Solving real problems with quadratics

Now let's use quadratic functions to model real things and solve problems. In this section, mostly a collection of worked examples, we'll divide problems into three loose classes. The key to solving them is to know the form and key points of a parabola (vertex, roots, y-intercept), and to figure out which you need to solve your problem.

  1. Problems for which some quadratic form is known or given

  2. Price or cost adjustment problems

  3. Area or volume optimization problems

Quadratic functions are very important. There's a lot to cover, so the material is broken up into sections.

  • This section on applications should be studied after working through:
  • Quadratic functions is the starting point for learning about polynomial functions with 2 as the largest power – quadratics.
  • Solving quadratic equations covers all methods of solving quadratic equations for x, i.e. finding roots.

Class 1 – Quadratic form given

In this type of problem, we are given a quadratic function that describes some phenomenon or property. Our job is to use its graph and equation to deduce properties and outcomes of the system it describes.

Example 1

The number of board feet (one board foot is 144 cubic inches) in a 16 ft. long tree can be approximated by the model $F(d) = 0.75 d^2 - 1.35 d - 9.4,$ where F is the number of board feet, and d is the diameter of the log. How many board feet are in a log with a diameter of 30 inches? Calculate the diameter that will produce the minimum number of board feet.

Solution: We ought to sketch a graph of this function to see how it works. First find the vertex. The x-coordinate is

$$\frac{-B}{2A} = \frac{1.3}{1.5} = 0.8667$$

The y-coordinate is just F(0.8667):

$$ \begin{align} f(0.8667) &= 0.75(0.8667)^2 \\[2pt] &\phantom{000} - 1.3(0.8667) - 9.4 \\[5pt] &= -9.9634 \end{align}$$

Now the roots. I'll use the handy calculator below to find those. You can use it for many of the problems on this page. The two calculated roots are

$$ \begin{align} r_1 &= 4.5529 \\[4pt] r_2 &= -2.7529 \end{align}$$

We won't be concerned with the negative root, because it represents a negative diameter. We can now sketch a graph of this function:

Now from the graph we can answer the second question first. Notice that the number of board feet is zero when the diameter is about 4.5 in. So our log needs to have a diameter > 4.5 in. in order for it to be of use.

The first question is just a simple matter of plugging 30 in. into our function and calculating the resulting number of board feet:

$$ \begin{align} F(30) &= 0.75(30)^2 \\[2pt] &\phantom{000} - 1.35(30) - 9.4 \\[5pt] &= 625.1 \; \text{ board feet} \end{align}$$

A quadratic solver

Enter A, B, C & hit [Calculate]
Root 1
Root 2

Here is a little quadratic solver tool that you can use to find the roots of any quadratic equation numerically. It might help you to solve these problems. It will calculate roots (real or imaginary), the vertex coordinates and the discriminant of any quadratic function.

Just enter the A, B and C constants and hit calculate.

Example 2

The drag on an automobile caused by air resistance (friction) increases quadratically with speed (in mi./h). The number of horsepower (a unit of power — energy divided by time) needed to overcome that drag is

$$H(s) = 0.005 s^2 + 0.008 s - 0.0035$$

for a certain make and model of car. How much power (in horsepower) is required to overcome wind drag if a car is traveling at 50 mi./h? at what speed will the car need to use 200 hp to overcome the wind drag?

Solution: The vertex of H(s) is (-0.8, -0.007). It's roots are $r_1 = 0.3576,$ and $r_2 = -1.9576.$   The graph looks like this:

For the first question, we just need H(50):

$$ \begin{align} H(50) &= 0.005(50)^2 \\[2pt] &\phantom{000} + 0.008(50) - 0.0035 \\[5pt] &= 12.9 \; \text{horsepower (hp)} \end{align}$$

For the second, we're solving $H(s) = 200:$

$$ \begin{align} 0.005 s^2 + 0.008 s - 0.0035 &= 200 \\[5pt] 0.005 s^2 + 0.008 s - 200.0035 &= 0 \end{align}$$

Using the calculator, our roots are

$$ \begin{align} r_1 &= 199.2 \; \text{mi./h} \\[5pt] r_2 &= -200.8 \end{align}$$

Notice that the second root gives us a negative speed, and we'll ignore that. This is common with problems involving quadratics like this. We need to restrict the domain to positive speeds.

