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Math Help - Optimization help needed..

  1. #16
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    I got a ' (x) =
    -4x^2 +2 - 1.41 -x(sqroot2)=0
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  2. #17
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    what happens when I multiply
    2(1-x^2)^0.5 -(x)(sqroot2) all by (1-x^2)^0.5/(1-x^2)^0.5 ??

    I think heres where i went wrong
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  3. #18
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    The negative answers are just algebraic answers from the graph,
    but we must work with positive values as we are dealing with lengths of the rectangle sides.

    After calculating the derivative, setting it to zero, rearranging the form i wrote, we end up with

    8x^4-7x^2+1=0

    This has 4 solutions, two of them are negative though.
    There is one correct positive answer corresponding to maximum x.

    The other solution is extraneous and is caused by squaring terms to end up with 8x^2-7x+1=0

    Don't struggle with it.
    If you post your calculations, they can be checked.
    I will add something later.
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  4. #19
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    Quote Originally Posted by wolfhound View Post
    what happens when I multiply
    2(1-x^2)^0.5 -(x)(sqroot2) all by (1-x^2)^0.5/(1-x^2)^0.5 ??

    I think heres where i went wrong
    From 2\sqrt{1-x^2}-\frac{2x^2}{\sqrt{1-x^2}}-\sqrt{2}=0

    we multiply by \sqrt{1-x^2} as the RHS is zero.

    2(1-x^2)-2x^2-\sqrt{2}\sqrt{1-x^2}=0

    2-2x^2-2x^2=\sqrt{2-2x^2}

    2-4x^2=\sqrt{2-2x^2}

    (2-4x^2)(2-4x^2)=2-2x^2

    4-8x^2-8x^2+16x^2=4-16x^2+16x^4=2-2x^2

    16x^4-14x^2+2=0

    8x^4-7x^2+1=0

    One of the positive solutions of this gives the maximum x
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  5. #20
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    You can then try completing the square to solve the resulting equation.

    8x^4-7x^2+1=0

    8x^4-7x^2=

    We first find "q".

    (\sqrt{8}x^2-q)(\sqrt{8}x^2-q)=8x^4-q\sqrt{8}x^2-q\sqrt{8}x^2+q^2=8x^4-2\sqrt{8}qx^2+q^2

    Therefore

    7=2\sqrt{8}q

    q=\frac{7}{2\sqrt{8}},\ q^2=\frac{49}{4(8)}=\frac{49}{32}

    q=\frac{7}{\sqrt{32}}

    Hence,

    8x^4-7x^2+\frac{49}{32} is a square

    8x^4-7x^2+\frac{32}{32}+\frac{17}{32}=\frac{17}{32}

    (\sqrt{8}x^2-\frac{7}{\sqrt{32}})^2=\frac{17}{32}

    \sqrt{8}x^2-\frac{7}{\sqrt{32}}=\pm\frac{\sqrt{17}}{\sqrt{32}}

    \sqrt{8}x^2=\frac{7+\sqrt{17}}{\sqrt{32}},\ \frac{7-\sqrt{17}}{\sqrt{32}}

    x^2=\frac{7+\sqrt{17}}{\sqrt{32}\sqrt{8}},\ \frac{7-\sqrt{17}}{\sqrt{32}\sqrt{8}}

    x=\sqrt{\frac{7-\sqrt{17}}{\sqrt{32}\sqrt{8}}},\ \sqrt{\frac{7+\sqrt{17}}{\sqrt{32}\sqrt{8}}}

    One of these is the correct x maximum.
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  6. #21
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    Here's a sketch of solving for the derivative = 0
    Attached Thumbnails Attached Thumbnails Optimization help needed..-x-maximum.jpg  
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  7. #22
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    Hello archie
    thanks again for your help and time,
    tomorrow I will do it again and use your methods, because I have been doing it again, and this time I got 0.166 ft^2 absolute max

    I dont know why I didnt get 0.168 but my way isnt the quadratic
    please look at my pic View image: recprob1.jpg
    thanks
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  8. #23
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    Well done, wolfhound,

    0.166 is close enough (it's within 2 thousandths of a square foot of 0.168).

