c´o diˆe`u kiˆe.n chung quy l`a gia’i hˆe... c´o diˆe`u kiˆe.n.
Trang 19.3 Cu c tri cu’a h`am nhiˆe`u biˆe´n
9.3.1 Cu c tri.
H`am f (x, y) c´o cu c da.i di.a phu.o.ng (ho˘a.c cu c tiˆe’u di.a phu.o.ng) b˘a`ng
f (x0, y0) ta.i diˆe’m M0(x0, y0) ∈ D nˆe´u tˆ` n ta.i δ-lˆan cˆa.n cu’a diˆe’m Mo 0
sao cho v´o.i mo.i diˆe’m M 6= M0 thuˆo.c lˆan cˆa.n ˆa´y ta c´o
f (M ) < f (M0) (tu.o.ng ´u.ng : f (M ) > f (M0)).
Go.i chung cu c da.i, cu c tiˆe’u cu’a h`am sˆo´ l`a cu c tri cu’a h`am sˆo´
Diˆ`u kiˆe.n cˆae ` n dˆe’ tˆo` n ta.i cu c tri di.a phu.o.ng: Nˆe´u ta.i diˆe’m M0 h`am
f (x, y) c´o cu c tri di.a phu.o.ng th`ı ta.i diˆe’m d´o ca’ hai da.o h`am riˆeng cˆa´p
1 (nˆe´u ch´ung tˆ` n ta.i) dˆe`u b˘a`ng 0 ho˘a.c ´ıt nhˆa´t mˆo.t trong hai da.o h`amo
riˆeng khˆong tˆ` n ta.i (d´o l`a nh˜u.ng diˆe’m t´o.i ha.n ho˘a.c diˆe’m d`u.ng cu’ao
h`am f (x, y)) Khˆong pha’i mo.i diˆe’m d`u.ng dˆe`u l`a diˆe’m cu c tri
Diˆ`u kiˆe.n du’: gia’ su.’e
f xx00 (M0) =, f xy00 (M0) = B, f yy00 (M0) = C.
Khi d´o:
i) Nˆe´u ∆(M0) =
A B
> 0 v` a A > 0 th`ı ta.i diˆe’m M0 h`am f c´o
cu c tiˆe’u di.a phu.o.ng
ii) Nˆe´u ∆(M0) =
A B
B C
> 0 v` a A < 0 th`ı ta.i diˆe’m M0 h`am f c´o cu c da.i di.a phu.o.ng
iii) Nˆe´u ∆(M0) =
A B
< 0 th`ı M0 l`a diˆe’m yˆen ngu a cu’a f, t´u.c l`a ta.i M0 h`am f khˆong c´o cu c tri
iv) Nˆe´u ∆(M0) =
A B
B C
= 0 th`ı M0 l`a diˆe’m nghi vˆa´n (h`am f c´o thˆe’ c´o v`a c˜ung c´o thˆe’ khˆong c´o cu c tri ta.i d´o)
Trang 29.3.2 Cu c tri c´o diˆe`u kiˆe.n
Trong tru.`o.ng ho p do.n gia’n nhˆa´t, cu c tri c´o diˆe`u kiˆe.n cu’a h`am f(x, y)
l`a cu c da.i ho˘a.c cu c tiˆe’u cu’a h`am d´o da.t du.o c v´o.i diˆe`u kiˆe.n c´ac biˆe´n
x v` a y tho’a m˜ an phu.o.ng tr`ınh ϕ(x, y) = 0 (phu.o.ng tr`ınh r`ang buˆo c).
Dˆe’ t`ım cu c tri c´o diˆe`u kiˆe.n v´o.i diˆe`u kiˆe.n r`ang buˆo.c ϕ(x, y) ta lˆa.p
h`am Lagrange (h`am bˆo’ tro..)
