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VJEBE IZ
MATEMATIKE 2Ivana Baranovi
Miroslav Jerkovi
Lekcija 9Lokalni ekstremi funkcije vievarijabla
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f
(x0, y0) (x, y)
f(x0, y0) f(x, y) f (x0, y0)
f(x0, y0) f(x, y) (x, y) f
f
(x0, y0) (x, y) f(x0, y0) f(x, y)
f
(x0, y0) f(x0, y0) f(x, y) (x, y) f
f
f
f
f
f
(x0, y0)
fx(x0, y0) = 0
fy(x0, y0) = 0
fx(x, y) =fy(x, y) = 0
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(x0, y0)
A:= fxx(x0, y0), B := fxy(x0, y0) = fyx(x0, y0), C:= fyy(x0,
y0).
:=ACB2
(x0, y0)
> 0 (x0, y0)
A 0
< 0
f
(x0, y0) (x0, y0)
= 0
f
(x0, y0)
(2)
< 0
(x0, y0) f
f
(x0
, y0)
H :=
fxx(x0, y0) fyx(x0, y0)fxy(x0, y0) fyy(x0, y0)
=
A B
B C
= det
A B
B C
fxy(x0, y0) = fyx(x0, y0)
B2
f(x, y) = 4xy x4 y4
fx(x, y) = fy(x, y) =
0
fx(x, y) = 4y 4x3 = 0fy(x, y) = 4x 4y3 = 0
y= x3 x= y3 y = x3 x= x9
x(x8 1) = 0x(x4 1)(x4 + 1) = 0x(x2 1)(x2 + 1)(x4 + 1) = 0x(x
1)(x + 1)(x2 + 1)(x4 + 1) = 0
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x
x1= 0
x2= 1
x3= 1
y= x3
y1= 0
y2= 1
y3= 1
(0, 0) (1, 1) (1,1)
f
fxx(x, y) = 12x2fxy(x, y) =fyx(x, y) = 4
fyy(x, y) = 12y2
x y
A
B
C
(0, 0)A= fxx(0, 0) = 0 B = fxy(0, 0) = 4 C=fyy(0, 0) = 0 =ACB2 =
16< 0
(1, 1)A= fxx(1, 1) = 12 B = fxy(1, 1) = 4 C=fyy(1, 1) = 12 =ACB2
= 128> 0 (1, 1)
A =12 < 0
(1, 1)
f(1, 1) = 2
(1,1)
A= fxx(1,1) = 12 B = fxy(1,1) = 4 C= fyy(1,1) = 12
= 128 > 0
A =12 < 0 (1,1)
f(1,1) = 2
f(x, y) = y sin x
fx(x, y) = y cos x = 0
fy(x, y) = sin x = 0.
x = k
k Z
cos x
k
k
y = 0
(k, 0)
k Z
fxx(x, y) = y sin xfxy(x, y) = fyx(x, y) = cos x
fyy(x, y) = 0.
k
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k
k = 2l
l Z
A= fxx(2l, 0) = 0
B = fxy(2l, 0) = cos(2l) = 1
C=fyy(2l, 0) = 0
=ACB2 = 012 = 1< 0 (k, 0) k
k
k= 2l+ 1
l Z
A = fxx((2l+ 1), 0) = 0 B = fxy((2l+ 1), 0) = cos((2l+ 1)) =1
C=fyy((2l+ 1), 0) = 0 =ACB2 = 0 (1)2 = 1< 0 (k, 0) k
(k, 0) k Z
f(x, y) = 3x2 2xy+ y2 8y
f(x, y) = x3 3xy y3
f(x, y) = y2 + xy+ 3y+ 2x + 3
f(x, y) = xy x3 y2
f(x, y) = x2 + y2 + 2
xy
f(x, y) = x3 + y3 3x 3y
f(x, y) = x2 + y
ey
f(x, y) = xey
f(x, y) = ex sin y
f(x, y) = y
x y2 x + 6y
z =z (x, y)
x2 + 2y2 + xz+ z2
3 = 0
F(x, y , z) := x2 + 2y2 + xz+ z2 3
Fx(x, y , z) = 2x + z
Fy(x, y , z) = 4y
Fz(x, y , z) =x + 2z
zx(x, y) =2x+zx+2z zy(x, y) = 4yx+2z
zx(x, y) = zy(x, y) = 0 2x+z = 0 = 4y
y = 0
z =2x
x2 + 2y2 +xz +z2 3 = 0
x2 = 1
x1 = 1 x2 = 1 z1= 2 z2= 2 (1, 0) (1, 0)
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zxx(x, y) = (2+zx)(x+2z)(2x+z)(1+2zx)
(x+2z)2
zyx(x, y) = zy(x+2z)(2x+z)2zy(x+2z)2 =zxy(x, y)zyy(x, y) =
4(x+2z)4y2zy(x+2z)2
zx(1, 0) = zy(1, 0 ) = 0 zx(1, 0) =
zy(1, 0) = 0
(1, 0)
z = 2
A = zxx(1, 0) = 23
B = zxy(1, 0 ) = 0 C =zyy(1, 0) =
43
=ACB2 = 23 43
= 89
>0 (1, 0) z(x, y)
A = 2
3 > 0
z = 2
(1, 0, 2)
z = 2
A = zxx(1, 0) =23 B = zxy(1, 0) = 0 C = zyy(1, 0) = 43 = AC B2 =
89 > 0 (1, 0)
z(x, y)
A= 23
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(3, 3)
3
54
(3, 3) f
2
x2 y2 z2 = 0 T(0, 1, 4)
x2 +y2 + 4z2 = 16
x + y+ 2z = 12
x= 0
y = 0
x y+ 1 = 0
T(3, 2, 1)
R
x2 +y2 +z2 16
O(0, 0, 0)
A(1, 0, 0)
B(0, 2, 0)
C(0, 0, 3) z