    The image is difficult to make out unfortunately...

    Here are my calculations for the derivative

    w=\sqrt{1-x^2}-\frac{1}{\sqrt{2}}

    Area=2xw=2x[\sqrt{1-x^2}-\frac{1}{\sqrt{2}}]=2x\sqrt{1-x^2}-\frac{2x}{\sqrt{2}}=2x\sqrt{1-x^2}-\sqrt{2}x

    \frac{d}{dx}A=2x(0.5[1-x^2]^{-0.5}(-2x))+2\sqrt{1-x^2}-\sqrt{2}=\frac{-2x^2}{\sqrt{1-x^2}}+2\sqrt{1-x^2}-\sqrt{2}

    This is zero for maximum area, so the value of x that makes this zero must be found.
    Last edited by Archie Meade; January 31st 2010 at 11:41 AM.
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  9. #24
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    hello, sorry file was too big to upload here,
    yes that what I got on the last line
    -2x^2 +2(1_x^2) -sqr2=0
    so -4x^2 +2 - 1.41 =0
    -4x^2=0.59
    x^2=-0.59/-4
    sqr => x=0.383

    a(0.383)= 0.166 ft^2
    Last edited by wolfhound; January 31st 2010 at 12:20 PM.
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  10. #25
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    Ok,

    you neglected to multiply \sqrt{2} by \sqrt{1-x^2}.

    you must multiply all 3 terms of the equation by \sqrt{1-x^2}

    It does involve much more calculations, unfortunately!
    and the value for "x max" is a bit off.

    it does give a very close answer, however.
    Well, i hope the exercise was beneficial, wolfhound.
    God loves a tryer!
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  11. #26
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    Quote Originally Posted by Archie Meade View Post
    Ok,

    you neglected to multiply \sqrt{2} by \sqrt{1-x^2}.

    you must multiply all 3 terms of the equation by \sqrt{1-x^2}

    It does involve much more calculations, unfortunately!
    and the value for "x max" is a bit off.

    it does give a very close answer, however.
    Well, i hope the exercise was beneficial, wolfhound.
    God loves a tryer!
    Hello again archie, ok so now I have sorted out this bit above, and worked everything down to
    8x^4 - 7x^2 + 1 =0
    so x^2 can be y
    8y^2 -7y + 1 =0
    .... y = 1/8 or y =1 so x = sqr 1/8 or x = sqr 1?? Help please?
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  12. #27
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    Quote Originally Posted by wolfhound View Post
    Hello again archie, ok so now I have sorted out this bit above, and worked everything down to
    8x^4 - 7x^2 + 1 =0
    so x^2 can be y
    8y^2 -7y + 1 =0
    .... y = 1/8 or y =1 so x = sqr 1/8 or x = sqr 1?? Help please?
    that's a lot faster than what i did!

    However, you can't factorise it to (8y-1)(y-1)

    as it will give you a -9y term in the middle.

    Instead, as 8y^2-7y+1 doesn't factorise conveniently, use

    For ay^2+by+c=0,\ y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    Hence y=\frac{7\pm\sqrt{49-32}}{16}

    x=\sqrt{y}=\sqrt{\frac{7+\sqrt{17}}{16}},\ \sqrt{\frac{7-\sqrt{17}}{16}}

    One of these is the correct x max
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  13. #28
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    Smile

    Quote Originally Posted by Archie Meade View Post
    that's a lot faster than what i did!

    However, you can't factorise it to (8y-1)(y-1)

    as it will give you a -9y term in the middle.

    Instead, as 8y^2-7y+1 doesn't factorise conveniently, use

    For ay^2+by+c=0,\ y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    Hence y=\frac{7\pm\sqrt{49-32}}{16}

    x=\sqrt{y}=\sqrt{\frac{7+\sqrt{17}}{16}},\ \sqrt{\frac{7-\sqrt{17}}{16}}

    One of these is the correct x max
    Hello archie, thanks for all your help, its such a relief to get it finished, I would have been lost without your help, I enjoyed it
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