F (x, y) = f (x, y) λ ϕ(x, y)
trong d´o λ l`a h˘a`ng sˆo´ nhˆan chu.a du.o c x´ac di.nh v`a di t`ım cu c tri thˆong thu.`o.ng cu’a h`am bˆo’ tro n`ay Dˆay l`a phu.o.ng ph´ap th`u.a sˆo´ bˆa´t di.nh Lagrange
T`ım diˆ`u kiˆe.n cˆae ` n dˆe’ tˆo` n ta.i cu c tri c´o diˆe`u kiˆe.n chung quy l`a gia’i
hˆe phu.o.ng tr`ınh
∂F
∂x =
∂f
∂x + λ
∂ϕ
∂x = 0
∂F
∂y =
∂f
∂y + λ
∂ϕ
∂y = 0 ϕ(x, y) = 0
(9.15)
T`u hˆe n`ay ta c´o thˆe’ x´ac di.nh x, y v`a λ.
Vˆa´n dˆ` tˆoe ` n ta.i v`a d˘a.c t´ınh cu’a cu c tri di.a phu.o.ng du.o c minh di.nh trˆen co so.’ x´et dˆa´u cu’a vi phˆan cˆa´p hai cu’a h`am bˆo’ tro
d2F = ∂
2
F
∂x2dx2+ 2 ∂
2
F
∂x∂y dxdy +
∂2F
∂y2dy2
du.o c t´ınh dˆo´i v´o.i c´ac gi´a tri x, y, λ thu du.o c khi gia’i hˆe (9.15) v´o.i diˆe`u kiˆe.n l`a
∂ϕ
∂x dx +
∂ϕ
∂y dy = 0 (dx
2
+ dy2 6= 0).
Cu thˆe’ l`a:
Trang 3i) Nˆe´u d2F < 0 h` am f (x, y) c´o cu c da.i c´o diˆe`u kiˆe.n.
ii) Nˆe´u d2F > 0 h` am f (x, y) c´o cu c tiˆe’u c´o diˆe`u kiˆe.n
iii) Nˆe´u d2F = 0 th`ı cˆ` n pha’i kha’o s´at.a
Nhˆa n x´et
i) Viˆe.c t`ım cu c tri cu’a h`am ba biˆe´n ho˘a.c nhiˆe`u ho.n du.o c tiˆe´n h`anh
tu.o.ng tu nhu o.’1
ii) Tu.o.ng tu c´o thˆe’ t`ım cu c tri c´o diˆe`u kiˆe.n cu’a h`am ba biˆe´n ho˘a.c
nhiˆ`u ho.n v´o.i mˆo.t ho˘a.c nhiˆe`u phu.o.ng tr`ınh r`ang buˆo.c (sˆo´ phu.o.nge
tr`ınh r`ang buˆo c pha’i b´e ho.n sˆo´ biˆe´n) Khi d´o cˆ` n lˆa.p h`am bˆo’ tro v´o.ia
sˆo´ th`u.a sˆo´ chu.a x´ac di.nh b˘a`ng sˆo´ phu.o.ng tr`ınh r`ang buˆo.c
iii) Ngo`ai phu.o.ng ph´ap th`u.a sˆo´ bˆa´t di.nh Lagrange, ngu.`o.i ta c`on
d`ung phu.o.ng ph´ap khu.’ biˆe´n sˆo´ dˆe’ t`ım cu c tri c´o diˆe`u kiˆe.n
9.3.3 Gi´ a tri l´ o.n nhˆ a ´t v` a b´ e nhˆ a ´t cu’a h` am
H`am kha’ vi trong miˆ`n d´ong bi ch˘a.n da.t gi´a tri l´o.n nhˆa´t (nho’ nhˆa´t)e
ho˘a.c ta.i diˆe’m d`u.ng ho˘a.c ta.i diˆe’m biˆen cu’a miˆe`n
C ´ AC V´ I DU . V´ ı du 1 T`ım cu..c tri di.a phu.o.ng cu’a h`am
f (x, y) = x4+ y4− 2x2+ 4xy − 2y2.
Gia’i i) Miˆ`n x´ac di.nh cu’a h`am l`a to`an m˘a.t ph˘a’ng Re 2
ii) T´ınh c´ac da.o h`am riˆeng f0
x v`a f0
y v`a t`ım c´ac diˆe’m t´o.i ha.n Ta c´o
f x0 = 4x3− 4x + 4y, f y0 = 4y3+ 4x − 4y.
