Question

Find the dimensions of a rectangle with area 1000 m² whose perimeter is as small as possible.

Short answer

\(10\sqrt {10} \)mi is the length and \(\frac{{1000}}{{10\sqrt {10} }}\)mi is the width.

Step-by-step solution

Given Data

1)Area of the rectangle is 1000 m².

2)Perimeter is as small as possible.

Determination of the dimensions of the rectangle

Let A be the area and

Length of the rectangle be x in miles and width be y in miles.

Now, \({\rm{Area}} = xy\)

Therefore, \(1000 = xy\)

Eliminating y from above equation,

\(y = \frac{{1000}}{x}\)

\({\rm{Area}} = xy\)

Let the perimeter be P.

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

Substitute \(y = \frac{{1000}}{x}\)in equation (1).

\(P = 2\left( {x + \frac{{1000}}{x}} \right)\)

Find \(P'\left( x \right)\)as:

\(P'\left( x \right) = 2 - \frac{{2000}}{{{x^2}}}\)

Put \(P'\left( x \right) = 0\) to find critical points.

\(\begin{aligned}{c}2 - \frac{{2000}}{{{x^2}}} &= 0\\\frac{{2000}}{{{x^2}}} &= 2\\2{x^2} &= 2000\\{x^2} &= 1000\\x &= 10\sqrt {10} \end{aligned}\)

Now find \(P''\left( x \right)\)as:

\(P'\left( x \right) = 2 - \frac{{2000}}{{{x^2}}}\)

\(P''\left( x \right) = \frac{{4000}}{{{x^3}}}\)……..(2)

Substitute \(x = 10\sqrt {10} \)equation (2)

\(\begin{aligned}P''\left( {10\sqrt {10} } \right) &= \frac{{4000}}{{100000\sqrt {10} }} > 0\\\end{aligned}\), therefore, perimeter is minimum.

Now, substitute \(x = 10\sqrt {10} \)in \(y = \frac{{1000}}{x}\) as:

\(y = \frac{{1000}}{{10\sqrt {10} }}\)

Thus, the required dimensions of the rectangle is\(x = 10\sqrt {10} \)miles in length and \(y = \frac{{1000}}{{10\sqrt {10} }}\)miles in width.