Quadratic Applications

This section covers:

Quadratic applications are very helpful in solving several types of word problems (other than the bouquet throwing problem), especially where optimization is involved.  Again, we can use the vertex to find the maximum or the minimum values, and roots to find solutions to quadratics.

Note that we did a Quadratic Inequality Real World Example here.

Quadratic Projectile Problem:

Jennifer hit a golf ball from the ground and it followed the projectile Golf Ball Trajectory, where t is the time in seconds, and h is the height of the ball.   Find the highest point that her golf ball reached and also when it hits the ground again.  Find a reasonable domain and range for this situation.


Note that in this example, we are using the generic equation  \(h\left( t \right)=-16{{t}^{2}}+{{v}_{0}}t+{{h}_{0}}\),  where, in simplistic terms, the –16 is the gravity (in feet per seconds per seconds), the  \({{v}_{0}}\)   is the initial velocity (in feet per seconds) and the  \({{h}_{0}}\)   is the initial height (in feet).  (If units are in meters, the gravity is –4.9 meters per second per second).

Since we need to find the highest point of the ball, we need to get the vertex of the parabola.  We also need to know when the height h is back to 0 again.  Then we can use these two values to find a reasonable domain and range:

Path of Ball Quadratic Problem

(We will discuss projectile motion using parametric equations here in the Parametric Equations section.)

Note that the independent variable represents time, not  distance; sometimes parabolas represent the distance on the x axis and the height on the y axis, and the shapes are similar.   Height versus distance would be the path or trajectory of the bouquet, as in the following problem.

Quadratics Trajectory (Path) Problem:

Audrey throws a ball in the air, and the path the ball makes is modeled by the parabola ,  measured in feet .   What is the maximum height the ball reaches, and how far (horizontally) from Audrey does is the ball at its maximum height?    How far does the ball travel before it hits the ground?


Note that in this problem, the x axis is measuring the horizontal distance of the path of the ball, not the time, so when we draw the parabola, it’s a true indication of the trajectory or path of the ball.

Note also that the equation given is in vertex form (if we add 8 to each side), so we can easily get the maximum x and y values.

To find out how far the ball travels before it hits the ground, we want to make the y value 0, so we’ll have  .   To solve this, we should not expand the square out, but solve using the square root method; this Is much easier.

So let’s solve this problem algebraically:

Quadratic Word Problem Finding the Path of a Ball

Optimization of Area Problem:

Let’s say we are building a cute little rectangular rose garden against the back of our house with a fence around it, but we only have 120 feet of fencing available.  What would be the dimensions (length and width) of the garden (with one side attached to the house) to make the area of the garden as large as possible?  Also, what is this area?  Also, what is a reasonable domain for the width of the garden?


Ok, like we did before, let’s set the variables to what the problem is asking.   And, as always, let’s draw a picture first when we have any sort of “geometry” word problem.

So let w equal the width of the garden, and now let’s draw a picture, using the fact that the perimeter is 120.  We could have also made the variable the length of the garden, depending on how we draw the picture, but since we have the width on two sides, I thought it might be easier to make the width “w”:

Picture of Rose Garden for Optimization

Now, we want to maximize the area, and (from Geometry), we know that the area (the y) is width times length, or w(120 – 2w).   We want to maximize the area, and when we learned about Quadratics, we learned that the vertex is the maximum y point, given an x point, (in our case, a w point).  Area depends on length and width – which makes sense.

So let’s get the vertex for  both algebraically and using a Graphing Calculator:

Solving Optimization Problem Using Quadratics

Now, to get the reasonable domain or appropriate domain, we have to think about the values that the width could ever be – to make that garden at all.  We know the width has to be positive, which means it has to be greater than zero.  But we also know that there is a minus sign in one of the expressions containing the width (the length, which is “120 – 2w”), and this also must be positive.  Since 120 – 2w > 0,  w < 60.  So a reasonable domain for the width is 0 to 60 feet or, in interval notation, (0, 60).

In this example, if we wanted to find the reasonable range for the area, we would look at the graph, and see that it is (0, 1800).  So typically a reasonable range for these types of problems is 0 to the y portion of the vertex.

Maximum Profit Problem:

The profit from selling local ballet tickets depends on the ticket price.  Using past receipts, we find that the profit can be modeled by the function , where x is the price of each ticket.  We want to find the ticket price that gives the maximum profit, and also find that maximum profit.


This problem is actually much easier since we are given the formula for the profit, given the price of each ticket.

We simply either graph the function to get the vertex, or use (, plug  into the x to get the y) to find the coordinates of the vertex: Using negative b over 2 a to find Vertex.

Since the vertex is (20, 6060), the ticket price should be $20 to maximize profit, and that maximum profit is $6060.

Maximum Revenue Problem:

A popular designer purse sells for $500 and 45,000 are sold a month.  The company did some research and realized that for each $20 decrease in price, they can sell 5000 more purses per month.  How much should the company charge for the purse so they can maximize monthly revenues?


This problem is a little trickier since we can’t really tell from the question what we should set the variables equal to.

But we can think of monthly revenue (what a company makes each month) as “price times number of purses sold”.   For example, without making any changes, the monthly revenue is $500 times 45,000 = $22,500,000 (a lot for purses, but definitely worth it!)

But, with the new information, we know that for each $20 decrease in price, they can sell an increase of 5000 purses per month.  For example, if they sell the purses for $500 – 20 = $480, they would sell 45,000 + 5000 = 50,000 purses, for a monthly revenue of $24,000,000, which is more!

So instead of doing all this by hand to find out what we should do to maximize the monthly revenues, we can use algebra to find the maximum monthly revenue by letting x = the number of $20 decreases (and hence sales of 5000 more purses) per month.  Then we can find the maximum of our quadratic to get our answers.

