炘涛 發問時間: 科學數學 · 4 年前

請問這幾題微積分怎麼算?

1.find the local maximum values,minimum values and saddle points

f(x,y)=(x-4)ln(xy)

2.find the absolute max and min

(a).f(x,y)=x^2+y^2+(x^2)y+4 on D={(x,y)屬於R^2 | |x|<=1,|y|<=1}

(b).on the disk D={(x,y):x^2+y^2<=25}consider the function f(x ,y)= x^2+y^2-12x+16y.

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  • 最佳解答

                                               20160529

    1.

    Find the local maximum values, minimum values and saddle points of f(x,y) = (x - 4) ln(xy)

    f(x,y) = (x - 4) ln(xy)

    fₓ = ln(xy) + (x - 4)/x = ln(xy) - 4/x + 1

    fᵧ = x/(xy) = (x - 4)/y

    fₓₓ = 1/x - 4/x²

    fᵧᵧ = (4 - x)/y²

    fₓᵧ = fᵧₓ = 1/y

    When fₓ = fᵧ = 0,

    { ln(xy) - 4/x + 1 = 0

    { (x - 4)/y = 0

    x = 4, y = 1/4

    The critical point is (4,1/4,0)

    fₓₓ(4,1/4) = 0.5

    fᵧᵧ(4,1/4) = 0

    fₓᵧ(4,1/4) = 4

    fₓₓ fᵧᵧ - (fₓᵧ)² = 0 - 4² = -16 < 0

    The local maximum values or local minimum values do not exist

    The saddle points is (4,1/4,0)

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    2. Find the absolute max and min

    (a) f(x,y) = x² + y² + x²y + 4 on D = {(x,y)∈R² | |x| ≤ 1, |y| ≤ 1}

    (b) on the disk D = {(x,y) : x² + y² ≤ 25}, consider the function f(x,y) = x² + y² - 12x + 16y

    2a.

    f(x,y) = x² + y² + x²y + 4 on D = {(x,y)∈R² | |x| ≤ 1, |y| ≤ 1}

    fₓ = 2x + 2xy

    fᵧ = 2y + x²

    When fₓ = fᵧ = 0,

    { 2x + 2xy = 0 ...... ①

    { 2y + x² = 0 ...... ②

    From ①, x = 0 or y = -1

    If x = 0, y = 0

    If y = -1, x = ±√2

    f(0,0) = 4

    f(±√2,-1) = 5

    On the boundary,

    When x = 1, f(1,y) = y² + y + 5 = (y + 1/2)² + 19/4

    f(1,-1) = 5

    f(1,-1/2) = 19/4

    f(1,1) = 7

    When x = -1. f(-1,y) = y² + y + 5 = (y + 1/2)² + 19/4

    f(-1,-1) = 5

    f(-1,-1/2) = 19/4

    f(-1,1) = 7

    When y = 1, f(x,1) = 2x² + 5

    f(0,1) = 5

    When y = -1, f(x,-1) = 5

    The absolute maximum = f(1,1) = f(-1,1) = 7

    The absolute minimum = f(0,0) = 4

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    2b.

    f(x,y) = x² + y² - 12x + 16y on D = {(x,y) : x² + y² ≤ 25}

    fₓ = 2x - 12

    fᵧ = 2y + 16

    When fₓ = fᵧ = 0,

    { 2x - 12 = 0 ...... ①

    { 2y + 16 = 0 ...... ②

    x = 6, y = -8

    (6,-8) is not on the disk

    On the boundary ( x² + y² = 25 ),

    Let x = 5 cosθ, y = 5 sinθ, 0 ≤ θ < 2π

    f(x,y) = 25 cos²θ + 25 sin²θ - 60 cosθ + 80 sinθ

    f(x,y) = 25 - 100 sin(θ + π - sin⁻¹(3/5))

    When θ = π/2 + sin⁻¹(3/5), x = -3, y = 4, f(x,y) = 125

    When θ = 3π/2 + sin⁻¹(3/5), x = 3, y = -4, f(x,y) = -75

    The absolute maximum = f(-3,4) = 125

    The absolute minimum = f(3,-4) = -75

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    Second Derivative Test

    Suppose f(x,y) is C² function and fₓ(x₀,y₀) = fᵧ(x₀,y₀) = 0, then

    1. f(x₀,y₀) is a local minimum if fₓₓ < 0 and fₓₓ fᵧᵧ - (fₓᵧ)² > 0 at (x₀,y₀)

    2. f(x₀,y₀) is a local maximum if fₓₓ > 0 and fₓₓ fᵧᵧ - (fₓᵧ)² > 0 at (x₀,y₀)

    3. f(x₀,y₀) is a saddle point if fₓₓ fᵧᵧ - (fₓᵧ)² < 0 at (x₀,y₀)

    4. the test is inconclusive if fₓₓ fᵧᵧ - (fₓᵧ)² = 0 at (x₀,y₀)

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