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Essential Calculus: Early Transcendentals
Found in: Page 239
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

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

The radius of semi-circular part is \(\frac{{30}}{{4 + \pi }} \approx 4.2\) ft.

Dimensions of rectangular window is \(8.4 \times 4.2\) ft.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: To sketch the figure and mention given data

Let \(a\) be the half of the width of the rectangle (radius of the semicircle) and \(h\) be the height of the rectangle.

The sketch is shown in the figure below:

Then the perimeter of the window is given by,

\(2a + 2h + \frac{1}{2}\left( {2\pi a} \right) = 2h + a\left( {2 + \pi } \right)\).

Given that the perimeter of the window is 30 ft.

Thus,

\(\begin{aligned}{c}2h + a\left( {2 + \pi } \right) &= 30\\2h &= 30 - a\left( {2 + \pi } \right)\\h &= \frac{{30 - a\left( {2 + \pi } \right)}}{2}\\h &= 15 - \left( {1 + \frac{\pi }{2}} \right)a\,\,\,\, \cdots \cdots \left( 1 \right)\end{aligned}\)

Step 2: To find the area

Area of the window is,

\(A\) = area of rectangle + area of semicircle

\(A = 2ah + \frac{1}{2}\pi {a^2}\,\,\, \cdots \cdots \left( 2 \right)\)

Substituting \(\left( 1 \right)\) in \(\left( 2 \right)\), we get,

\(\begin{aligned}{c}A &= 2a\left( {15 - \left( {1 + \frac{\pi }{2}} \right)a} \right) + \frac{1}{2}\pi {a^2}\\A &= 30a - 2{a^2}\left( {1 + \frac{\pi }{2}} \right) + \frac{1}{2}\pi {a^2}\\A &= 30a - 2{a^2} - \pi {a^2} + \frac{1}{2}\pi {a^2}\\A &= 30 - \left( {2 + \frac{\pi }{2}} \right){a^2}\end{aligned}\)

We need to maximize the \(A\) to find \(a\), for which it is maximum.

Differentiating \(A\), we get,

\(\begin{aligned}{l}A' &= 30 - 2\left( {2 + \frac{\pi }{2}} \right)\\A' &= 30 - \left( {4 + \pi } \right)a\end{aligned}\)

Thus,

\(\begin{aligned}{c}A' &= 0\\30 - \left( {4 + \pi } \right)a &= 0\\\left( {4 + \pi } \right)a &= 30\\a &= \frac{{30}}{{4 + \pi }}\\a \approx 4.2\end{aligned\)

Step 3:  To find the dimensions of the window

The domain conditions are,\(a > 0,\,h > 0,\,A > 0\).

Therefore,

\(a \in \left( {0,\,\frac{{15}}{{1 + {\pi \mathord{\left/

{\vphantom {\pi 2}} \right.

\kern-\nulldelimiterspace} 2}}}} \right)\) .

Now, \(A\left( 0 \right) = 0\),

\(\begin{array}{c}A\left( {\frac{{30}}{{4 + \pi }}} \right) \approx A\left( {4.2} \right)\\ \approx 63\end{array}\) ,

\(A\left( {\frac{{15}}{{1 + {\pi \mathord{\left/

{\vphantom {\pi 2}} \right.

\kern-\nulldelimiterspace} 2}}}} \right) \approx 53.5\).

Hence, area of window is maximum when \(a = \frac{{30}}{{4 + \pi }}\).

Equation \(\left( 1 \right)\) becomes,

\(\begin{array}{l}h = 15 - \left( {1 + \frac{\pi }{2}} \right) \times \left( {\frac{{30}}{{4 + \pi }}} \right)\\ \Rightarrow h = \frac{{30}}{{4 + \pi }}\end{array}\)

Hence, maximum light is admitted through the window, when:

The radius of semi-circular part is \(\frac{{30}}{{4 + \pi }} \approx 4.2\) ft.

Dimensions of rectangular window is \(8.4 \times 4.2\) ft.

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