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Chapter 8: Problem 34

A Norman window has the shape of a rectangle surmounted by a semicircle of diameter equal to the width of the rectangle (see the accompanying figure). If the perimeter of the window is 20 feet (including the semicircle), what dimensions will admit the most light (maximize the area)? (Hint: Express \(L\) in terms of \(r\). Recall that the circumference of a circle \(=2 \pi r,\) and the area of a circle \(=\pi r^{2},\) where \(r\) is the radius of the circle.)

Short Answer

Expert verified
The width is \( \frac{40}{4 + \pi} \) feet and the height is \( \frac{20}{4 + \pi} \) feet.

Step by step solution

01

Understand the structure of the window

The Norman window consists of a rectangle with a semicircle on top. Let the width of the rectangle be denoted as \(w\) and its height as \(h\). The radius of the semicircle is then \(r = \frac{w}{2}\).
02

Write the equations for the perimeter and area

The perimeter of the window includes the perimeter of the rectangle (minus the top width) plus the circumference of the semicircle. This can be written as: \[ P = w + 2h + \frac{1}{2} (2 \pi r) \] Given that \(P = 20\) feet, substitute \(r = \frac{w}{2}\) to get: \[ 20 = w + 2h + \frac{1}{2} (2 \pi \frac{w}{2}) \]\[ 20 = w + 2h + \pi \frac{w}{2} \]\[ 20 = w + 2h + \pi \frac{w}{2} \].
03

Solve for \(h\) in terms of \(w\)

Rearrange the perimeter equation to solve for \(h\):\[ 20 = w + 2h + \pi \frac{w}{2} \]\[ 20 = w + \pi \frac{w}{2} + 2h \]\[ 2h = 20 - w - \pi \frac{w}{2} \]\[ h = 10 - \frac{1}{2} w - \frac{\pi}{4} w \].
04

Write the area equation

The area of the window is the sum of the area of the rectangle and the area of the semicircle.\[ A = w * h + \frac{1}{2} \pi r^2 \]Substitute \(h = 10 - \frac{1}{2} w - \frac{\pi}{4} w \) and \( r = \frac{w}{2} \):\[ A = w(10 - \frac{1}{2} w - \frac{\pi}{4} w) + \frac{1}{2} \pi \left(\frac{w}{2}\right)^2 \].
05

Differentiate the area equation to find the maximum

Differentiate \(A\) with respect to \(w\):\[ A = w(10 - \frac{1}{2} w - \frac{\pi}{4} w) + \frac{1}{8} \pi w^2 \]\[ A = 10w - \frac{1}{2}w^2 - \frac{\pi}{4} w^2 + \frac{1}{8} \pi w^2 \]Combine like terms to get:\[ A = 10w + ( \frac{1}{8} \pi - \frac{\pi}{4} - \frac{1}{2}) w^2 \]\[ \frac{dA}{dw} = 10 - (1 + \frac{\pi}{4}) w \].Set \( \frac{dA}{dw} = 0 \) and solve for \(w\):\[ 10 - (1 + \frac{\pi}{4}) w = 0 \]\[ w = \frac{10}{1 + \frac{\pi}{4}} \]Calculate \(w\):\[ w = \frac{40}{ 4+ \pi} \].
06

Calculate height \(h\)

Using \( w = \frac{40}{4+\pi}\), substitute back into the equation for \(h\):\[ h = 10 - \frac{1}{2} (\frac{40}{4+\pi}) - \frac{\pi}{4} (\frac{40}{4+\pi}) \].
07

Simplify and solve

Solve for \(h\):\[ h = 10 - \frac{20}{4 + \pi} - \frac{10\pi}{4 + \pi} \]\[ h = 10 - \frac{20 + 10\pi}{4 + \pi} \]\[ h = 10 - \frac{2 (10 + 5\pi)}{4 + \pi} \]\[ h = 10 - \frac{2}{4 + \pi}(10 + 5\pi) \]\[ h = \frac{40 + 10\pi - 20 - 10\pi}{4 + \pi} = \]\[ h = \frac{20}{4 + \pi}, w = \frac{40}{4 + \pi} \]. Thus, the dimensions are h, w.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Perimeter Calculation
To find the dimensions of a Norman window that maximize the area, we need to understand its structure and calculate its perimeter. A Norman window consists of a rectangle with a semicircle on top. Let the width of the rectangle be denoted as \(w\) and its height as \(h\). The radius of the semicircle is then half the width of the rectangle: \(r = \frac{w}{2}\).
The total perimeter of the window is the perimeter of the rectangle (excluding the top width) plus the circumference of the semicircle. The perimeter equation is:
\[ P = w + 2h + \frac{1}{2} (2 \pi r) \] Given that the total perimeter is 20 feet, substituting \(r\) with \(\frac{w}{2}\) gives us:
\[ 20 = w + 2h + \pi \frac{w}{2} \] Simplifying further:
\[ 20 = w + 2h + \pi \frac{w}{2} \] This is the perimeter equation. Rearranging it will help you to solve for \(h\) in terms of \(w\) later on.
Area Optimization
To maximize the light admitted through the Norman window, we must maximize its area. The area of the window is the sum of the area of the rectangular part and the semicircular part.
The area of the rectangle is \(width \times height = w \times h\), and the area of the semicircle is half the area of a full circle: \(\frac{1}{2} \pi r^{2}\).
Substituting \(r = \frac{w}{2}\) into the area equation and using \(h\) in terms of \(w\) from the perimeter calculation, we get:
\[ A = w (10 - \frac{1}{2} w - \frac{\pi}{4} w) + \frac{1}{2} \pi (\frac{w}{2})^{2} \] Simplifying further gives:
\[ A = 10w - \frac{1}{2}w^{2} - \frac{\pi}{4}w^{2} + \frac{1}{8} \pi w^{2} \] Combining like terms in the area equation, we uncover the expression that captures the total window area.
Differentiation
To find the values of \(w\) and \(h\) that maximize the area, we must use differentiation. This allows us to find where the area function reaches its highest point. Differentiating the area equation with respect to \(w\) gives:
\[ \frac{dA}{dw} = 10 - (1 + \frac{\pi}{4}) w \] Setting the derivative equal to zero to find critical points:
\[ 10 - (1 + \frac{\pi}{4}) w = 0 \] Solving for \(w\) gives:
\[ w = \frac{10}{ 1 + \frac{\pi}{4}} = \frac{40}{4 + \pi} \] Substituting \(w\) back into the equation for \(h\) from the perimeter calculation:
\[ h = 10 - \frac{1}{2} (\frac{40}{4 + \pi}) - \frac{\pi}{4} (\frac{40}{4 + \pi}) \] Simplifying this, we get:
\[ h = \frac{20}{4 + \pi} \] Hence, the dimensions that allow the window to admit the most light are:
\( w = \frac{40}{4 + \pi} \) and \( h = \frac{20}{4 + \pi} \)

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