– 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. 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:
\[P=2L+2\left(2L\right)\]
\[P=2L+4L\]
\[P=6L\]