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