Applications of Quadratic Functions

 

Learning Objective(s)

·         Apply quadratic functions to real world situations in order to solve problems.

 

 

Quadratic functions are more than algebraic curiosities—they are widely used in science, business, and engineering. The U-shape of a parabola can describe the trajectories of water jets in a fountain and a bouncing ball, or be incorporated into structures like the parabolic reflectors that form the base of satellite dishes and car headlights. Quadratic functions help forecast business profit and loss, plot the course of moving objects, and assist in determining minimum and maximum values. Most of the objects we use every day, from cars to clocks, would not exist if someone, somewhere hadn't applied quadratic functions to their design.

 

We commonly use quadratic equations in situations where two things are multiplied together and they both depend of the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, we use a quadratic equation. Because the quantity of a product sold often depends on the price, we sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge.

 

 

A very common and easy-to-understand application of a quadratic function is the trajectory followed by objects thrown upward at an angle. In these cases, the parabola represents the path of the ball (or rock, or arrow, or whatever is tossed).  If we plot distance on the x-axis and height on the y-axis, the distance of the throw will be the x value when y is zero. This value is one of the roots of a quadratic equation, or x-intercepts, of the parabola.  We know how to find the roots of a quadratic equation—by either factoring, completing the square, or by applying the quadratic formula.

 

Let’s look at a throw made by a shot-putter. Notice that x = 0 when the shot-putter has the shot (a heavy metal ball) in his hand—the shot hasn't gone anywhere yet. The shot-putter usually starts with the shot at his shoulder, so y (height) is not 0 when x = 0:

 

 

 

 

Although a stadium field of synthetic turf appears to be flat, its surface is actually shaped like a parabola. This is so that rainwater runs off to the sides. If we take a cross section of the turf, the surface can be modeled by , where x is the distance from the left end of the field and y is the height of the field. What is the width of the field?     A) 80 ft B) 1.5 ft C) 234 ft D) 160 ft   Show/Hide Answer A) Incorrect. The width of the field is the distance between the roots of the equation. Set the equation equal to 0:  –0.000234(x – 80)2 + 1.5 = 0. Subtracting 1.5 from both sides and then dividing both sides by -0.000234 gives (x – 80)2 = 6410 (approximately). Taking the square root of both sides gives x – 80 = ±80, so the roots are x = 0 and x = 160. The correct answer is 160 feet.   B) Incorrect. The width of the field is the distance between the roots of the equation. Set the equation equal to 0:  –0.000234(x – 80)2 + 1.5 = 0. Subtracting 1.5 from both sides and then dividing both sides by -0.000234 gives (x – 80)2 = 6410 (approximately). Taking the square root of both sides gives x – 80 = ±80, so the roots are x = 0 and x = 160. The correct answer is 160 feet.   C) Incorrect. The width of the field is the distance between the roots of the equation. Set the equation equal to 0:  –0.000234(x – 80)2 + 1.5 = 0. Subtracting 1.5 from both sides and then dividing both sides by -0.000234 gives (x – 80)2 = 6410 (approximately). Taking the square root of both sides gives x – 80 = ±80, so the roots are x = 0 and x = 160. The correct answer is 160 feet.   D) Correct. The width of the field is the distance between the roots of the equation. Set the equation equal to 0:  –0.000234(x – 80)2 + 1.5 = 0. Subtracting 1.5 from both sides and then dividing both sides by -0.000234 gives (x – 80)2 = 6410 (approximately). Taking the square root of both sides gives x – 80 = ±80, so the roots are x = 0 and x = 160. The correct answer is 160 feet.

 

 

Another common use of the quadratic equation in real world applications is to find maximum (the most or highest) or minimum (the least or lowest) values of something.  Recall that the vertex is the turning point of a parabola. For a parabola opening downward, the vertex is the high point, which occurs at the maximum possible y value. For a parabola opening upward, the vertex is the lowest point of the parabola, and occurs at the minimum y value.

 

To use a quadratic equation to find a maximum or minimum, we usually want to put the quadratic equation into the vertex form of a quadratic equation, . This lets us quickly identify the coordinates of the vertex (h, k).

 

Let's see how this works by trying a motion problem. The equation  is commonly used to model an object that is launched or thrown. The variable h represents the height in feet, and t represents the time in seconds. The other two values are usually given numbers: h₀ is the initial height in feet and v₀ is the initial velocity in feet/second.

 

When working with this equation, we assume the object is in “free fall”, which means it is moving under the sole influence of gravity.  There is no air resistance or any other interference of any kind (not so likely in the real world, but still, these equations are quite helpful).

 

 

We could have found the vertex by using other methods, such as by graphing or by using the formula  to find the x-coordinate of the vertex, then substitute that x-value into the original formula to get the y-value of the vertex).

