 # Rectangle has area 16 m^2. Express the perimeter of the rectangle as a function of the length of one of its sides. – If the length of the rectangle is assumed to be larger than its width, calculate the domain of the Perimeter $P$ in terms of interval notation.

The purpose of this guide is to derive an expression for the perimeter $P$ of the given rectangle in terms of the length of one of its sides and find the domain of Perimeter $P$ in terms of the upper and lower limits.

The basic concept behind this guide is the substitution method for solving simultaneous equations, and the limit function to find the domain of a certain function.

The Substitution method is used to find the value of variables involved in two or more simultaneous linear equations. If a function has a fixed value and consists of $2$ variable i.e. $x$ and $y$, we can use the substitution method to find the value of variables by expressing them in the form of a single variable.

The domain of any function is defined as the set or range of minimum and maximum input values for which the given function is completely solved.

Given that:

Area of the rectangle $A=16\ {\mathrm{ft}}^2$

The Length of the Rectangle is $L$.

The Width of the Rectangle is $W$.

We have to find the Perimeter $P$ of the rectangle in terms of one of its sides. Let’s assume it as the Length $L$ of the rectangle.

The Area of rectangle is defined as follows:

$A=L\times W$

$16=L\times W$

As we are given the value of Area $A=16\ {\mathrm{ft}}^2$, we will express it in terms of a single parameter $L$ as follows:

$W=\frac{16}{L}$

Now, the Perimeter $P$ of a rectangle are:

$P=2L+2W$

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

$P=2L+\frac{32}{L}$

For the domain of perimeter, we have assumed that the length of the rectangle is larger than its width.

So, the minimum value of Length can be $L=W$:

$A=L\times W$

$16=L\times L$

$L=4$

As we have assumed that $L=W$, so:

$W=4$

But as it is given that Length is larger than Width, the lower limit will be $L=4$.

$\lim_{L\to 4}{P(L)}=\lim_{L\to 4}{2L\ +2\left(\frac{16}{L}\right)}$

$\lim_{L\to 4}{P(4)}=2(4)+2\left(\frac{16}{4}\right)=16$

Hence the perimeter $P$ has a lower limit of $16$.

Now for the upper limit of length, consider the area of the rectangle:

$A=L\times W$

$16=L\times\frac{16}{L}$

Length $L$ will cancel out which means that its value will be very high and approaching infinity $\infty$ and the width $W$ will approach zero. Hence:

$L\rightarrow\infty$

$\lim_{L\to\infty}{P(L)}=\lim_{L\to\infty}{2L\ +2\left(\frac{16}{L}\right)}$

$\lim_{L\to\infty}{P(\infty)}=2(\infty)+2\left(\frac{16}{\infty}\right)=\infty$

Hence, the perimeter $P$ have an upper limit infinity $\infty$.

Hence, the perimeter of the rectangle has the domain $(4,\ \infty)$.

## Numerical Result

The Perimeter of the Rectangle in terms of one side is:

$P=2L+\frac{32}{L}$

The Perimeter of the Rectangle has the domain $(4,\ \infty)$

## Example

If the length of a rectangle is half of its width, find an expression that represents the perimeter of the rectangle in terms of its length.

Solution

Given that:

$L=\frac{1}{2}W$

$W=2L$

We have to find the Perimeter $P$ of the rectangle in terms of its length $L$.

The Perimeter $P$ of a rectangle are:

$P=2L+2W$

Substituting the value of $W$ in the above equation:

$P=2L+2\left(2L\right)$

$P=2L+4L$

$P=6L$