Question

What is the smallest perimeter possible for a rectangle of an area $16\text{ }i{{n}^{2}}$?

Answer 1

Hint: In this problem we need to calculate the dimensions of the rectangle that satisfies the given condition. For this we will first assume the length and width of the rectangle as $l$ and $b$ respectively. Now we have the area of the rectangle as $16\text{ }i{{n}^{2}}$. We have the formula for the area of rectangle of length $l$ and width $b$ as $lb$. Now we will equate both the values and write the expression for either length or width. After we will calculate the perimeter of the rectangle which is given by $2\left( l+b \right)$. We will substitute the calculated value of length or width in the previous step and simplify the equation. To calculate the extremities of the obtained function we need to differentiate the obtained function and equate it to zero. On simplification we will get the value of length or width of the rectangle.

Complete step-by-step solution:
Let the length and width of the rectangle are $l$ and $b$ respectively.
Now the area of rectangle having length as $l$ and width as $b$ is given by
$A=lb$.
In the problem they have mentioned that the area of the rectangle is $16\text{ }i{{n}^{2}}$.
Now equating the both the values, then we will get
$lb=16$
Dividing the above equation with $l$ on both sides, then we will get the width of the rectangle as
$b=\dfrac{16}{l}$
Now the perimeter of the rectangle having length $l$ and width $b$ is given by
$p=2\left( l+b \right)$
Substituting the value of width $b=\dfrac{16}{l}$ in the above equation, then we will get
$p=2\left( l+\dfrac{16}{l} \right)$
Simplifying the above equation by using mathematical operations, then we will get
$p=2l+\dfrac{32}{l}$
To find the extremities of the above equation we are going to differentiate the above equation with respect to $l$ and equating it to zero, then we will get
$\dfrac{d}{dl}\left( 2l+\dfrac{32}{l} \right)=0$
Simplifying the above equation by using differentiation formulas, then we will have
$\begin{align}
  & \dfrac{d}{dl}\left( 2l \right)+\dfrac{d}{dl}\left( \dfrac{32}{l} \right)=0 \\
 & \Rightarrow 2\dfrac{dl}{dl}+32\dfrac{d}{dl}\left( \dfrac{1}{l} \right)=0 \\
\end{align}$
We have the differentiation formulas $\dfrac{d}{dx}\left( x \right)=1$, $\dfrac{d}{dx}\left( \dfrac{1}{x} \right)=-\dfrac{1}{{{x}^{2}}}$. Substituting these values in the above equation, then we will get
$2-\dfrac{32}{{{l}^{2}}}=0$
Simplifying the above equation by using basic mathematical operations to get the value of $l$, then we will get
$\begin{align}
  & 2{{l}^{2}}-32=0 \\
 & \Rightarrow 2{{l}^{2}}=32 \\
 & \Rightarrow {{l}^{2}}=\dfrac{32}{2} \\
\end{align}$
Applying square root on both sides of the above equation, then we will get
$\sqrt{l}=\sqrt{16}$
We have the value of $\sqrt{16}$ as $\pm 4$. Substituting this value in the above equation, then we will get
$l=\pm 4$
Here $l$ is the length which can not be in negatives hence the length of the rectangle is
$l=4$
Substituting this value in the equation $b=\dfrac{16}{l}$ to get the value of width, then we will get
$\begin{align}
  & b=\dfrac{16}{4} \\
 & \Rightarrow b=4 \\
\end{align}$
We have the length and width of the rectangle as $4\text{ in}$ and $4\text{ in}$ respectively.

Note: In this problem we have the length and width of the rectangle as same. You may think that only a square has the same length and width. But remember that a square is also a rectangle with the same length and width.