Do d´o
4x3 − 4x + 4y = 0 4y3 + 4x − 4y = 0
Trang 4v`a t`u d´o
(
x1 = 0
y1 = 0
(
x2 = −
√ 2
y2 =
√ 2
(
x3 =
√ 2
y3 = −
√
2.
Nhu vˆa.y ta c´o ba diˆe’m t´o.i ha.n V`ı f x0, f y0 tˆ` n ta.i v´o.i mo.i diˆe’mo
M (x, y) ∈ R2 nˆen h`am khˆong c`on diˆe’m t´o.i ha.n n`ao kh´ac
iii) Ta t´ınh c´ac da.o h`am riˆeng cˆa´p hai v`a gi´a tri cu’a ch´ung ta.i c´ac diˆe’m t´o.i ha.n
f xx00 (x, y) = 12x2 = 4, f xy00 = 4, f yy00 = 12y2 − 4.
Ta.i diˆe’m O(0, 0): A = −4, B = 4, C = −4
Ta.i diˆe’m M1(−
√
2, +
√
2): A = 20, B = 4, C = 20 Ta.i diˆe’m M2(+
√
2, −
√
2): A = 20, B = 4, C = 20.
iv) Ta.i diˆe’m O(0, 0)ta c´o
A B
B C
=
= 16 − 16 = 0.
Dˆa´u hiˆe.u du’ khˆong cho ta cˆau tra’ l`o.i Ta nhˆa.n x´et r˘a`ng trong lˆan
cˆa.n bˆa´t k`y cu’a diˆe’m O tˆo` n ta.i nh˜u.ng diˆe’m m`a f(x, y) > 0 v`a nh˜u.ng
diˆe’m m`a f (x, y) < 0 Ch˘ a’ng ha.n do.c theo trung c Ox (y = 0) ta c´o
f (x, y)
y=0 = f (x, 0) = x4− 2x2 = −x2(2 − x2) < 0
ta.i nh˜u.ng diˆe’m du’ gˆa` n (0, 0), v`a do.c theo du.`o.ng th˘a’ng y = x
f (x, y)
y=x = f (x, x) = 2x4 > 0
Nhu vˆa.y, ta.i nh˜u.ng diˆe’m kh´ac nhau cu’a mˆo.t lˆan cˆa.n n`ao d´o cu’a diˆe’m O(0, 0) sˆo´ gia to`an phˆ` n ∆f (x, y) khˆong c´o c`a ung mˆo.t dˆa´u v`a do d´o ta.i O(0, 0) h`am khˆong c´o cu c tri di.a phu.o.ng.
Ta.i diˆe’m M1(−
√
2,
√ 2) ta c´o
A B
B C
=
20 4
4 20
= 400 − 16 > 0
Trang 5v`a A > 0 nˆ en ta.i M1(−
√
2,
√ 2) h`am c´o cu c tiˆe’u di.a phu.o.ng v`a
fmin = −8
Ta.i diˆe’m M2(√2, −√2) ta c´o AC − B2 > 0 v` a A > 0 nˆen ta.i d´o
h`am c´o cu c tiˆe’u di.a phu.o.ng v`a fmin = −8
V´ ı du 2 Kha’o s´at v`a t`ım cu c tri cu’a h`am
f (x, y) = x2+ xy + y2− 2x − 3y.
Gia’i i) Hiˆe’n nhiˆen D f ≡R
ii) T`ım diˆe’m d`u.ng Ta c´o
f0
x = 2x + y − 2
f0
y = x + 2y − 3 ⇒
2x + y − 2 = 0,
x + 2y − 3 = 0.
