Question

A rectangular garden has a perimeter of $120ft$.

How do you find an equation for the area of the rectangle as a function of the width, then determine the length and width of the rectangle which provide the maximum area?

Answer 1

Finding the length and width of the given rectangular garden:

Step-1: Assumption:

Let the length and width of the rectangular garden be $lft$ and $wft$ respectively.

Step-2: Finding the relation between the length and the width of the rectangle using its perimeter

We know that the perimeter of a rectangle of length $l$ unit and width $w$ unit is $S=2\left(l+w\right)$ unit.

So, the perimeter of the given rectangular garden is $S=2\left(l+w\right)ft$.

Now, given that the perimeter is $120ft$.

So, we must get:

$$\begin{gathered} S=120 \\ \Rightarrow 2\left(l+w\right)=120 \\ \Rightarrow l+w=\frac{120}{2} \\ \Rightarrow l=60-w \end{gathered}$$

Step-3: Finding the area as a function of the width

We know that the area of a rectangle of length $l$ unit and width $w$ unit is $A=lw$ square unit.

Here, the width is $wft$ and the length is $(60-w)ft$.

So, the area will be: $w\left(60-w\right)=60w-w^{2}f{t}^2$.

Suppose, $f\left(w\right)=60w-w^2$.

This is required function of $w$ which calculates the area of the rectangle and it is the function to be maximized.

Step-4: Finding the critical points of $f\left(w\right)$

We know that the critical points of $f\left(w\right)$ will be the solution of $f\text{'}\left(w\right)=0$.

Now, differentiating $f\left(w\right)$ with respect to $w$, we get: $f\text{'}\left(w\right)=60-2w$.

Now,

$$\begin{gathered} f\text{'}\left(w\right)=0 \\ \Rightarrow 60-2w=0 \\ \Rightarrow 2w=60 \\ \Rightarrow w=\frac{60}{2} \\ \therefore w=30 \end{gathered}$$

Hence, $w=30$ is the critical point of $f\left(w\right)$.

Step-5: Checking if $w=30$ is a point of maximum of the function $f\left(w\right)$

We know that $w=a$ is a point of maximum of the function $f\left(w\right)$ if $f\text{'}\text{'}\left(a\right)<0$.

Now, differentiating $f\text{'}\left(w\right)$ with respect to $w$, we get: $f\text{'}\text{'}\left(w\right)=-2$.

Then, at $w=30$, we have: $f\text{'}\text{'}\left(30\right)=-2<0$.

Hence, $w=30$ is a point of maximum of the function $f\left(w\right)$.

Step-6: Finding the length and the width

Thus, the width of the rectangle for which the area will be maximum is $w=30ft$.

Now, we can find the length using the relation $l=60-w$. So, the length will be:

$$\begin{gathered} l=60-w \\ =60-30 \\ =30ft \end{gathered}$$

Therefore, the length and the width of the given rectangular garden are: $l=30ft$ and $w=30ft$ respectively i.e. in fact the garden will be square shaped.