Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Show that of all rectangles with a given area, the one with smallest perimeter is a square.
Show that of all rectangles with a given perimeter, the one with greatest area is a square.

a. \(x = y = \sqrt A \)and rectangle is a square.

b. A is maximum when \(x = y\) i.e., rectangle is a square.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Data

Assume the data as shown below:

Let x be the length and y be the width of the rectangle .

Let area is A and perimeter is P.

(a)Step 2: To show that of all rectangles with a given area, the one with smallest perimeter is a square

Perimeter of rectangle is \(P = 2\left( {x + y} \right)\) ……..(1)

Area of the rectangle is \(A = xy\)

Therefore, \(y = \frac{A}{x}\) ……...(2)

Substitute value of (2) in (1) as:

\(P = 2\left( {x + y} \right)\)

\(P = 2\left( {x + \frac{A}{x}} \right)\)

For maximum or minimum values of perimeter P, differentiate P and equate to 0.

\(\begin{array}{c}\frac{{dP}}{{dx}} = 0\\2\left( {1 - \frac{A}{{{x^2}}}} \right) = 0\\\left( {1 - \frac{A}{{{x^2}}}} \right) = 0\\\frac{A}{{{x^2}}} = 1\\{x^2} = A\\x = \sqrt A \end{array}\)

(Dimensions of the rectangle is always positive)

Now, find second derivative of P

\(\begin{array}{c}P' = 2\left( {1 - \frac{A}{{{x^2}}}} \right)\\P'' = \frac{{4A}}{{{x^3}}}\end{array}\)

Find second derivative of P at \(x = \sqrt A \)

\(\begin{array}{c}{\left( {\frac{{{d^2}P}}{{d{x^2}}}} \right)_{x = \sqrt A }} = \frac{{4A}}{{\sqrt {{A^3}} }}\\ = \frac{4}{{\sqrt A }} > 0\end{array}\)

Therefore, for \(x = \sqrt A \) Perimeter, P or rectangle is the smallest.

\(y = \frac{A}{x}\)

\(y = \frac{A}{{\sqrt A }} = \sqrt A \)

Thus, length and width of rectangle are equal I,e., \(x = y = \sqrt A \)for smallest rectangle and so rectangle is a square.

(a)Step 3: To show that of all rectangles with a given perimeter, the one with greatest area is a square.

Perimeter of rectangle is \(P = 2\left( {x + y} \right)\)

Area of the rectangle is \(A = xy\)

Now, \(P = 2\left( {x + y} \right)\)…...(1)

Therefore, \(y = \frac{P}{2} - x\)

Substitute \(y = \frac{P}{2} - x\)in \(A = xy\)

\(A = xy = x\left( {\frac{P}{2} - x} \right) = \frac{{Px}}{2} - {x^2}\)

For maximum or minimum values of perimeter A, differentiate Aand equate to 0.

\(\frac{{dA}}{{dx}} = \frac{P}{2} - 2x\)

\(\begin{array}{c}\frac{P}{2} - 2x = 0\\\frac{P}{2} = 2x\\P = 4x\end{array}\)

Put \(P = 4x\)in equation (1) as:

\(\begin{array}{c}4x = 2\left( {x + y} \right)\\4x = 2x + 2y\\2x = 2y\\x = y\end{array}\)

And \({\left( {\frac{{{d^2}A}}{{d{x^2}}}} \right)_{x = y}} = - 2 < 0\)

Thus, A is maximum when \(x = y\)i.e., rectangle is a square.

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Show that of all rectangles with a given area, the one with smallest perimeter is a square.
Show that of all rectangles with a given perimeter, the one with greatest area is a square.

Step by Step Solution

Step 1: Given Data

(a)Step 2: To show that of all rectangles with a given area, the one with smallest perimeter is a square

(a)Step 3: To show that of all rectangles with a given perimeter, the one with greatest area is a square.

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Show that of all rectangles with a given area, the one with smallest perimeter is a square.Show that of all rectangles with a given perimeter, the one with greatest area is a square.

Step by Step Solution

Step 1: Given Data

(a)Step 2: To show that of all rectangles with a given area, the one with smallest perimeter is a square

(a)Step 3: To show that of all rectangles with a given perimeter, the one with greatest area is a square.

Most popular questions for Math Textbooks

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Show that of all rectangles with a given area, the one with smallest perimeter is a square.
Show that of all rectangles with a given perimeter, the one with greatest area is a square.