**– 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**.

## Expert Answer

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**. Letâ€™s assume it as the **Length** $L$ of the **rectangle**.

The **Area** ofÂ **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:

\[P=2L+2\left(2L\right)\]

\[P=2L+4L\]

\[P=6L\]