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Math Help - Light through a window

  1. #1
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    Light through a window

    I have a problem that I have solution to but not the proof. Can anyone help?

    The problem is this:

    A window, shaped as a rectangle with a semicircle on top, has a perimeter of 5m. The semicircle has a radius r and the rectangle has sides which are effectively 2r x h (height) what value of r gives the maximum light (area).

    Simple question I know, but have problems with the details.
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  2. #2
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    Quote Originally Posted by Anthony Hodson
    I have a problem that I have solution to but not the proof. Can anyone help?

    The problem is this:

    A window, shaped as a rectangle with a semicircle on top, has a perimeter of 5m. The semicircle has a radius r and the rectangle has sides which are effectively 2r x h (height) what value of r gives the maximum light (area).

    Simple question I know, but have problems with the details.
    The perimeter is:

    2h\ +\ 2r\ +\ \pi r\ =\ 5

    rearrange this to give h in terms of r.

    The area is:

    Area\ =\ 2r.h\ +\ \pi .r^2/2.

    Substitute the representation you have for h in terms of r into
    this, then find the r that maximises the area.

    You do this by differentiating Area w.r.t. r and
    setting the derivative to zero, and solving for r.

    RonL
    Last edited by CaptainBlack; December 9th 2005 at 05:45 AM.
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  3. #3
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    Thanks,

    I have got that far, although I see you have 2*pi*r for the perimeter, although i used pi*r (semicircle). However, Its the last part I can't get to. I cant see the max and min values of r?

    can you help?
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  4. #4
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    Quote Originally Posted by Anthony Hodson
    Thanks,

    I have got that far, although I see you have 2*pi*r for the perimeter, although i used pi*r (semicircle).
    Opps, typo, you are right

    However, Its the last part I can't get to. I cant see the max and min values of r?

    can you help?
    Last edited by CaptainBlack; December 9th 2005 at 06:00 AM.
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  5. #5
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    Quote Originally Posted by Anthony Hodson
    Thanks,

    Its the last part I can't get to. I cant see the max and min values of r?

    can you help?
    We have <you will need to check this - NOT GUARANTEED TYPO FREE >:<Corrected Now - I hope>

    h\ =\ \frac{5-r(2+\pi)}{2},

    so substituting the RHS of this for h in the Area
    equation gives:

    Area\ =\ 5r\ -\ (2+\pi/2)r^2

    Then:

    \frac{d\ Area}{dr}\ =\ 5\ -\ 2(2+\pi/2)r.

    Now set

    \frac{d\ Area}{dr}\ =\ 0

    and solve for r

    That is solve:

    5\ -\ 2(2+\pi/2)r\ =\ 0

    for r.

    RonL
    Last edited by CaptainBlack; December 9th 2005 at 08:44 AM.
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  6. #6
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    Thanks again, brilliant response,

    not sure if its a typo or not, your area equation seems to be missing a semicircle?

    It would appear i am on the right track, however I am expecting two values for r = 0.707 and a - figure?

    I am sure that this a simple equation, but I must be making silly mistakes, I am still thinking that the Derivative should a quadratic equation but I may be wrong?

    Regards
    Tony
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  7. #7
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    Quote Originally Posted by Anthony Hodson
    Thanks again, brilliant response,

    not sure if its a typo or not, your area equation seems to be missing a semicircle?
    <br />
Area\ =\ 2r.h\ +\ \pi .r^2/2<br />

    First term is the area of the rectangular part, the second the
    area of the semi-circle.

    h\ =\ \frac{5-r(2+\pi)}{2}

    so,

    Area\ =\ 2r.\frac{5-r(2+\pi)}{2}\ +\ \pi .r^2/2,

    Area\ =\ r.5-r^2(2+\pi)\ +\ \pi .r^2/2,

    Area\ =\ r.5-2.r^2-\pi .r^2/2,

    Area\ =\ r.5-r^2.(2+\pi/2)

    which might now be right

    I will correct the previous post when I get a chance.

    RonL
    Last edited by CaptainBlack; December 9th 2005 at 07:52 AM.
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  8. #8
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    Many Thanks,
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  9. #9
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    Quote Originally Posted by Anthony Hodson
    I have a problem that I have solution to but not the proof. Can anyone help?

    The problem is this:

    A window, shaped as a rectangle with a semicircle on top, has a perimeter of 5m. The semicircle has a radius r and the rectangle has sides which are effectively 2r x h (height) what value of r gives the maximum light (area).

    Simple question I know, but have problems with the details.
    Here is one way.

    Perimeter, P = (pi)r +h +2r +h = 5
    2h +r(2+pi) = 5
    Differentiate both sides with respect to r,
    2(dh/dr) +(2+pi) = 0
    dh/dr = -(2+pi)/2 -------------***

    Area, A = pi(r^2)/2 +2r(h)
    Differentiate both sides with respect to r,
    dA/dr = (pi)r +2[r(dh/dr) +h]
    Set dA/dr to zero, for max A, and plug in the value of dh/dr,
    0 = (pi)r +2[r{-(2+pi)/2} +h]
    0 = (pi)r -r(2+pi) +2h
    0 = (pi)r -2r -(pi)r +2h
    r = h ---------for max A.

    Back to the perimeter,
    2h +r(2+pi) = 5
    Substitute r for h,
    2r +r(2+pi) = 5
    r(2 +2+pi) = 5
    r = 5/(4+pi) = 0.70 m. -------for max A, answer.
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  10. #10
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    Quote Originally Posted by ticbol
    Here is one way...
    And here is another, which unfortunately will be all Greek to Antony
    but I am posting anyway for a bit of fun .

    Maximise:

    Area(r,h)\ =\ 2r.h\ +\ \pi.r^2/2,

    subject to the constraint:

    Perimeter(r,h)\ =\ 2h\ +\ r(2+\pi)\ =\ 5.

    We will use the Lagrange Multipliers, so we let:

    H(r,h,\lambda)\ =\  2r.h\ +\ \pi.r^2/2\ +\ \lambda[2h\ +\ r(2+\pi)\ -\ 5]

    and we seek solutions of:

     \frac{\partial H}{\partial r}\ =\ 2h\ +\ \pi r\ +\ \lambda (2+\pi)\ =\ 0

     \frac{\partial H}{\partial h}\ =\ 2r\ +\ 2 \lambda\ =\ 0

    (the constraint also must hold - it expressed the condition that
     \partial H/ \partial \lambda\ =\ 0).

    The second of these equations tells us that:

    \lambda \ =\ -r,

    which when substituted into the first of the equations gives:

    h\ =\ r,

    which when substituted back into the constraint gives:

    r\ =\ \frac{5}{4+\pi}

    RonL
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  11. #11
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    your right greek, well done to all.

    Thanks
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