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How to Identify Min/Max/Critical or Saddle Points?

Hello, currently I'm on my last attempt on the following question.

Consider the function

f(x,y) =(x^2)y +y^3 - 48y

f has __ at 4sqrt(3), 0)

f has __ at (0,0)

f has __ at (0, -4)

f has __ at (4sqrt(3), 0)

f has __ at (0,4)

I have tried (respectively)

Saddle Saddle

Critical No Critical Point

Minimum Minimum

Saddle Saddle

Maximum Maximum

The option are "Minimum, Maximum, Some Critical Point, No Critical Point, Saddle Point or ?"

Any help is greatly appreciated.

Thank you in advance!

1 Answer

Relevance
  • ?
    Lv 7
    3 years ago

    I’ll use the method described in this link (see blue ‘Fact’ box ¾-way down):

    http://tutorial.math.lamar.edu/Classes/CalcIII/Rel...

    f(x,y) =x²y + y³ – 48y

    ∂f/∂x = 2xy

    ∂²f/∂x² = 2y

    ∂f/∂y = x² + 3y² – 48

    ∂²f/∂y² = 6y

    ∂²f/(∂x∂y) = 2x

    D = (∂²f/∂x²)(∂²f/∂y²) - (∂²f/(∂x∂y))² = (2y)(6y) – (2x)² = 12y² - 4x²

    _______________________

    Identify the critical points: if ∂f/∂x=0 and ∂f/∂y=0, this is a critical point.

    For each critical point, find D and classify point (using D and ∂²f/∂x²) as a rel.max./rel.min//saddle.

    _______________________

    Point A(4√3, 0)

    ∂f/∂x = 2xy = 2*4√3*0 = 0

    ∂f/∂y = x² + 3y² – 48 = (4√3)² + 3(0)² – 48 = 0

    So this is a critical point.

    D = 12y² – 4x² = 12(0)² – 4(4√3)² = negative value. D<0

    So (4√3, 0) is a saddle point.

    _______________________

    Point B(0,0)

    ∂f/∂x = 2xy = 2*0*0 = 0

    ∂f/∂y = x² + 3y² – 48 = (0)² + 3(0)² – 48 = negative

    So this is a not a critical point.

    _______________________

    Point C(0, -4)

    ∂f/∂x = 2xy = 2*0*(-4) = 0

    ∂f/∂y = x² + 3y² – 48 = (0)² + 3(-4)² – 48 = 0

    So this is a critical point.

    D = 12y² – 4x² = 12(-4)² – 4(0)² = positive value. D>0

    ∂²f/∂x² = 2y = 2*(-4) = negative

    So this is a relative maximum.

    _______________________

    Point D(4√3, 0)

    This is the same as point A. Maybe you have a typing error.

    _______________________

    Point E(0, 4)

    ∂f/∂x = 2xy = 2*0*(4) = 0

    ∂f/∂y = x² + 3y² – 48 = (0)² + 3(4)² – 48 = 0

    So this is a critical point.

    D = 12y² – 4x² = 12(4)² – 4(0)² = positive value. D>0

    ∂²f/∂x² = 2y = 2*4 = positive

    So this is a relative minimum.

    _______________________

    Summary:

    Critical: saddle

    Not critical

    Critical: relative maximum

    Probably error as point is same as 1st point

    Critical: relative minimum

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