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Homework Help: Optimization in Several Variables

  1. Dec 10, 2008 #1
    F(x,y) = (2*y+1)*e^(x^2-y)
    Find critical point and prove there is only one.
    Use second derivative test to determine nature of crit. pt.


    I know the procedure in solving it: set partial derivatives to zero and solve resulting equations. And by second derivative test, if D>0, f(a,b) is local min/max; D<0, (a,b) is saddle point. if f_xx(a,b)>0, f(a,b) is min
    where D=D(a,b)=f_xx(a,b)f_yy(a,b)-f_xy(a,b)^2

    I have no idea how to get the partial derivatives and start the problem. Any help will be appreciated, thanks.
     
  2. jcsd
  3. Dec 10, 2008 #2

    Office_Shredder

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    To take a partial derivative, you just treat one of either x or y as a constant, and differentiate with respect to the other one. So for example if the function was f(x,y) = x2y2 + x3 the partial derivative with respect to x is

    2xy2 + 3x2

    and with respect to y

    2yx2
     
  4. Dec 10, 2008 #3
    Well yea I know that.. I meant I did not know how to get the partials specifically for the function I posted above : F(x,y) = (2*y+1)*e^(x^2-y).
    Is dF/dx = (2*y+1)*e^(x^2-y)*2*x ? and dF/dy = ?
     
  5. Dec 10, 2008 #4

    HallsofIvy

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    Yes. To find dF/dy, use the product rule. What are d(2y+1)/dy and de^(x^2-y)/y separately?
     
  6. Dec 10, 2008 #5
    d(2y+1)/dy= 2
    de^(x^2-y)/dy = -e^(x^2-y)
    so by product rule, dF/dy= 2*e^(x^2-y) ?
     
  7. Dec 10, 2008 #6

    Office_Shredder

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    The product rule is d(f*g)/dy = df/dy*g + dg/dy*f not df/dy*dg/dy
     
  8. Dec 10, 2008 #7
    So dF/dy= (2y+1)*(-e^(x^2-y) + 2*(e^(x^2-y)) ?
     
  9. Dec 10, 2008 #8

    Mark44

    Staff: Mentor

    Technically speaking, the two first derivatives are partial derivatives [tex]\frac{\partial{F}}{\partial{x}}[/tex] and [tex]\frac{\partial{F}}{\partial{y}}[/tex], also written as [tex]F_x[/tex] and [tex]F_y[/tex].
     
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