Question

A rectangle has area\({\bf{16}}\;{{\bf{m}}^{\bf{2}}}\). Express the perimeter of the rectangle as a function of the length of one of its sides.

Short answer

The expression for the perimeter of a rectangle as a function of length is \({\bf{p}}\left( {\bf{L}} \right){\bf{ = 2}}\left( {{\bf{L + }}\frac{{{\bf{16}}}}{{\bf{L}}}} \right)\) and domain of the function is\({\bf{(0,4]}}\).

Step-by-step solution

Determine the equation for the area of a rectangle

The area of a rectangle is given as;

\(A = L \cdot B\)

Here,\(L\)and\(B\)are the length and breadth of the rectangle,\(A\)is an area of the rectangle.

Substitute all the values in the above equation.

\(\begin{array}{c}16\;{{\rm{m}}^2} = L \cdot B\\B = \frac{{16}}{L}\end{array}\)

Determine the expression for the perimeter of the rectangle as a function of length

The perimeter of a rectangle is given as;

\(p\left( L \right) = 2\left( {L + B} \right)\)

Substitute all the values in the above equation.

\(p\left( L \right) = 2\left( {L + \frac{{16}}{L}} \right)\)

Substitute\(p\left( L \right) = 0\)to find the domain of the area of the rectangle.

\(\begin{array}{c}p\left( L \right) = 0\\2\left( {L + \frac{{16}}{L}} \right) = 0\\2L + \frac{{32}}{L} = 0\end{array}\)

For domain

\(L > 0\)because the length of the rectangle will be positive.

The length of the rectangle will be greater than the width of the rectangle because we are calculating the domain for perimeter as a function of the length of the rectangle and this will fix the upper limit of the length of the rectangle, so the domain is given as:

\(\begin{array}{c}\frac{{32}}{L} > 2L\\\frac{{16}}{L} > L\\{L^2} < 16\\ - 4 < L < 4\end{array}\)

The negative side is not possible so

\(0 < L \le 4\)

The domain for the perimeter of a rectangle is\((0,4]\)

Therefore, the expression for the perimeter of a rectangle as a function of length is \(p\left( L \right) = 2\left( {L + \frac{{16}}{L}} \right)\) and the domain of the function is\((0,4]\).