Optimization is one of the most common and important uses of calculus as it offers us concrete evidence to make well-informed decisions.

Using calculus to identify a function's extreme values has many practical applications to real life.

For example, suppose you wanted to build a wooden deck in your backyard but had a limited budget for purchasing the materials.

Calculus can help you determine the best dimensions for the deck that will allow you to stay on budget while maximizing the area of your deck space.

In This Module

• We will work through a variety of optimization problems to see the methods we can use to solve them.

At a local lake, a lifeguard has \(500\) meters of buoyant rope to mark the boundary lines of a rectangular swimming area. She does not need a boundary line along the side of the swimming area where there is beach. What are the dimensions of the largest swimming area that this lifeguard can make?

Solution

The problem tells us that the length of the three sides of the rectangular swimming area totals \(500\) m.

We will use the formula \(A=lw\) to calculate the area of the rectangle.

We begin by sketching a diagram to match this problem.

As the lifeguard changes the length and width of the rectangular boundary, the swimming area changes in size.

So, let's label the length, \(l\), and width, \(w\), of the swimming area on the diagram.

The lifeguard has a limited amount of rope. We know that

At a local lake, a lifeguard has \(500\) meters of buoyant rope to mark the boundary lines of a rectangular swimming area. She does not need a boundary line along the side of the swimming area where there is beach. What are the dimensions of the largest swimming area that this lifeguard can make?

Solution

\(l+2w=500\), which implies that \(l=500-2w\).

Note: We assume that both \(l\) and \(w\) are non-negative, so to make a swimming area with this rope, \(0\leq w\leq250\) and \(0\leq l\leq500\).

Since the lifeguard wants to maximize the swimming area, \(A\), we will use the derivative to determine the maximum value of the function

Before differentiating, we need to express the area, \(A\), in terms of only one variable.

This can be achieved by substituting \(l=500-2w\) into \(A=lw\) giving a function for the area \(A(w)\) in terms of the width \(w\).

At a local lake, a lifeguard has \(500\) meters of buoyant rope to mark the boundary lines of a rectangular swimming area. She does not need a boundary line along the side of the swimming area where there is beach. What are the dimensions of the largest swimming area that this lifeguard can make?

Solution

Since \(A(w)\) is a polynomial, which tells us that \(A'(w)\) exists for all \(w\), the extreme value theorem tells us that a maximum and/or minimum value will occur when \(A'(w)=0\) or at the endpoints of the interval \(0\leq w\leq250\).

Differentiating the area function, we have

Let's now solve for \(w\) when \(A'(w)=0\).

At a local lake, a lifeguard has \(500\) meters of buoyant rope to mark the boundary lines of a rectangular swimming area. She does not need a boundary line along the side of the swimming area where there is beach. What are the dimensions of the largest swimming area that this lifeguard can make?

Solution

The maximum swimming area occurs when \(w=125\) m.

We can find the length, \(l\), of the swimming area by substituting \(w=125\) m into the function \(l=500-2w.\)

Thus, the rectangular swimming area's dimensions should be \(125\) meters wide and \(250\) meters long to have an optimal area.