Chapter 1.1, Problem 54E

To determine

To find: The function of the perimeter of the rectangle in terms of the length of one of the sides of rectangle and the function domain.

Answer to Problem 54E

The formula for the function of perimeter of the rectangle in terms of its length is $P\left(l\right)=\frac{2l^2+32}{l}$.

The domain of the function of the perimeter (P) of the rectangle in terms of its length is $l>0$.

In case of length is to be larger than breadth, the domain of P is $l>4$.

Explanation of Solution

Given:

The area of the rectangle is $16{\text{m}}^2$.

Formula used:

Area of the rectangle, $A=l\times b$, where ‘l’ is the length of the rectangle and ‘b’ is the breadth of the rectangle.

Perimeter of the rectangle, $P=2\left(l+b\right)$, where ‘l’ is the length of the rectangle and ‘b’ is the breadth of the rectangle.

Calculation:

Let the length of the rectangle be l and the breadth be b. Since the area of the rectangle is $16{\text{m}}^2$,

$l\times b=16{\text{m}}^2$ (1)

Express the equation (1) in terms of length.

$\begin{array}{c}l\times b=16\\ b=\frac{16}{l}\end{array}$

Therefore, the function of perimeter of the rectangle in terms of l is $P\left(l\right)=2\left(l+\frac{16}{l}\right)$.

That is, $P\left(l\right)=2l+\frac{32}{l}$.

Therefore, the function of perimeter of the rectangle in terms of its length is $P\left(l\right)=\frac{2l^2+32}{l}$.

The domain of P is $l>0$ as the length l cannot take the negative values.

In the case of length is to be larger than breadth, the domain of P is $l>4$.