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Homework Help: Norman Window - maximum light

  1. Sep 7, 2008 #1
    1. The problem statement, all variables and given/known data

    a norman window has the shape of a rectangle surmounted by a semicircle. A norman window with the perimeter 30 ft. is to be constructed.

    a.) find a function that models the area of the window.
    b.) find the dimensions of the window that admit the greatest amount of light

    2. Relevant equations
    part a.) area= LW (this problem, it will be xy, with x as the width)
    semicircle circumference= 1/2[tex]\Pi[/tex]x s
    part b.) i dont know and I cannot use derivatives or calculus, which is why im having
    trouble here.

    3. The attempt at a solution

    P= x + 2y + 1/2[tex]\Pi[/tex]x = 30
    2y = 30 - 1/2[tex]\Pi[/tex]x - x
    y = 15 - 1/2x - 1/4[tex]\Pi[/tex]x

    A= (x)(15 - 1/2x - 1/4[tex]\Pi[/tex]x) + 1/2[tex]\Pi[/tex](1/2x)2
    (x)(15 - 1/2x - 1/4[tex]\Pi[/tex]x) + 1/8[tex]\Pi[/tex]x2
    hence A= 15x - 1/2x2 - 1/8[tex]\Pi[/tex]2

    i know the area equation is correct. I have no idea as to how to begin to figure out the max dimensions for the light. i would appreciate any clues. and sorry if the pi isnt looking right.
  2. jcsd
  3. Sep 7, 2008 #2
    Assuming your area equation is correct,

    This type polynomial has a special name; what is it?

    What do we call the graph of this type of polynomial?

    What do we know about the number of (in this case) maximums this graph has, and how do we find it/them?
  4. Sep 7, 2008 #3
    well, on the area equation, i did write it wrong, it was supposed to be:

    A = 15x - 1/2x2 - 1/8[tex]\Pi[/tex]x2

    anyway, there are 2 x2, so i dont know. i can only see there being one maximum. i dont know, maybe its because the pi is in there and there are 2 numbers of the same degree, i just dont know because i have never done a problem like this. i put the area equation in the calculator, like i should have done in the beginning, with the pi and everything and the max is x=8.40. i know that is also correct and i dont know why i didnt try that before.

    so, putting it into the y (or height) formula, 15 - 1/2(8.40) - 1/4[tex]\Pi[/tex](8.40) = 4.20

    and the max area would be 465.51ft2

    my book gives a long, completely different way of going about the solution that i cant figure out by looking at it. im not sure if any of this is what you meant, but thanks for taking the time to reply and get me thinking.
  5. Sep 7, 2008 #4
    The last two terms in your equation for A can be written

    - (1/2 + (1/8)pi) x2

    The graph of your equation is a parabola which opens down (the coefficient of x2 is negative) and thus has one maximum.

    You were on the right track! This is not an easy problem.
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