So the car will have to travel at about 200 mi./h in order to expend 200 hp of power fighting wind resistance.

In the 1970s, under President Jimmy Carter, an engineer, US national speed limits were set to 55 mi/h. Beyond that speed, the amount of energy consumed by any car increases rapidly. Carter and the math were right, but in our culture, speed is valued, it seems, over energy consumption.

Example 3

The height of a projectile (like a thrown ball) launched at an angle of 45˚ to the ground is

$$h(t) = v_o t - \frac{1}{2} gt^2,$$

where $v_o$ is the initial velocity, t is the time in seconds (s) and $g = 9.8 \,\frac{m}{s^2}$ is the acceleration of gravity on Earth. If a ball is thrown upward at a 45˚ angle with an initial velocity of 30 m/s, how long will it be in the air?

Solution: With the help of our calculator, let's go ahead and graph this function, too. That amounts to sketching the arc of the flight of the ball.

$$h(t) = 30t - 4.9 t^2$$

The vertex is (3.061, 45.918), and the roots are t = 0 and t = 6.12 s. Here is the path:

The form of the graph gives us our solution. The right-hand root is at 6.12 s. That's how long the ball was in the air. We can also see that its maximum height was 45.9 m, and that half of the time of flight was going up, the other half down. That's because of the mirror symmetry of the quadratic function.

You can learn more about projectiles in the projectile motion section.

Class 2 – Price adjustment problems

In this class of problems, we have an amount (such as a sale price or a manufacturing cost), and the number of items sold or made. We generally have some research that suggests that if a price were reduced/increased, the income would decrease/increase. We seek to find the optimal pricing to produce the highest sales or lowest cost.

Our independent variable will be the number of increases or decreases in price of a certain amount, and we'll use it in a cost or price function that we'll write.

Example 4

A manufacturer produces widgets in a factory. She sells 1000 widgets per week at a cost of \$20.00 each. Some marketing research suggests that the price is too low, and that for each \$1 increase in price, the number of sales will drop by 20 sales. That is, if the price is increased to \$21.00, she should expect sales to drop to 980 units. Calculate the best price for a widget – the one that yields the most profit.

Solution: Our sales function will be then cost per item multiplied by the number of items sold.

$$ \begin{align} S(x) &= (20 + 1x)(1000 - 20x) \\[5pt] &= -20x^2 + 600x + 20,000 \end{align}$$

We can calculate the vertex and roots to get a picture of this function. The vertex is (15, 24,500) and the roots are x = -20, 50.

Now it's easy to see that this function has a maximum, and that it represents the maximum number of sales dollars. Here's how we interpret the results. The value of x at the maximum is 15, which represents 15 \$1.00 increases in price, to a total price of \$20 + \$15 to \$35. That means the total number of sales will drop to 1000 - 15(20) = 700 sales.

The original sales total was \$20,000, and the optimal value is \$24,500. Now of course, this is a pretty simplistic view of sales and marketing. Still, it does suggest that for this business, some increase in price, perhaps one that increases sales without sacrificing too many customers, might be in order.

Example 5

If a car dealership sets the price of their cars at \$28,000, they will sell 54 cars in a month. Every time they drop the price by \$1000, as during a sale, two more cars are sold. Calculate the price of cars that will maximize every-day sales revenue (revenue is another word for income).

Solution: We'll let x be the number of \$1000 price drops, and write a revenue function that looks like this:

$$R(x) = (28,000 - $1000x)(54 + 2x),$$

where revenue is (dollars per car) × (number of cars). Our quadratic function is

$$ \begin{align} R(x) = -2000x^2 &- 26,000x \\[5pt] &+ 1,512,000 \end{align}$$

Now once you're familiar with this kind of problem, you'll realize that you don't really need to graph the function.

