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8 June, 11:59

A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. The goal is to find the area of the largest possible Norman window with a perimeter of 35 feet?

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  1. 8 June, 13:31
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    A (max) = 72,13 ft²

    Step-by-step explanation:

    Total area of the Norman window is area of the semicircle plus area of the rectangle

    Let call x radius of the semicircle, then the width of the rectangle is 2*x

    Area of semicircle A (sc) = π*r²/2 ⇒ A (sc) = π*x²/2

    Area of rectangle (sides 2*x and y) A (r) = 2*x*y

    Area of window A (w) = π*x²/2 + 2 * x*y

    Perimeter of the window is:

    Three sides of the rectangle P (r) = 2*y + 2*x

    Length of the semicircle P (sc) = π*x

    Then 35 = 2*y + 2*x + π*x ⇒ 2*y = 35 - 2*x - π*x

    y = 35 - x * (2 + π) / 2

    Area of window as a function of x

    A (x) = π*x²/2 + 2 * [ 35 - x (2 + π) / 2 ]*x

    A (x) = π*x²/2 + x * (35 - 2*x - π*x)

    A (x) = π*x²/2 + 35*x - 2*x² - π*x²

    Taking derivatives on both sides of the equation

    A' (x) = π*x + 35 - 4*x - 2 * π*x

    A' (x) = 0 - π*x - 4*x = - 35 ⇒ - x (4 + π) = - 35 ⇒ x = 4,90 ft

    And y = (35 - x - π*x) / 2 ⇒ y = (35 - 4,90 - 15,38) / 2

    y = 7,36 ft

    A (max) = 2*x*y * + π*x²/2

    A (max) = 72,13 ft²
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