Quadratic Functions: Apply It

Lumen Learning, OpenStax

Quadratic Functions: Apply It

Finding the Maximum and Minimum Value of a Quadratic Function

There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.

The vertex of a parabola is the highest (maximum) or lowest (minimum) point, depending on the direction the parabola opens. Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).

Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).

A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased [latex]80[/latex] feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side. Diagram of the garden and the backyard.

Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length [latex]L[/latex]. Then, use the formula to answer: What dimensions should she make her garden to maximize the enclosed area?

Let’s use a diagram to record the given information.Let [latex]L[/latex] represents the length of the fence and [latex]W[/latex] represents the width of the fence.We know we have only [latex]80[/latex] feet of fence available, and [latex]L+W+L=80[/latex], or more simply, [latex]2L+W=80[/latex].

This allows us to represent the width, [latex]W[/latex], in terms of [latex]L[/latex].

[latex]W=80 - 2L[/latex]

Now, we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width.

[latex]A=L \cdot W=L \cdot (80 - 2L)=80L - 2{L}^{2}[/latex].

This formula represents the area of the fence in terms of the variable length [latex]L[/latex].

Thus, the function, written in general form, is[latex]A\left(L\right)=-2{L}^{2}+80L[/latex].

Now, to answer the question: What dimensions should she make her garden to maximize the enclosed area?

The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area.

Based on the function above, we have [latex]a=-2,b=80[/latex], and [latex]c=0[/latex].

Vertex:

[latex]h= -\dfrac{b}{2a} = -\dfrac{80}{2(-2)} = -\dfrac{80}{-4} = 20[/latex]

and

[latex]\begin{align}k&=A\left(20\right) \\&=80\left(20\right)-2{\left(20\right)}^{2}\\&=800 \end{align}[/latex]

Thus, the vertex is [latex](20, 800)[/latex].

Graph of the parabolic function A(L)=-2L^2+80L, which the x-axis is labeled Length (L) and the y-axis is labeled Area (A). The vertex is at (20, 800). This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function below.

Interpretation of the vertex: The maximum value of the function is an area of [latex]800[/latex] square feet, which occurs when [latex]L=20[/latex] feet. When the shorter sides are [latex]L = 20[/latex] feet, there is [latex]W = 80-2(20) = 40[/latex] feet of fencing left for the longer side.

So, to maximize the area, she should enclose the garden so the two shorter sides have length [latex]20[/latex] feet and the longer side parallel to the existing fence has length [latex]40[/latex] feet.

The problem we solved above is called a constrained optimization problem. We can optimize our desired outcome given a constraint, which in this case was a limited amount of fencing materials.

Finding Maximum Revenue

Quadratic functions aren’t just abstract math concepts—they’re super useful in real-life situations, especially in business! One cool way we can use quadratic functions is to figure out how to maximize revenue.

Think of it this way: if you’re running a business and selling a product, you want to find the best price to sell that product so you make the most money. Revenue (the money you bring in) is calculated by multiplying the price per unit by the number of units sold. But here’s the catch—the number of units sold usually changes with the price, creating a quadratic relationship.

Let’s see how to use quadratic functions to find that sweet spot—the price that brings in the most revenue! By looking at the graph of the quadratic function, especially the vertex, we can figure out the optimal price to charge.

How To: Given an application involving revenue, use a quadratic equation to find the maximum.

Write a quadratic equation for revenue.
Find the vertex of the quadratic equation.
Determine the [latex]y[/latex]-value of the vertex.

The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease.

A local newspaper currently has [latex]84,000[/latex] subscribers at a quarterly charge of [latex]\$30[/latex]. Market research has suggested that if the owners raise the price to [latex]\$32[/latex], they would lose [latex]5,000[/latex] subscribers.

Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?

Revenue (the money you bring in) is calculated by multiplying the price per unit by the number of units sold.

In this scenario, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity.

Let’s introduce variables: Let [latex]p[/latex] for price per subscription and [latex]Q[/latex] for quantity, giving us the equation [latex]\text{Revenue}=pQ[/latex].

Subscriptions are linearly related to the price. Let’s find the linear function represent the subscription.

Slope

We know that currently [latex]p=30[/latex] and [latex]Q=84,000[/latex]. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, [latex]p=32[/latex] and [latex]Q=79,000[/latex]. From this we can find a linear equation relating the two quantities. The slope will be

[latex]\begin{align}m&=\dfrac{79,000 - 84,000}{32 - 30} \\ &=\dfrac{-5,000}{2} \\ &=-2,500 \end{align}[/latex]

This tells us the paper will lose 2,500 subscribers for each dollar they raise the price.

[latex]y[/latex]-intercept

[latex]\begin{align}&Q=-2500p+b &&\text{Substitute in the point }Q=84,000\text{ and }p=30 \\ &84,000=-2500\left(30\right)+b &&\text{Solve for }b \\ &b=159,000 \end{align}[/latex]

This gives us the linear equation [latex]Q=-2,500p+159,000[/latex] relating cost and subscribers.

Back to the revenue equation.

[latex]\begin{align}&\text{Revenue}=pQ \\ &\text{Revenue}=p\left(-2,500p+159,000\right) \\ &\text{Revenue}=-2,500{p}^{2}+159,000p \end{align}[/latex]

Revenue Function: [latex]R(p) = -2,500{p}^{2}+159,000p[/latex]

We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.

Graph of the parabolic function which the x-axis is labeled Price (p) and the y-axis is labeled Revenue ($). The vertex is at (31.80, 258100).

[latex]\begin{align}h&=-\dfrac{159,000}{2\left(-2,500\right)} \\ &=31.8 \end{align}[/latex]

The model tells us that the maximum revenue will occur if the newspaper charges [latex]\$31.80[/latex] for a subscription.

To find what the maximum revenue is, we evaluate the revenue function.

[latex]\begin{align}\text{maximum revenue}&=-2,500{\left(31.8\right)}^{2}+159,000\left(31.8\right) \\ &=\$2,528,100\hfill \end{align}[/latex]

The price should the newspaper charge for a quarterly subscription is [latex]\$31.80[/latex] to obtain the maximum revenue of [latex]\$2,528,100[/latex].

In the example above, we knew the number of subscribers to a newspaper and used that information to find the optimal price for each subscription. What if the price of subscriptions is affected by competition?

Previously, we found a quadratic function that modeled revenue as a function of price.

[latex]\text{Revenue}-2,500{p}^{2}+159,000p[/latex]

We found that selling the paper at [latex]\$31.80[/latex] per subscription would maximize revenue. What if your closest competitor sells their paper for [latex]\$25.00[/latex] per subscription? What is the maximum revenue you can make you sell your paper for the same?