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Thread: find extreeemes

  1. September 11th 2012, 11:09 PM #1
    jacob93
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    find extreeemes

    f(x,y)=x+xy^2
    only def for x^2+y^2 \leq 4
    find the highest and lowest value for f.

    Attempt: https://dl.dropbox.com/u/586465/Phot...2 09 04 42.jpg
    I find (0,0) and (±√4 ; ±2/3*√6)

    The book says the answer is: Skärmavbild 2012-09-12 kl. 09.08.23.png
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  2. September 12th 2012, 12:21 AM #2
    kalyanram
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    Re: find extreeemes

    In your post g'(x) calculation is incorrect. In case you want to verify see the solution below.
    Spoiler:

    Quote Originally Posted by jacob93 View Post
    f(x,y)=x+xy^2
    only def for x^2+y^2 \leq 4
    find the highest and lowest value for f.
    Consider
    f(x,y)=x+xy^2 = x(1+y^2) f(x,y) \ge 0 x \ge 0 and f(x,y) < 0 x < 0 one conclusion that can be drawn is f(x,y) reaches its extremes under the restriction x^2+y^2 \leq 4 at precisely on the circle x^2+y^2 = 4. So substitute y^2 = 4 -x^2 in f(x,y). We have g(x) = x(5-x^2) Equating the derivative to zero we have g'(x) = 5- 3x^2 = 0 \implies x = \pm \sqrt{\frac{5}{3}}
    Last edited by kalyanram; September 12th 2012 at 12:27 AM.
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  3. September 13th 2012, 12:53 AM #3
    jacob93
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    Re: find extreeemes

    thanks
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