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Math Help - Optimize the expression subject to the given constraint.

  1. #1
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    Optimize the expression subject to the given constraint.

    If x^2 + y^2 = 4x + 6y,

    find the maximum value of

    \sqrt{8x+6y+4} + \sqrt{6x+4}.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mathwizard View Post
    If x^2 + y^2 = 4x + 6y,

    find the maximum value of

    \sqrt{8x+6y+4} + \sqrt{6x+4}.
    You know what, wouldn't someone be an absolute genius if these are questions on an exam or something and they submitted them as challenge questions because they knew that the best on the site would step up. Then after we had done our best you claimed the credit. I am absolutely positive that this is not the case with you (you seem like a good guy) but man would you be smart to do so.

    This is a lengthy case of numerical calculations

    All you have to do is complete the square in the original, solve for x or y, imput it into the second equation, then differentiate it, solve for zero, and then test the results in the second derivative to make sure they are maxs or mins. If there turns out to be more than one relative, just plug in the values into the initial equation and the largest value is the absolute max.

    If someone would care to take the time to do this problem they can have all the credit.

    Mathstud
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    You know what, wouldn't someone be an absolute genius if these are questions on an exam or something and they submitted them as challenge questions because they knew that the best on the site would step up. Then after we had done our best you claimed the credit. I am absolutely positive that this is not the case with you (you seem like a good guy) but man would you be smart to do so.

    This is a lengthy case of numerical calculations

    All you have to do is complete the square in the original, solve for x or y, imput it into the second equation, then differentiate it, solve for zero, and then test the results in the second derivative to make sure they are maxs or mins. If there turns out to be more than one relative, just plug in the values into the initial equation and the largest value is the absolute max.

    If someone would care to take the time to do this problem they can have all the credit.

    Mathstud
    Haha; I like your brute force approach. Try generalizing it.

    NOTE: This can be solved without calculus.
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  4. #4
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    I'd say to use a Lagrange multiplier, rather than brute force. As for doing it without calculus altogether, I'm not sure how you'd do that, but I imagine the fact that the constraint is the circle with center (2,3) and radius \sqrt{13} (and thus passes through the origin) may be of some use.

    --Kevin C.
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