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Math Help - Finding critical numbers of a product involving an exponential function

  1. #1
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    Finding critical numbers of a product involving an exponential function

    Ok, so the original function is f(x,y)=(x^2+4y^2)e^{1-x^2-y^2} and the problem is to find the derivative with respect to x, respect to y, respect to xx, respect to yy, and respect to xy/yx and find the critical numbers as well as any relative extrema or saddle point if applicable.

    I found that the derivative with respect to x is f'x(x,y)=2xe^{1-x^2-y^2}(1-x^2-4y^2), the derivative with respect to y is f'y(x,y)=2ye^{1-x^2-y^2}(4-x^2-4y^2), the derivative with respect to xx is f'xx(x,y)=2e^{1-x^2-y^2}(1-5x^2+2x^4+8x^2y^2-4y^2), the derivative with respect to yy is f'yy(x,y)=2e^{1-x^2-y^2}(4-x^2-20y^2+2x^2y^2+8y^4) and finally the derivative with respect to xy is f'xy(x,y)=-4xye^{1-x^2-y^2}(5-x^2-4y^2).

    Now, I tried setting 2xe^{1-x^2-y^2} from f'x equal to 2ye^{1-x^2-y^2} from f'y but that didn't work. I tried doing substitution for 4-x^2-4y^2=0 and 1-x^2-4y^2=0 but that ends up being 1=4 which obviously isn't true nor is it a CN...so how would I find the CN's for this problem? I'm sure I'm overlooking something simple, however, I can't find out what that simple thing is.
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    Quote Originally Posted by emttim84 View Post
    Ok, so the original function is f(x,y)=(x^2+4y^2)e^{1-x^2-y^2} and the problem is to find the derivative with respect to x, respect to y, respect to xx, respect to yy, and respect to xy/yx and find the critical numbers as well as any relative extrema or saddle point if applicable.

    I found that the derivative with respect to x is f'x(x,y)=2xe^{1-x^2-y^2}(1-x^2-4y^2), the derivative with respect to y is f'y(x,y)=2ye^{1-x^2-y^2}(4-x^2-4y^2), the derivative with respect to xx is f'xx(x,y)=2e^{1-x^2-y^2}(1-5x^2+2x^4+8x^2y^2-4y^2), the derivative with respect to yy is f'yy(x,y)=2e^{1-x^2-y^2}(4-x^2-20y^2+2x^2y^2+8y^4) and finally the derivative with respect to xy is f'xy(x,y)=-4xye^{1-x^2-y^2}(5-x^2-4y^2).

    Now, I tried setting 2xe^{1-x^2-y^2} from f'x equal to 2ye^{1-x^2-y^2} from f'y but that didn't work. I tried doing substitution for 4-x^2-4y^2=0 and 1-x^2-4y^2=0 but that ends up being 1=4 which obviously isn't true nor is it a CN...so how would I find the CN's for this problem? I'm sure I'm overlooking something simple, however, I can't find out what that simple thing is.
    When you put each of your (correct ) partial derivatives equal to zero you get:

    x = 0 or 1 - x^2 - 4y^2 = 0

    y = 0 or 4 - x^2 - 4y^2 = 0

    Note that the factor e^{1 - x^2 - y^2} \neq 0 for real values of x and y.

    You're quite right about 1 - x^2 - 4y^2 = 0 and 4 - x^2 - 4y^2 = 0 being simultaneously inconsistent.

    But it looks like you overlooked the simple solution (0, 0) ........
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    Quote Originally Posted by mr fantastic View Post
    When you put each of your (correct ) partial derivatives equal to zero you get:

    x = 0 or 1 - x^2 - 4y^2 = 0

    y = 0 or 4 - x^2 - 4y^2 = 0

    Note that the factor e^{1 - x^2 - y^2} \neq 0 for real values of x and y.

    You're quite right about 1 - x^2 - 4y^2 = 0 and 4 - x^2 - 4y^2 = 0 being simultaneously inconsistent.

    But it looks like you overlooked the simple solution (0, 0) ........
    Hmm, fair enough, but the solutions manual states there are the critical numbers 0,0....0,1...0,-1...1,0...-1,0 so I'm mainly unsure of where they get the plus or minus 1 for the X and then the Y values from.
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    Quote Originally Posted by mr fantastic View Post
    When you put each of your (correct ) partial derivatives equal to zero you get:

    x = 0 or 1 - x^2 - 4y^2 = 0

    y = 0 or 4 - x^2 - 4y^2 = 0

    Note that the factor e^{1 - x^2 - y^2} \neq 0 for real values of x and y.

    You're quite right about 1 - x^2 - 4y^2 = 0 and 4 - x^2 - 4y^2 = 0 being simultaneously inconsistent.

    But it looks like you overlooked the simple solution (0, 0) ........
    The other possibilities for the partial derivatives being equal to zero are:

    Case 1. x = 0 and 4 - x^2 - 4y^2 = 0.

    Case 2. y = 0 and 1 - x^2 - 4y^2 = 0.

    Case 1: Substitute x = 0 into 4 - x^2 - 4y^2 = 0 and you have 4 - 4y^2 = 0 ......

    So (0, 1) and (0, -1) are critical points.

    Case 2: Substitute y = 0 into 1 - x^2 - 4y^2 = 0 and you have .......
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    Quote Originally Posted by mr fantastic View Post
    The other possibilities for the partial derivatives being equal to zero are:

    Case 1. x = 0 and 4 - x^2 - 4y^2 = 0.

    Case 2. y = 0 and 1 - x^2 - 4y^2 = 0.

    Case 1: Substitute x = 0 into 4 - x^2 - 4y^2 = 0 and you have 4 - 4y^2 = 0 ......

    So (0, 1) and (0, -1) are critical points.

    Case 2: Substitute y = 0 into 1 - x^2 - 4y^2 = 0 and you have .......
    Ah-ha....I see. So if I get a problem like this where a little digging must be done to find the CN's, assume x=0 for the derivative of a function with respect to x and plug x=0 into the derivative to find some of the CN's and assume y=0 for the derivative of a function with respect to y and plug in y=0 into the derivative to find the rest of the CNs? Thanks!
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    Quote Originally Posted by emttim84 View Post
    Ah-ha....I see. So if I get a problem like this where a little digging must be done to find the CN's, assume x=0 for the derivative of a function with respect to x and plug x=0 into the derivative to find some of the CN's and assume y=0 for the derivative of a function with respect to y and plug in y=0 into the derivative to find the rest of the CNs? Thanks!
    f_x = 0 when x = 0, f_y = 0 when y = 0. There's no assumption .....

    When f_x = 0 it's a matter of looking for when f_y = 0 .......

    Ditto when f_y = 0 .....
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