Question

A rectangular area is to be enclosed by a wall on one side and fencing on the other three sides. If $18m$ of fencing are used, what is the maximum area that can be enclosed?

Answer 1

Finding the equation according to the given information:

Step 1: Write the relation between the length and width of the wall.

Considering $x$ is the length of the side opposite to the wall,

And $y$ is the width of the other two sides.

We are given;

$$\begin{gathered} x+2y=18 \\ \Rightarrow x=18-2y....\left(i\right) \end{gathered}$$

Step 2: Find the area of the rectangle.

The area is a rectangle $A=x\times y$,

Replace $x$ using the equation $\left(i\right)$:

$$\begin{gathered} \Rightarrow A=(18-2y)y \\ \Rightarrow A=-2y^2+18y \end{gathered}$$

Step 3: Find the maximum area

Taking the derivative of the above equation

$\frac{dA}{dy}=-4y+18$

To find the maximum area, set the derivative equal to zero,

$$\begin{gathered} \Rightarrow -4y+18=0 \\ \Rightarrow y=\frac{18}{4} \\ \Rightarrow y=\frac{9}{2}=4.5m \end{gathered}$$

Put the value of $y$into the equation $\left(i\right)$

$$\begin{gathered} \Rightarrow x=18-2\times 4.5 \\ \Rightarrow x=18-9 \\ \Rightarrow x=9m \end{gathered}$$

Therefore, the maximum area is $A=x\times y=4.5\times 9=40.5m^2$.

Hence, the maximum area that can be enclosed is $40.5m^2$.