A Norman window is a rectangle with a semicircle on top. Suppose that the perimeter of a particular

A Norman window is a rectangle with a semicircle on top. Suppose that the perimeter of a particular Norman window is to be 25 feet. What should the rectangle's dimensions be in order to maximize the area of the window and, therefore, allow in as much light as possible? (Round your answers to two decimal places.)width ftheight ft

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  1. 0.85 or 85%

    step-by-step explanation:

    1) divide 23 by 27

    2) round the decimal to the nearest 10

    3) so your answer would be 0.85 or 85%

  2. [tex]Length =\frac{25}{4 + \pi}[/tex] and [tex]Width = \frac{50}{4+\pi}[/tex]

    Step-by-step explanation:

    This question is better understood with an attachment.

    See attachment for illustration.

    Given

    Represent Perimeter with P

    [tex]P = 25ft[/tex]

    Required

    Determine the dimension of the rectangle that maximizes the area

    First, we calculate the perimeter of the rectangular part of the window.

    From the attachment, the rectangle is not closed at the top.

    So, The perimeter would be the sum of the three closed sides

    Where

    [tex]Width = 2x[/tex]

    [tex]Length = y[/tex]

    So:

    [tex]P_{Rectangle} = y + y + 2x[/tex]

    [tex]P_{Rectangle} = 2y + 2x[/tex]

    Next, we determine the circumference of the semi circle.

    Circumference of a semicircle is calculated as:

    [tex]C = \frac{1}{2}\pi r[/tex]

    From the attachment,

    [tex]Radius (r) = x[/tex]

    So, we have:

    [tex]C = \frac{1}{2}2\pi * x[/tex]

    [tex]C = \pi x[/tex]

    So, the perimeter of the window is:

    [tex]P = P_{Rectangle} + C[/tex]

    [tex]P =2y + 2x + \pi x[/tex]

    Recall that: [tex]P = 25[/tex]

    So, we have:

    [tex]25 =2y + 2x +\pi x[/tex]

    Make 2y the subject

    [tex]2y = 25 - 2x - \pi x[/tex]

    Make y the subject:

    [tex]y = \frac{25}{2} - \frac{2x}{2} - \frac{\pi x}{2}[/tex]

    [tex]y = \frac{25}{2} - x - \frac{\pi x}{2}[/tex]

    Next, we determine the area (A) of the window

    A = Area of Rectangle + Area of Semicircle

    [tex]A = 2x * y + \frac{1}{2}\pi r^2[/tex]

    [tex]A = 2xy + \frac{1}{2}\pi r^2[/tex]

    Recall that

    [tex]Radius (r) = x[/tex]

    [tex]A = 2xy + \frac{1}{2}\pi x^2[/tex]

    Substitute [tex]\frac{25}{2} - x - \frac{\pi x}{2}[/tex] for y in [tex]A = 2xy + \frac{1}{2}\pi x^2[/tex]

    [tex]A = 2x(\frac{25}{2} - x - \frac{\pi x}{2}) + \frac{1}{2}\pi x^2[/tex]

    Open Bracket

    [tex]A = 2x * \frac{25}{2} - 2x * x - 2x * \frac{\pi x}{2} + \frac{1}{2}\pi x^2[/tex]

    [tex]A = 25x - 2x^2 - \pi x^2 + \frac{1}{2}\pi x^2[/tex]

    [tex]A = 25x - 2x^2 - \frac{1}{2}\pi x^2[/tex]

    To maximize area, we have to determine differentiate both sides and set A' = 0

    Differentiate

    [tex]A' = 25 - 4x - \pi x[/tex]

    [tex]A' = 0[/tex]

    So, we have:

    [tex]0 = 25 - 4x - \pi x[/tex]

    Factorize:

    [tex]0 = 25 -x(4 + \pi)[/tex]

    [tex]-25 =-x(4 + \pi)[/tex]

    Solve for x

    [tex]x = \frac{-25}{-(4+\pi)}[/tex]

    [tex]x = \frac{25}{4+\pi}[/tex]

    Recall that

    [tex]Width = 2x[/tex]

    [tex]Width = 2(\frac{25}{4+\pi})[/tex]

    [tex]Width = \frac{50}{4+\pi}[/tex]

    Recall that:

    [tex]y = \frac{25}{2} - x - \frac{\pi x}{2}[/tex]

    Substitute [tex]\frac{25}{4+\pi}[/tex] for x

    [tex]y = \frac{25}{2} - (\frac{25}{4+\pi}) - \frac{\pi (\frac{25}{4+\pi})}{2}[/tex]

    [tex]y = \frac{25}{2} - (\frac{25}{4+\pi}) - \frac{\frac{25\pi}{4+\pi}}{2}[/tex]

    [tex]y = \frac{25}{2} - (\frac{25}{4+\pi}) - \frac{25\pi}{4+\pi} * \frac{1}{2}[/tex]

    [tex]y = \frac{25}{2} - \frac{25}{4+\pi} - \frac{25\pi}{2(4+\pi)}[/tex]

    [tex]y = \frac{25(4+\pi) - 25 * 2 - 25\pi}{2(4 + \pi)}[/tex]

    [tex]y = \frac{100+25\pi - 50 - 25\pi}{2(4 + \pi)}[/tex]

    [tex]y = \frac{100- 50+25\pi - 25\pi}{2(4 + \pi)}[/tex]

    [tex]y = \frac{50}{2(4 + \pi)}[/tex]

    [tex]y = \frac{25}{4 + \pi}[/tex]

    Recall that:

    [tex]Length = y[/tex]

    So:

    [tex]Length =\frac{25}{4 + \pi}[/tex]

    Hence, the dimension of the rectangle is:

    [tex]Length =\frac{25}{4 + \pi}[/tex] and [tex]Width = \frac{50}{4+\pi}[/tex]

    [tex]A Norman window is a rectangle with a semicircle on top. Suppose that the perimeter of a particular[/tex]

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