do thi da ve)

Ta CO max y : khong c6 vi lim y = +QO

BAI TAP TLT LUYEN

1 Tim GTLN va GTNN cua ham so : y = 4 cos^ x + sVs sin x + 7 sin^ x

Tim

TL

N v

a GTN

N cu

a y

= slx

+ 1 +

V4-

x

4.

Tim

TL

N v

a GTN

X

5.

Tim

TL

N v

a GTN

TL

N v

a GTN

TN

N cii

a y = -xlnx-xln

TN

N cii

a y = sin^

^ x + co

s ^"

* x ' ~

9.

Tim mien gia t

ri cii

a ham

TL

N v

a GTN

N cu

a ham

1 + cosx

y = cos -2x + co

s 4x

11.

Cho phirong

trin

h :

x^ (2 si

-n a

- l)x + 6 sin^ a

- si

n a

- 1 = 0

Goi x^

, X

2

la cac nghie

m cii

a phifOn

g trin

h 1)

Tim GTL

N v

a GTN

N cu

a u = x

^ + x|

12.

Tim

TN

N cu

a ham

so

y = 2x^ +15x

^ +

36 X

-3

0 tron

g [-3: 2

13.

Goi

Xj , X 2

la nghiem

ciia phiion

g trin

h

12x2 -6ax + a

^ _

4 + :

t GTL

N v

a GTN

N

14.

Tim

TL

N v

a GTN

§2 ifNG DVNG B A I T O A N TIM G T L N , GTNN V A O V I E C

G I A I C A C B A I T O A N K H A C

Noi dung phan nay chi neu cac ufng dung sau :

1. Chufng minh cac bat dang thufc

2. Tim gia t r i ciia tham so de phiiong trinh c6 nghiem

3 Giai cac phi/ang trinh c6 dang "khong mau miTc"

trong do m, M la cac hSng so ; fix) la ham so ciia bien so x c6 khi (thay

vi fix) la ham ciia x) la bieu thufc chufa nhieu dai lifcJng tham gia nhiTng diTo'c xem la ham ciia m6t trong cac dai liio'ng do

II PhUdng phap giai

Bai toan chufng minh cac bat dang thufc da cho diTo'c xem la hoan thanh neu ta chufng minh difo'c :

max fix) = M (cho a))

min fix) = m (cho b))

maxf(x) = M

min f (x) = m (cho c))

Khi do tuy tCrng loai ham so, khi tim GTLN hoac GTNN cua fix) c6 the dung cac phtfcfng phap so" cap (khong diing dao ham) hoac phtfcfng phap cao cap (dung dao ham)

Chii y : Trong cac bat dang thufc a), b) va c) deu phai c6 dang thufc (dau =)

2

1 ^

Chufng min

h rSn

g vdr

i mo

i x t

hi ba

t din

Ta

CO

x^ + (1 - x)^

= x

^ +

1 - 5

x + lOx^

- lOx

^ + 5x^ - x^

= Sx^

- lOx

^ + lOx^

"*

-lOx^

+ 10x2-5

+ 32x

2 I6x + 3 > 0

-« (lex^

^ 32x3 ^ 24x2

_ 8

x + 1) + (Sx^ - 8x + 2) >

0

(2x 1)^ + 2(2x -

Dau = xa

y r

a k

hi 2x-l = 0ox = — 2

Cdch 2.

Goi f (x) = x^ + (1 - x)^

= > f(x) >

c chufn

g min

h ne

u t

a chufn

g min

+5(l-x)'^

.(

-l) =

5

f (x) =

5 x4

-(

x)

l-4

x2 + (1 - x)2 x2-(

x)

1 >

0 vd

i mo

i x, ne

n

f (x) = 0c^

2x-l = 0

<:^

x = -

2

Bien thien ciia ham fix) cho bdi bang difdfi day

l i m f(x) = lim(5x'' - lOx^ + lOx^ - 5x + 1)

Bai t o a n 22

3x^ — 4xy Cho ham so f ( x , y ) = — ^ ~ - trong do x va y khong dong thcfi bSng 0

(x + 2v)^

g thof

i bkn

g 0, dSn

-4xy+ y^

) _ _ (2x-y)

g thcf

i bSn

g 0, dan

ma x + 2

Ba

i toa

n 23

Hay chufn

g min

0 1)

dung vdt

i mo

i x, y trong d

o

P(x, y) = x^ +

xy + y^ - 3(x + y) +

h

min P(x, y) =

3

3"

y-2

2 - - + y + y 2 X + [ 2

2x + y-

3)

