炘涛 發問時間: 科學數學 · 4 年前

use lagrange multipliers to find the max and min subject to given constrain.?

1.f(x,y)=x^2+y^2;xy=1

2.f(x,y)=x^2+y^2;2x^2+2xy+3y^2=1

3.f(x,y)=(x^2)y;x^2+2y^2=6

4.f(x,y,z)=x^2+y^2+z^2;x-y=1,y^2-z^2=1

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                                               20160529

    1. f(x,y) = x² + y²;xy = 1

    Let g(x,y) = xy - 1 = 0

    ∇f = (2x,2y)

    ∇g = (y,x)

    (2x,2y) = λ(y,x)

    { 2x = λy ...... ①

    { 2y = λx ...... ②

    { xy = 1 ...... ③

    x ① - y ②:(x - y)(x + y) = 0,x = y or x = -y

    If x = y, x² = 1, x = ±1

    If x = -y. x² = -1, no solutions

    f(1,1) = 2

    f(-1,-1) = 2

    The min = f(1,1) = f(-1,-1) = 2

    Let y = 1/x

    As x→∞, f(x,y)→∞

    The max does not exist.

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    2. f(x,y) = x² + y²;2x² + 2xy + 3y² = 1

    Let g(x,y) = 2x² + 2xy + 3y² - 1 = 0

    ∇f = (2x,2y)

    ∇g = (4x + 2y,2x + 6y)

    (2x,2y) = λ(4x + 2y,2x + 6y)

    { 2x = λ(4x + 2y) ...... ①

    { 2y = λ(2x + 6y) ...... ②

    { 2x² + 2xy + 3y² = 1 ...... ③

    y ① - x ②:λ(2y² - 2x² - 2xy) = 0,λ(5y² - 1) = 0,λ = 0 or y = ±1/√5

    If λ = 0, x = y = 0, 2x² + 2xy + 3y² ≠ 1, no solutions

    If y = 1/√5, 2x² + 2x/√5 + 3(1/√5)² = 1, x = (-1/√5 ± 1)/2

    If y = -1/√5, 2x² - 2x/√5 + 3(1/√5)² = 1, x = (1/√5 ± 1)/2

    f( (-1/√5 + 1)/2,1/√5 ) = 1/2 - 1/(2√5)

    f( (-1/√5 - 1)/2,1/√5 ) = 1/2 + 1/(2√5)

    f( (1/√5 + 1)/2,-1/√5 ) = 1/2 + 1/(2√5)

    f( (1/√5 - 1)/2,-1/√5 ) = 1/2 - 1/(2√5)

    The min = f( (-1/√5 + 1)/2,1/√5 ) = f( (1/√5 - 1)/2,-1/√5 ) = 1/2 - 1/(2√5)

    The max = f( (-1/√5 - 1)/2,1/√5 ) = f( (1/√5 + 1)/2,-1/√5 ) = 1/2 + 1/(2√5)

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    3. f(x,y) = x²y;x² + 2y² = 6

    Let g(x,y) = x² + 2y² - 6 = 0

    ∇f = (2xy,x²)

    ∇g = (2x,4y)

    (2xy,x²) = λ(2x,4y)

    { 2xy = 2λx ...... ①

    { x² = 4λy ...... ②

    { x² + 2y² = 6 ...... ③

    2y ① - x ②:x(2y - x)(2y + x) = 0,x = 0 or x = 2y or x = -2y

    If x = 0,y = ±√3

    If x = 2y,y = 1 and x = 2   or   y = -1 and x = -2

    If x = -2y,y = 1 and x = -2   or   y = -1 and x = 2

    f(0,±√3) = 0

    f(2,1) = 4

    f(-2,-1) = -4

    f(-2,1) = 4

    f(2,-1) = -4

    The min = f(-2,-1) = f(2,-1) = -4

    The max = f(2,1) = f(-2,1) = 4

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    4. f(x,y,z) = x² + y² + z²;x - y = 1, y² - z² = 1

    Let g(x,y) = x - y - 1 = 0 and h(x,y,z) = y² - z² - 1

    ∇f = (2x,2y,2z)

    ∇g = (1,-1,0)

    ∇h = (0,2y,-2z)

    (2x,2y,2z) = λ(1,-1,0) + μ(0,2y,-2z)

    { 2x = λ ...... ①

    { 2y = -λ + 2μy ...... ②

    { 2z = -2μz ...... ③

    { x - y = 1 ...... ④

    { y² - z² = 1 ...... ⑤

    From ③, z = 0 or μ = -1

    If z = 0, y = 1 and x = 2   or   y = -1 and x = 0

    If μ = -1,

    { 2x = λ

    { 4y = -λ

    x = -2y,y = -1/3,x = 2/3,z² = -8/9 < 0, no solutions

    f(2,1,0) = 5

    f(0,-1,0) = 1

    The min = f(0,-1,0) = 1

    Let y = x - 1 and z = √(y² - 1)

    As x→∞, f(x,y,z)→∞

    The max does not exist

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    Lagrange Multipliers with one constraint

    Suppose f(x,y) and g(x,y) are differentiable and ∇g ≠ 0 when g(x,y) = 0

    The local maximum and local minimum point of f(x,y) (if exist) subject to the constraint g(x,y) = 0 are satisfied the equations

    { ∇f = λ∇g

    { g(x,y) = 0

    For the function of three independent variables, the condition is similar.

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    Lagrange Multipliers with two constraints

    Suppose f(x,y,z), g(x,y,z), h(x,y,z) are differentiable and ∇g is not parpallel to ∇h when g(x,y,z) = 0 and h(x,y,z) = 0

    The local maximum and local minimum point of f(x,y,z) (if exist) subject to the constraints g(x,y,z) = 0 and h(x,y,z) = 0 are satisfied the equations

    { ∇f = λ∇g + μ∇h

    { g(x,y,z) = 0

    { h(x,y,z) = 0

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