Hˆe thu du.o c c´o nghiˆe.m l`a x0 = 1
3, y0 =
4
3 Do d´o
1
3,
4 3
l`a diˆe’m d`u.ng v`a ngo`ai diˆe’m d`u.ng d´o h`am f khˆong c´o diˆe’m d`u.ng n`ao kh´ac v`ı
f0
x v`a f0
y tˆ` n tˆa.i ∀(x, y).o
iii) Kha’o s´at cu c tri Ta c´o A = f00
x2 = 2, B f00
xy = 1, C = f00
y2 = 2
Do d´o
∆(M0) =
2 1
1 2
= 3 > 0 v` a A = 2 > 0
nˆen h`am f c´o cu..c tiˆe’u ta.i diˆe’m M0(1
3,
4 3
N
V´ ı du 3 T`ım cu..c tri cu’a h`am f(x, y) = 6 − 4x − 3y v´o.i diˆe`u kiˆe.n l`a
x v` a y liˆen hˆe v´o.i nhau bo.’i phu.o.ng tr`ınh x2+ y2 = 1
Gia’i Ta lˆa.p h`am Lagrange
F (x, y) = 6 − 4x − 3y + λ(x2+ y2 − 1).
Ta c´o
∂F
∂x = −4 + 2λx,
∂F
∂y = −3 + 2λy
Trang 6v`a ta gia’i hˆe phu.o.ng tr`ınh
−4 + 2λx = 0
−3 + 2λx = 0
x2+ y2 = 1 Gia’i ra ta c´o
λ1 = 5
2, x1 =
4
5, y1 =
3 5
λ2 = −5
2, x2 = −
4
5, y2 = −
3 5 V`ı
∂2F
∂x2 = 2λ, ∂
2F
∂x∂y = 0,
∂2F
∂y2 = 2λ
nˆ
d2F = 2λ(dx2 + dy2).
Nˆe´u λ = 5
2, x =
4
5, y =
3
5 th`ı d
2
F > 0 nˆen ta.i diˆe’m 4
5,
3 5
h`am c´o cu c tiˆe’u c´o diˆe`u kiˆe.n
Nˆe´u λ = −5
2, x = −
4
5, y = −
3
5 th`ı d
2
F < 0 v`a do d´o h`am c´o cu c da.i c´o diˆe`u kiˆe.n ta.i diˆe’m−4
5, −
3 5
Nhu vˆa.y
fmax= 6 + 16
5 +
9
5 = 11,
fmin = 6 − 16
5 −
9
5 = 1. N
V´ ı du 4 T`ım cu..c tri c´o diˆe`u kiˆe.n cu’a h`am
1) f (x, y) = x2 + y2+ xy − 5x − 4y + 10, x + y = 4.
2) u = f (x, y, z) = x + y + z2
(
z − x = 1,
y − xz = 1.
Trang 7Gia’i 1) T`u phu.o.ng tr`ınh r`ang buˆo.c x + y = 4 ta c´o y = 4 − x v`a
f (x, y) = x2+ (4 − x)2+ x(4 − x) − 5x − 4(4 − x) + 10
= x2− 5x + 10,
ta thu du.o c h`am mˆo.t biˆe´n sˆo´
g(x) = x2− 5x + 10
v`a cu c tri di.a phu.o.ng cu’a g(x) c˜ung ch´ınh l`a cu c tri c´o diˆe`u kiˆe.n cu’a
h`am f (x, y) ´Ap du.ng phu.o.ng ph´ap kha’o s´at h`am sˆo´ mˆo.t biˆe´n sˆo´ dˆo´i
v´o.i g(x) ta t`ım du.o c g(x) c´o cu c tiˆe’u di.a phu.o.ng
gmin = g5
2
= 15
4 ·
cu c tiˆe’u c´o diˆ`ue kiˆe.n ta.i diˆe’m 5
2,
3 2
(y = 4 − x ⇒ y = 4 −5
2 =
3
2) v`a
fmin = f5
2,
3 2
= 15
4 · 2) T`u c´ac phu.o.ng tr`ınh r`ang buˆo.c ta c´o
z = 1 + x
y = x2+ x + 1
v`a thˆe´ v`ao h`am d˜a cho ta du.o c h`am mˆo.t biˆe´n sˆo´
u = f (x, y(x), z(x)) = g(x) = 2x2+ 4x + 2.