2+

-i

(y -l

Ro rang la P(x, y) > 0 \6i moi x, y, dau = xay r a k h i

cos'^ (D sin* (D + cos"* ( 0 4 4 4

t diigf

c k

hi x = ±

Ta

CO

2 x^) (l +4x

(l + x^)2(l +

x2).2x 4x(x

2-i)

y' = O ox =

0,

x = ±

l

hm

y

= li

n cu

a y cho bd

n t

a thu duftfc

(dpcm)

Ba

i t oa

n 25

Chufng min

1

^ (x

^ y^

-)(

xV ) ^

1

4

~ (l + x2)2(l +

y2)2 ~

4 1)

LOI GIAI

Goi u = ( x ^ - y ^ ) ( l - x V ) (l + x2)2(l + y2)2

Ta chi can chutog minh max u = — va min u = - —

4 4 (dung phep bien doi stf cap)

Ta bien doi difo'c

va ta thu diroc u(x,y) = z(x) - t(y)

TCr do ta CO :

max u = max z(x) - min t(y)

min u = min z(x) - max t(y) *^

(do z(x) va t(y) la hai ham so c6 bien so rieng biet)

Van de con lai la tim max va min cua ham so

h(a)=

(l + a2)2

Ta CO ngay : h(a) = 0, dau = xay ra khi a = 0

Tii do ta c6 dtfo'c min h(a) = h(0) = 0

De tim max h(a) (do a ^ 0 vi khi a = 0 ham h(a) dat GTNN, ta da xet) ta chia tuf va mau cua h(a) cho a^ va difoc

1 h(a) =

( 1^

a +

-do

1 — a +

a

> 2, da

u = xay r

a k

hi a = ±

1

\

a + -

V

a

> 4 =>

h(a) <

—, da

u = xay r

a k

hi a = ±

chi vie

c xe

m a = x roi a = y trong

ham

so h(a) v

a ch

u y t

di *) t

- 0 = — dat dtfof

c k

minu = 0

- — = -

h

Chii y :

Co the dun

a ,

y = tan

a dtfof

c ^ -

u = — (co

s 4p

- cos

Ba

i to an

26

Cho ha

m s

o u ^ x

^ + y^, bie

t ran

g x va

y tho

a ma

n phirorn

g trin

h :

(x2 y2 +1)' + 4x2y2 -

Sx^ 2y2 +

2 =

0 1)

Hay chufng

a ch

i ca

n chijfn

g min

h : vdi mo

i x

, y

thoa ma

n 1) t

hi :

max u = 3

va min

u = +

l

Bien doi

1) o (x -

yf +

4x2y2 5x2 _

x2+

3 =

0

(ta d

a lam xuat hie

n u = x

^ + y^)

bat dang thufc Cosi V i vay t a bien doi h a m u ve dang sau :

x + 1 y + 1 z + 1

= 3

- V

v& ma

x u = 3

V >

3 „, ^ =

, da

u = xay r

a kh

i x + l

^ = -

3 3 => x + l = z + l = y + l i x a kh xay r dau =

= y==

z = —

u = xay r

a kh

i x = y = z = -

4 4,

ta c

6

1^

(x + y)(l-xy)

^1

2 (l +

= (x

^y)(l-xy)

n chufn

g min

h ma

x u = —

va mi

n u = -

Chi ca

n d

e y den dSn

g thuf

c dun

g sa

u : x^Xl + 1 + = ( xy)2 (1 - yf + (X + y^)

Khi d

o u=

(x

-H y)(

l-xy)

(x +

r + (1 - x

dau = xay ra k h i (x + y) (1 - xy)

_x + y = xy - 1

Ta t h u di/ofc u < => i < u <

-2 -2 -2 TCr do ta ket luan dirorc

Goi y = V ( l - x)^ + + x)^ Ta chufng m i n h : max y = 4V2

Do 0 < X < 1, ta chon x = coscp vdi 0 < cp < —

2 2 2 2 COS — < COS —, dau = xay r a k h i cos — = 0 hoSc cos —

2 2 2 2 sin^ ^ + cos^ - < sin^ ^ + cos^ ^ = 1

dau = xa

y r

a k

hi sin — =

• 9 sin —

= 1

• 9 sin —

2

= 0

va chu

= 1

cos(p

= 1

- 2 sin

2 ^ = -1

y <

4>/2.1

, da

u = xay r

30

Cho cac s

o thii

c x va

y khong dong thor

i bkn

<^

j"

' y

'<

2V 2-

Nhan xet ra

ng

maxu

Ux 2^

4y2j

= 1

ta bieu dien ham y ve dang mdi sau :

Do cac phiio'ng t r i n h a) va b) chac ch&n c6 nghiem (p, tijf do suy r a x va y

nen ta ket luan difoc :

Ta can chuTng m i n h max A = 1

Tii dieu k i e n 1) t a chon difOc

a - cos a

6 - cosfi' v d i 0 < a, P < 71