Chapter 1.1, Problem 61PS

Solution

To find: The maximum area that swimming section can have.

  $\frac{P^2}{8}$

Given information:

A rope of length $P$ which is to be used to rope off a rectangular swimming section.

The width is given as $w$ and length as $l$ .

Concept used:

Formula for the area of a rectangle is given as $\text{Area}=\text{length}\times \text{ width}$ .

Vertex form a quadratic function is given as $y=a{\left(x-h\right)}^2+k$ , where $\left(h,k\right)$ is the vertex of the quadratic function.

The vertex of the quadratic function represents the maximum/minimum point of the function.

If $a$ positive, the vertex is the minimum point and if $a$ is negative, the vertex is the maximum point of the function.

Calculation:

The perimeter of the swimming section is $l+2w$ .

Since the length of the rope is $P$ , it gives the equation, $P=l+2w$ .

Solve the equation $P=l+2w$ for $l$ to get, $l=P-2w$ .

Now, the area of the swimming section can be given as follows:

  $\begin{array}{c}\text{Area}=\text{length}\times \text{ width}\\ =lw\end{array}$

Substitute the value of $l$ and simplify.

  $\begin{array}{c}\text{Area}=lw\\ =\left(P-2w\right)w\\ =Pw-2w^2\\ =-2w^2+Pw\end{array}$

So, the area function is a quadratic function in terms of $w$ .

Rewrite the equation vertex form.

  $\begin{array}{c}\text{Area}=-2w^2+Pw\\ =-2\left(w^2-\frac{Pw}{2}\right)\\ =-2\left(w^2-2\cdot w\cdot \frac{P}{4}+{\left(\frac{P}{4}\right)}^2-{\left(\frac{P}{4}\right)}^2\right)\\ =-2\left(w^2-2\cdot w\cdot \frac{P}{4}+{\left(\frac{P}{4}\right)}^2\right)+2{\left(\frac{P}{4}\right)}^2\\ =-2{\left(w-\frac{P}{4}\right)}^2+2{\left(\frac{P}{4}\right)}^2\\ =-2{\left(w-\frac{P}{4}\right)}^2+\frac{P^2}{8}\end{array}$

Here, $a=-2$ , which is negative. So, the vertex is the maximum point of the function.

Therefore, the maximum value of the function $\text{Area}=-2w^2+Pw$ is $\frac{P^2}{8}$ .

That means, the maximum area that swimming section can have in terms of $P$ is $\frac{P^2}{8}$ .

Conclusion:

The maximum area that swimming section can have in terms of $P$ is $\frac{P^2}{8}$ .