 

A farmer has 1000 feet of fencing and a very big field. She can enclose a rectangular area with dimensions x ft and 500 – x ft. What is the largest rectangular area she can create?   A) 62,500 ft2 B) 250,000 ft2 C) 1,000 ft2 D) 500 ft2   Show/Hide Answer A) Correct. The area y is y = x(500 – x), or y = -x2 + 500x (in ft2). Completing the square to find the vertex form of this equation gives y = -(x – 250)2 + 62,500. The vertex of the parabola will be (250, 62,500). The maximum area possible is the y coordinate of the vertex, or 62,500 ft2.   B) Incorrect. The area y is y = x(500 – x), or y = -x2 + 500x (in ft2). When completing the square to find the vertex form of this equation, the binomial to be squared is x – 250, not x – 500. You probably forgot that the number is , not b. The vertex form is y = -(x – 250)2 + 62,500 and the vertex of the parabola will be (250, 62,500). The maximum area possible is the y coordinate of the vertex, or 62,500 ft2.   C) Incorrect. The perimeter of the rectangular area will be 1,000 feet, because that’s the amount of fencing you have. The area y is y = x(500 – x), or y = -x2 + 500x (in ft2). Completing the square to find the vertex form of this equation gives y = -(x – 250)2 + 62,500. The vertex of the parabola will be (250, 62,500). The maximum area possible is the y coordinate of the vertex, or 62,500 ft2.   D) Incorrect. The area y is y = x(500 – x), or y = -x2 + 500x (in ft2). You can read the maximum area using the vertex form of the equation, but this not the vertex form. The vertex form of a quadratic equation is y = a(x – h)2 + k. Completing the square to find the vertex form of this equation gives y = -(x – 250)2 + 62,500. The vertex of the parabola will be (250, 62,500). The maximum area possible is the y coordinate of the vertex, or 62,500 ft2.

 

Quadratic equations are sometimes used to model situations and relationships in business, science and medicine. A common use in business is to maximize profit, that is, the difference between the total revenue (money taken in) and the production costs (money spent).

 

The relationship between the cost of an item and the quantity sold is often linear. In other words, for each $1 increase in price there is a corresponding decrease in quantity sold. (Think about it: if the price of something goes up, do you buy more or less? Hopefully less!) Once we determine a relationship between the selling price of an item and the quantity sold, we can think about how to generate the most profit.  At what selling price do we make the most money?

 

The amount of profit will be found by taking the total revenue (the quantity sold multiplied by the selling price) and subtracting the cost to produce the all the items: Profit = Total Revenue – Production Costs.  We can integrate the linear relationship of selling price to quantity and the Profit formula and create a quadratic equation, which we can then maximize.  Let’s look at an example:

 

Here is a sample data set:

 

 

 

In order to calculate the profit, we also need to know how much it costs to produce each item. For this example, the cost to produce each item is $10.

 

 

Here is a graph of the profit function, showing the vertex:

 

 

Next is a word problem you may not think is a quadratic equation.  The area problem below does not include a quadratic formula of any type and the problem seems to be something you have solved many times before by simply multiplying.  But in order to solve it, you will need to use a quadratic equation.

 

 

 

 

You are provided with the following price vs. quantity information.  Write an equation that will represent yearly profit P for a selling price s. The production cost per item is $30.   Selling Price s Quantity Sold q 100 7000 200 6000 500 3000 600 2000 800 0   A) P = -10s + 8000 B) P = sq – 30q C) P = D) P =   Show/Hide Answer A) Incorrect. q = -10s + 8000, which is an important part of finding the profit, but you are not done. The revenue (money taken in) is sq, and the cost is 30q, so the profit is P = sq – 30q. Using the equation for q that you found, P = s(-10s + 8000) – 30(-10s + 8000). Multiplying and simplifying gives P = -10s2 +8300s – 240,000.   B) Incorrect. While it’s true that P = sq – 30q, this equation represents yearly profit for selling price s and quantity sold q. You are not done. The graph showing the relationship between q and s is a line with slope -10 and y-intercept 8000, so q = -10s + 8000. This means P = s(-10s + 8000) – 30(-10s + 8000). Multiplying and simplifying gives P = -10s2 +8300s – 240,000.   C) Correct. The revenue (money taken in) is sq, and the cost is 30q, so the profit is P = sq – 30q. The graph showing the relationship between q and s is a line with slope -10 and y-intercept 8000, so q = -10s + 8000. This means P = s(-10s + 8000) – 30(-10s + 8000). Multiplying and simplifying gives P = -10s2 +8300s – 240,000.   D) Incorrect. You used some important quantities in this problem, but incorrectly. The revenue (money taken in) is sq, and the cost is 30q, so the profit is P = sq – 30q. The graph showing the relationship between q and s is a line with slope -10 and y-intercept 8000, so q = -10s + 8000. This means P = s(-10s + 8000) – 30(-10s + 8000). Multiplying and simplifying gives P = -10s2 +8300s – 240,000.

 

 

Quadratic functions are used in many types of real world situations. They are useful in describing the trajectory of a ball, determining the height of a thrown object and in optimizing profit for businesses.  When solving a problem using a quadratic function it may be necessary to find the vertex or to describe a section of the parabola.