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Math Help - 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  d(y)=\frac{1}{100}(y^4-200y^2+22500) where  d(y) is the distance from the origin to  (\frac{150-y^2}{10},y) squared.

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


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

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

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

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

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

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

    Also the point  (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 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 d(y)=\frac{1}{100}(y^4-200y^2+22500)
    Using the distance formula,  D=\sqrt{\left(\frac{150-y^2}{10}\right)^2+y^2} .
    Then I let  d=D^2 .
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  5. #5
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    ohh i see, but wouldn't letting 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  x^2 is a monotonically increasing function on  [0,\infty) .
    Since  D>0 ,  D^2 is minimized/maximized  \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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