Question

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

Answer 1

Answer:

$Length =\frac{25}{4 + \pi}$ and $Width = \frac{50}{4+\pi}$

Step-by-step explanation:

This question is better understood with an attachment.

See attachment for illustration.

Given

Represent Perimeter with P

$P = 25ft$

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

$Width = 2x$

$Length = y$

So:

$P_{Rectangle} = y + y + 2x$

$P_{Rectangle} = 2y + 2x$

Next, we determine the circumference of the semi circle.

Circumference of a semicircle is calculated as:

$C = \frac{1}{2}\pi r$

From the attachment,

$Radius (r) = x$

So, we have:

$C = \frac{1}{2}2\pi * x$

$C = \pi x$

So, the perimeter of the window is:

$P = P_{Rectangle} + C$

$P =2y + 2x + \pi x$

Recall that: $P = 25$

So, we have:

$25 =2y + 2x +\pi x$

Make 2y the subject

$2y = 25 - 2x - \pi x$

Make y the subject:

$y = \frac{25}{2} - \frac{2x}{2} - \frac{\pi x}{2}$

$y = \frac{25}{2} - x - \frac{\pi x}{2}$

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

A = Area of Rectangle + Area of Semicircle

$A = 2x * y + \frac{1}{2}\pi r^2$

$A = 2xy + \frac{1}{2}\pi r^2$

Recall that

$Radius (r) = x$

$A = 2xy + \frac{1}{2}\pi x^2$

Substitute $\frac{25}{2} - x - \frac{\pi x}{2}$ for y in $A = 2xy + \frac{1}{2}\pi x^2$

$A = 2x(\frac{25}{2} - x - \frac{\pi x}{2}) + \frac{1}{2}\pi x^2$

Open Bracket

$A = 2x * \frac{25}{2} - 2x * x - 2x * \frac{\pi x}{2} + \frac{1}{2}\pi x^2$

$A = 25x - 2x^2 - \pi x^2 + \frac{1}{2}\pi x^2$

$A = 25x - 2x^2 - \frac{1}{2}\pi x^2$

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

Differentiate

$A' = 25 - 4x - \pi x$

$A' = 0$

So, we have:

$0 = 25 - 4x - \pi x$

Factorize:

$0 = 25 -x(4 + \pi)$

$-25 =-x(4 + \pi)$

Solve for x

$x = \frac{-25}{-(4+\pi)}$

$x = \frac{25}{4+\pi}$

Recall that

$Width = 2x$

$Width = 2(\frac{25}{4+\pi})$

$Width = \frac{50}{4+\pi}$

Recall that:

$y = \frac{25}{2} - x - \frac{\pi x}{2}$

Substitute $\frac{25}{4+\pi}$ for x

$y = \frac{25}{2} - (\frac{25}{4+\pi}) - \frac{\pi (\frac{25}{4+\pi})}{2}$

$y = \frac{25}{2} - (\frac{25}{4+\pi}) - \frac{\frac{25\pi}{4+\pi}}{2}$

$y = \frac{25}{2} - (\frac{25}{4+\pi}) - \frac{25\pi}{4+\pi} * \frac{1}{2}$

$y = \frac{25}{2} - \frac{25}{4+\pi} - \frac{25\pi}{2(4+\pi)}$

$y = \frac{25(4+\pi) - 25 * 2 - 25\pi}{2(4 + \pi)}$

$y = \frac{100+25\pi - 50 - 25\pi}{2(4 + \pi)}$

$y = \frac{100- 50+25\pi - 25\pi}{2(4 + \pi)}$

$y = \frac{50}{2(4 + \pi)}$

$y = \frac{25}{4 + \pi}$

Recall that:

$Length = y$

So:

$Length =\frac{25}{4 + \pi}$

Hence, the dimension of the rectangle is:

$Length =\frac{25}{4 + \pi}$ and $Width = \frac{50}{4+\pi}$