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Homework Help: Why this maximization approach fails?

  1. Oct 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Find all points at which the direction of fastest change of the function [tex] f(x,y) = x^2 + y^2 -2x - 2y [/tex]is in the direction of <1,1>.

    2. Relevant equations
    [tex] <\nabla f = \frac{\delta f}{\delta x} , \frac{\delta f}{\delta y} , \frac{\delta f}{\delta z}>[/tex]

    3. The attempt at a solution
    [tex] \frac{\nabla f}{|\nabla f|}[/tex] = <1,1>

    This doesn't work but

    [tex]\frac{\delta f}{\delta x} = \frac{\delta f}{\delta y}[/tex]

    does. This latter approach makes sense. The partial derivatives at x and y have the same value. The former approach should work, however. In general, [tex] \nabla f [/tex] gives the direction of maximum change.
  2. jcsd
  3. Oct 14, 2015 #2

    set [tex] \frac{\nabla f}{|\nabla f|}[/tex] to <root(2)/2, root(2)/2>

  4. Oct 14, 2015 #3


    Staff: Mentor

    Your function is one with two variables, so the gradient should be a vector with two components, not three.
    $$\nabla f = <\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}>$$
    Do you have a feel for what the surface for this function looks like? The graph of the surface is a paraboloid that opens upward.
  5. Oct 15, 2015 #4


    User Avatar
    Science Advisor

    Just a note on Latex. First, your [itex]\nabla F[/itex] belongs outside the < > denoting the vector. Second you can use "\partial" to indicate partial derivatives rather than "\delta":
    [tex] \nabla f = <\frac{\partial f}{\partial x} , \frac{\partial f}{\partial y} , \frac{\partial f}{\partial z}>[/tex]

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