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Thread: Help on a word problem please!

  1. #1
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    Help on a word problem please!

    I am having a lot of trouble trying to solve this problem. I don't really know how to start it off. Any kind of help would be great!! Thanks a lot!

    Let S denote the set of points with coordinates (x,y) such that 0 ≤ x ≤ 15 and y ^2 = 150 - 10x. Find the coordinates of the point(s) of S nearest to the origin. Also find the coordinates of the point(s) of S farthest from the origin. Justify your answer.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by kaiba4192 View Post
    I am having a lot of trouble trying to solve this problem. I don't really know how to start it off. Any kind of help would be great!! Thanks a lot!

    Let S denote the set of points with coordinates (x,y) such that 0 ≤ x ≤ 15 and y ^2 = 150 - 10x. Find the coordinates of the point(s) of S nearest to the origin. Also find the coordinates of the point(s) of S farthest from the origin. Justify your answer.
    We want to minimize and maximize the function $\displaystyle d(y)=\frac{1}{100}(y^4-200y^2+22500) $ where $\displaystyle d(y) $ is the distance from the origin to $\displaystyle (\frac{150-y^2}{10},y) $ squared.

    The reason we can minimize the distance squared is because $\displaystyle x^2 $ is a monotonically increasing function on $\displaystyle [0,\infty) $.


    Let's solve $\displaystyle d\;'(y) = 0 $.
    $\displaystyle 4y^3-400y=0 \implies y(y+10)(y-10)=0\implies y=-10,0,10 $.

    $\displaystyle d(\pm10) = 125 $ and $\displaystyle d(0) = 225 $.

    This means $\displaystyle y=-10,10 $ is a local minimum and $\displaystyle y=0 $ is a local maximum.

    Now to check the endpoints. $\displaystyle 0\leq x\leq 15 \implies -5\sqrt{6}\leq y \leq 5\sqrt{6} $.

    $\displaystyle d(\pm5\sqrt{6}) = 150 $.

    Therefore $\displaystyle (5, 10) $ and $\displaystyle (5, -10) $ are the points in $\displaystyle S $ that minimize the distance to the origin.

    Also the point $\displaystyle (15, 0) $ maximizes our desired distance.
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  3. #3
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    thanks a lot for replying.

    How did you you get the function $\displaystyle d(y)=\frac{1}{100}(y^4-200y^2+22500)$
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by kaiba4192 View Post
    thanks a lot for replying.

    How did you you get the function $\displaystyle d(y)=\frac{1}{100}(y^4-200y^2+22500)$
    Using the distance formula, $\displaystyle D=\sqrt{\left(\frac{150-y^2}{10}\right)^2+y^2} $.
    Then I let $\displaystyle d=D^2 $.
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  5. #5
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    ohh i see, but wouldn't letting $\displaystyle d=D^2$ change the result because we're not deriving the original function?
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    The reason we can minimize the distance squared is because $\displaystyle x^2 $ is a monotonically increasing function on $\displaystyle [0,\infty) $.
    Since $\displaystyle D>0 $, $\displaystyle D^2 $ is minimized/maximized $\displaystyle \iff D $ is minimized/maximized.
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  7. #7
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    ohh ok, i get it now, thanks a lot!!
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