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Extrema of multivariable function

  1. Jun 11, 2007 #1
    Hi

    I'm studying for a calculus exam and I'm a little stuck on finding the extrema for multivariable functions.

    For the particular question I'm trying to do now I need to find and classify the extrema for the function f(x,y) = (4x^2)(e^y) - 2x^4 - e^4y.

    I can find the first derivatives, being fx = 8xe^y - 8x and fy = (4x^2)(e^y) - 4e^4y and I know I have to let them be equal to 0 to find where the extrema are located but I'm not sure how to do that. I guess it's just the exponentials that are throwing me off.

    Any help would be appreciated.
     
    Last edited: Jun 11, 2007
  2. jcsd
  3. Jun 11, 2007 #2

    D H

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    You have done well so far. The next step is to simultaneously set those partial derivatives to zero. So try that and show us the results.
     
  4. Jun 11, 2007 #3
    Thanks

    Well it's 0 = 8xe^y - 8x and 0 = (4x^2)(e^y) - 4e^4y. But I'm not sure where to go from there. I'm not sure what to do with the exponentials, I know they can never equal zero, but I'm not sure what that means for my equations.
     
  5. Jun 11, 2007 #4

    D H

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    Simplify the fx=0 equation.
     
  6. Jun 11, 2007 #5
    0 = 8x(e^y - 1). But I still don't know what to do with the exponential.
     
  7. Jun 11, 2007 #6

    D H

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    Any equation of the form a*b=0 tells you that either a=0 or b=0 (or both). This is obviously of that form. x=0 doesn't lead to any extrema. (Why not?) What happens when you set e^y-1 to zero?
     
  8. Jun 11, 2007 #7
    Thanks for that. I now realise that y can be set to zero in the equation 8x(e^y - 1). And then if you sub y = 0 into the other equation you get 1 and -1 which yields the points (-1,0) and (1,0) (which is the answer in the back of the book :smile:). And if x = 0 then the other equation doesn't work. Thanks heaps. I now see how stupid I was initially lol.
     
  9. Jun 13, 2007 #8
    I wish when I learned calculus (or any other courses) I could have such instruction.
     
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