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Math Help - [SOLVED] Optimization in Several Variables

  1. #1
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    [SOLVED] Optimization in Several Variables

    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.
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  2. #2
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    Quote Originally Posted by coe236 View Post
    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.
     f(x,y) = 2y.exp(x^2-y) + exp(x^2-y)

     \frac{\partial f}{\partial x} = 2x.2y.exp(x^2-y) + 2x.exp(x^2-y) = exp(x^2-y)(4xy+2x) = 0

    The exponential function is never zero, hence 4xy+2x = 0

     \frac{\partial f}{\partial y} = 2exp(x^2-y) + 2y(-1)exp(x^2-y) -exp(x^2-y) = exp(x^2-y)(2-2y-1) = 0

    Again exponential never zero, so:  2-2y-1 = 0

    Surely ou can continue from there?
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  3. #3
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    yup, that simplified things a lot. Thank you
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