# matematika - lokalni ekstremi

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• 8/12/2019 matematika - lokalni ekstremi

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VJEBE IZ

MATEMATIKE 2Ivana Baranovi

Miroslav Jerkovi

Lekcija 9Lokalni ekstremi funkcije vievarijabla

• 8/12/2019 matematika - lokalni ekstremi

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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

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