Post a New Question


posted by .

A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 30ft, find the dimensions of the window so that the greatest possible amount of light is admitted.

I keep screwing up in creating my equation.

p = perimeter of window
l = length of rectangle
d = w = width of rectangle
circumference of a semi circle = 2(PI*r)/2 = rPi

I get p = rPi + 2L + 2W

I don't know what I am doing wrong since I am unable to get a proper derivative.

  • calculus -

    How about starting with simpler definitions.
    Let the radius of the semicircle be r
    then the length of the rectangle is 2r.
    let the width be x.
    2x + 2r + (1/2)2πr = 30
    2x + 2r + πr = 30
    x = (30-2r-πr)/2

    area = (1/2)πr^2 + 2xr
    = (1/2)πr^2 + 2r(30-2r-πr)
    =(1/2)πr^2 + 30r - 2r^2 - πr^2
    d(area)/dr = πr + 30 - 4r - 2πr
    = 0 for a max of area
    r(π-4-2π) = -30
    r = 30/(π+4) = appr. 4.2
    then x = 4.2 also
    so the rectangle is 8.4 long and 4.2 high and the semicircle sitting on top has a diameter of 8.4

  • calculus -

    Wouldn't the equation for the perimeter be

    2x + 4r + πr = 30?

    Since there are two lengths?

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question