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Q13E

Expert-verifiedFound in: Page 238

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.

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*.

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.

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 *A*and 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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