Dˆ˜ d`ang thˆa´y r˘a`ng h`am g(x) c´o cu c tiˆe’u ta.i x = −1 (khi d´o y = 1,e
z = 0) v`a do d´o h`am f (x, y, z) c´o cu c tiˆe’u c´o diˆe`u kiˆe.n ta.i diˆe’m
(−1, 1, 0) v`a
f = f (−1, 1, 0) = 0 N
Trang 8V´ ı du 5 B˘a`ng phu.o.ng ph´ap th`u.a sˆo´ bˆa´t di.nh Lagrange t`ım cu c tri c´o diˆ`u kiˆe.n cu’a h`ame
u = x + y + z2
v´o.i diˆ`u kiˆe.ne
(
z − x = 1
(xem v´ı du 4, ii))
Gia’i Ta lˆa.p h`am Lagrange
F (x, y, z) = x + y + z2+ λ1(z − x − 1) + λ2(y − zx − 1)
v`a x´et hˆe phu.o.ng tr`ınh
∂F
∂x = 1 − λ1− λ2z = 0
∂F
∂y = 1 + λ2 = 0
∂F
∂z = 2z + λ1− λ2x = 0
ϕ1 = z − x − 1 = 0
ϕ2 = y − xz − 1 = 0.
Hˆe n`ay c´o nghiˆe.m duy nhˆa´t x = −1, y = 1, z = 0, λ1 = 1 v`a
λ2 = −1 ngh˜ıa l`a M0(−1, 1, 0) l`a diˆe’m duy nhˆa´t c´o thˆe’ c´o cu c tri cu’a h`am v´o.i c´ac diˆ`u kiˆe.n r`ang buˆo.c ϕe 1 v`a ϕ2
T`u c´ac hˆe th´u.c
z − x = 1
y − xz = 1
ta thˆa´y r˘a`ng (9.16) x´ac di.nh c˘a.p h`am ˆa’n y(x) v`a z(x) (trong tru.`o.ng ho p n`ay y(x) v`a z(x) dˆe˜ d`ang r´ut ra t`u (9.16)) Gia’ su.’ thˆe´ nghiˆe.m
Trang 9y(x) v` a z(x) v`ao hˆe (9.16) v`a b˘a`ng c´ach lˆa´y vi phˆan c´ac dˆo` ng nhˆa´t
th´u.c thu du.o c ta c´o
(
dz − dx = 0
dy − xdz − zdx = 0 ⇒
(
dz = dx
dy = (x + z)dx. (9.17)
Bˆay gi`o t´ınh vi phˆan cˆa´p hai cu’a h`am Lagrange
d2F = 2(dz)2 − 2λ2dxdz. (9.18) Thay gi´a tri λ2 = −1 v`a (9.17) v`ao (9.18) ta thu du.o c da.ng to`an
phu.o.ng x´ac di.nh du.o.ng l`a
d2F = 4dx2.
T`u d´o suy ra h`am d˜a cho c´o cu c tiˆe’u c´o diˆe`u kiˆe.n ta.i diˆe’m
M0(−1, 1, 0) v` a fmin= 0 N
V´ ı du 6 T`ım gi´a tri l´o.n nhˆa´t v`a nho’ nhˆa´t cu’a h`am
f (x, y) = x2+ y2− xy + x + y
trong miˆ`ne
D = {x 6 0, y 6 0, x + y > −3}.
Gia’i Miˆ`n D d˜a cho l`a tam gi´ac OAB v´o.i dı’nh ta.i A(−3, 0),e
B(0, −3) v` a O(0, 0).
i) T`ım c´ac diˆe’m d`u.ng:
f x0 = 2x − y + 1 = 0
f y0 = 2y − x + 1 = 0
T`u d´o x = −1, y = −1 Vˆa.y diˆe’m d`u.ng l`a M(−1, −1).
Ta.i diˆe’m M ta c´o:
f (M ) = f (−1, −1) = −1.
... 4x3− 4x + 4y, f y0 = 4y3+ 4x − 4y.Do d´o
4x3 − 4x + 4y = 4y3 + 4x − 4y = 0...
∂x2dx2+ ∂
2
F
∂x∂y dxdy +< /sup>
∂2F
∂y2dy2... l`a
∂ϕ
∂x dx +< /sup>
∂ϕ
∂y dy = 0 (dx
2
+ dy2 6= 0).
Cu