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Gradients and hessian problem

  1. Aug 26, 2007 #1
    1. The problem statement, all variables and given/known data

    The height of a certain hill (in feet) is given by h(x,y) = 10(2xy-3x^2-4y^2-18x+28y+12)
    where y is the distance (in miles) north, x the distance east of South Hadley.

    a) Where is the top of the hill located?
    b) How high is the hill?
    c) How steep is the slope (in feet per mile) at a point 1 mile north and one mile east of South Hadley? In what direction is the slope steepest, at that point?

    2. Relevant equations

    dh = grad h dot dr = |grad h| |dr| cos (theta)

    3. The attempt at a solution

    I found the gradient of h = 20(y-3x-9)x + 20(x-4y+14)y
    I realize setting the gradient of h to zero will give you a max, min, saddle point, or shoulder. I am pretty lost other than that. Any help? Thanks
     
  2. jcsd
  3. Aug 26, 2007 #2

    dextercioby

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    Evualuate the hessian at each critical point.
     
  4. Aug 26, 2007 #3

    arildno

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    Actually, he should evaluate the determinant of the Hessian at every critical point..
     
  5. Aug 27, 2007 #4

    dextercioby

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    The Hessian is the determinant, that's the way i learnt in school and it's a substantive. While in the formulation "the hessian matrix", it's an adjective.
     
  6. Aug 27, 2007 #5

    arildno

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    Then my school taught it differently. :smile:
     
  7. Aug 27, 2007 #6
    To get a better feel for the equation, note that it is quadratic in both variables. By completing the square you will be able to rewrite this equation as one that describes an ellipse. The center of this ellipse is the top of the hill.
     
  8. Jan 6, 2011 #7
    Re: Gradients

    Though it is clearly way too late to help, I figured I would give what, to me, is the clearest way to solve this.

    a) Since your gradient must be equal to zero for a max/min/saddle/etc. and because x components do not add with y components, both your x and y components must be equal to zero. This gives you two equations and two unknowns and you can solve for x and y.

    y=3x+9
    x=4y-14

    Solving these yields only one answer. Since this is a real life problem, common sense tells you that our hill needs a maximum at dh=0, since the hill does not continue into outer space forever or dive deep into the Earth, and we have only one place to put it; at the solution to the two equations above.

    b) Plug the location of your hills maximum back into the original equation.

    c) The gradient can be defined as a vector with direction in the direction of greatest change and magnitude equal to the slope of the function. You have your gradient already. Plug your given values of x=1 and y=1 into your gradient and treat it like a vector. Its magnitude is your slope and its direction is the direction of greatest change.

    Unless this is a very common problem, I imagine we have the same textbook. All the information I have provided is available on the previous page.
     
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