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Found in: Page 238

### Essential Calculus: Early Transcendentals

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

# 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 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.