It's a downward-opening parabola, and we seek its maximum, which is its vertex. The x-coordinate is

$$\frac{-B}{2A} = \frac{26,000}{-4,000} = -6.5.$$

The fact that our optimal x is negative suggests that we ought to raise the prices rather than lower them. A raise of $6.5 × \$1000$ leads to an average car price of $\$28,000 + \$6500 = \$34,500.$

Class 3 – Optimizing area/volume

When we use the word "optimize," we mean to find something that's the greatest value over some range or the least, depending on the situation. Those might be things like greatest area surrounded by a fixed perimeter, or least cost of an item. If the system is modeled by a quadratic function, then we're always talking about the location of the vertex. Here are a couple of examples.

Example 6

Let's say you have a 500 ft. roll of fencing material and a large field. You want to construct a rectangular playground area. Calculate the dimensions of the largest such play yard by finding the maximum of the appropriate quadratic function.

Solution: It's always best to begin a problem like this by sketching a simple diagram.

Now we know two things about this area. First, its perimeter is 500 ft.,

$$2x + 2y = 500.$$

Second, the area is $A = xy.$

It is the area function that we want to optimize (find the maximum of). The problem is that it's a function of two variables, x and y. We'll use the perimeter to solve for one (say y) in terms of the other to make the area function A(x) – a function of one variable:

$$ \begin{align} 2y + 2x &= 500 \\[5pt] 2y &= 500 - 2x \\[5pt] y &= 250 - x \end{align}$$

So our area function is

$$ \begin{align} A(x) &= x(250 - x) \\[5pt] &= -x^2 + 250x \end{align}$$

Now this is a downward-opening parabola, so it has a maximum value at its vertex. The x-coordinate is:

$$\frac{-B}{2A} = \frac{-250}{-2} = 125$$

The y-coordinate can be found from our perimeter equation:

$$y = 250 - 125 = 125$$

So our optimal fenced area is a square. What we've calculated here is that the largest-possible area that can be surrounded by a rectangular perimeter is a square. That's not surprising.

Example 7

A 500 ft. roll of fencing material is available to form the largest pen (in terms of area) possible. One side of the pen is to be up against the side of a large building, so no fence is needed there. Calculate the dimensions of the largest pen that can be made in this way.

Solution: Here is the picture of this problem:

Penning this area involves only three sides of fence. The perimeter is

$$2x + y = 500.$$

As in the example above, the area is $A = xy,$ and that's the function we want to maximize. First we have to turn A(x, y) into A(x) by solving for y in the perimeter equation and using the result to eliminate y from A(x, y).

$$ \begin{align} y + 2x &= 500 \\[5pt] y &= 500 - 2x \end{align}$$

So our area function is

$$ \begin{align} A(x) &= x(500 - 2x) \\[5pt] &= -2x^2 + 500x \end{align}$$

Now this is a downward-opening parabola, so it has a maximum value at its vertex. The x-coordinate is:

$$\frac{-B}{2A} = \frac{-500}{-4} = 125$$

The y-coordinate can be found from our perimeter equation:

$$y = 500 - 2(125) = 250$$

This time our optimal (maximal) area is a rectangle:

Example 8

An athletic field with a perimeter of 400 m consists of a rectangle with semicircular ends, as shown. Find the dimensions of the oval that yield the greatest-possible rectangular field.

Solution: Here's a diagram of the field with the critical dimensions:

Now the perimeter of the field is the distance around the oval. It consists of two segments of length x, and the perimeter of a circle of radius r, p = 2πr (the circle is split, one side on each end). So the perimeter is

$$2x + 2\pi r = 400$$

The area of the rectangular field, what we're trying to maximize, is

$$A = 2rx$$

If we solve the perimeter equation for r, we get

$$r = \frac{200 - x}{\pi}$$

By substituting that value into the area, we can eliminate r:

$$A(x) = -\frac{2}{\pi}x^2 + \frac{400}{\pi}x$$

That's the equation of a parabola that opens downward, and therefore has a maximum (its vertex). The x-coordinate of the maximum is

$$\frac{-B}{2A} = \frac{-400/\pi}{-4/\pi} = 100$$

Now we can find r using the perimeter formula

$$r = \frac{200 - 100}{\pi} \approx 32 \; m$$

Here's a new diagram with the dimensions:

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