Here is our equation:

Equation to Maximize Profit We want to find the vertex (maximum) of this quadratic equation, which we can get from a graphing calculator (8, 28900000).  So 8 is the number of $20 decreases the company can charge (and number of “5000 more purses” batches the company can sell) a month.  So this means to get the maximum revenues, the company should sell their purses at Calculate Price of Purses, and thus they should be able to sell 45,000 + 5000(8) or 85,000 more per month for a maximum profit of $28,900,000.

Bunny Rabbit Population Problem:

The observed bunny rabbit population on an island is given by the function , where is the time in months since they began observing the rabbits.  (a) When is the maximum population attained, (b) what is the maximum population, and (c) when does the bunny rabbit population disappear from the island?


This one isn’t too bad, since we are given the equation.  The last part of the question is a little different, though, so I’ll go through the graphing calculator steps for this.

(a) and (b) involved finding the vertex again, which will be a maximum.   Part (a) involves getting the x part of the vertex, since the “x” here is “t”, which is time.  Part (b), the maximum population, will be the y part of the vertex, since the “y” here is “p”, which is population.

So, from the graphing calculator, we see that the vertex is (162.5, 11762.5).  So the maximum rabbit population was 11762 rabbits (we can’t have half of a rabbit!) when it was 162.5 months after they began observing the rabbit population.  This answers (a) and (b) above.

For (c), we need to see when the graph goes back down to 0; this is when there are no rabbits left on the island.  Let’s review how to do this with the graphing calculator:

Finding Zeros for Quadratics Rabbit Problem So, to answer (c) above, the rabbit population will disappear from the island at around 334 months from when the observations started.

Linear Increase/Decrease Problem:

OK, use your imaginations on this one (sorry!):

Taylor and Miranda are performing on a magic dimension-changing stage that is 20 yards long by 15 yards wide.  The length is decreasing linearly (with time) at a rate of 2 yards per hour, and the width is increasing linearly (with time) at a rate of 3 yards per hour.  When will the stage have the maximum area, and when will the stage disappear (have an area of 0 square yards)?  (Better get off that stage, Taylor and Miranda!)


This one’s a little trickier, since we are asking when the stage will be the greatest area, and when it will have an area of 0, yet we are only given distances and rates.

But since we’re finding areas, we need to work with distances only.  And we know that Distance = Rate  x Time.   So do you see how at time t, the length of the stage is (20 – 2t) and the width is (15 + 3t)?  Think about it:  after one hour, the length of the stage will have decreased by 2 yards, and the width will have increased by 3 yards, so the new stage will be 18 by 18 yards.  After two hours, the length will be 16 yards, and the width will be 21 yards, and so on.

Now, let’s find the answers, with and without using the graphing calculator:

Solve for Maximum Area

Now we need to find when the stage will have no area left.  We need to set the equation to 0, or find the rightmost root with the calculator:

Finding Zeros for Quadratic Problem

Pythagorean Theorem Quadratic Application:

OK, here’s one where you’ll use a bit of Geometry.  You probably learned the Pythagorean Theorem awhile back – it’s the one with the right triangle and all the squares in it!

Here is the type of problem you may get:

The hypotenuse of a right triangle is 4 inches longer than one leg and 2 inches longer than the other.  Find the dimensions of the triangle.  Also, find a reasonable domain for the hypotenuse.


This really isn’t an optimization problem, but we’ll see how easy it is to solve with quadratics.  Let’s draw a picture first and then use the Pythagorean Theorem:

Pythagorean Theorem Quadratic Problem

To get the reasonable domain for the hypotenuse, we know it has to be greater than 0, and since we have minus signs in the expressions for the legs, we have to look at those, too.  Both of the legs must have values that are positive, so “x – 2″, and “x – 4″ both must be positive.  So x has to be greater than 4 (do you see why?).  So the reasonable domain for the hypotenuse is x > 4, or 4 to Infinity.

Quadratic Inequality Problem:

You may encounter a problem like this – which is really not too difficult.

You have to remember that the value of the discriminant of quadratics in standard form a x squared plus b x plus c has three possibilities:

  1. Discriminant Equals Zero means there is only one real solution.
  2.  means there are two real solutions.
  3. Discriminant Less Than Zero means there are no real solutions, only imaginary solutions.

Here is the problem:

Given Solve for b in Quadratic, find the values of b so there are no real solutions to the quadratic.


We see that a = 2 and c = 8.  So we need to find b such that the discriminant is  <  0:

Quadratic Inequality Problem So b would have to be between –8 and 8 (but can’t include –8 or 8) so there are no real solutions to .  Try some numbers for b to convince yourself that this is correct!

Finding a Quadratic Equation from Points or a Graph

Emmy throws a dog toy up in the air from 5 feet above the ground.  When the toy is 2 feet from the her, the toy reaches a maximum height of 9 feet, and then lands back on the ground 5 feet from her.  Find the “a” (the coefficient of the \({{x}^{2}}\)) for the parabola of the flight of the toy, and write this quadratic equation in vertex formstandard form, and factored form.

Learn these rules, practice, practice, practice, and you’ll rock at math!

On to Solving Absolute Value Equations and Inequalities – you’re ready!

44 thoughts on “Quadratic Applications

  1. William and Stephanie are playing a rather dangerous game of catch on the rooftops of skyscrapers in downtown Chicago. William is preparing to throw a tennis ball off the roof of the John Hancock Center,1130 feet above street level Stephanie is waiting to catch the ball on the rooftop of the neighboring Water Tower Place,830 feet above street level. William throws the ball,which leaves his hand with an upward velocity of 46 feet per second, and travels in a parabolic path. Stephanie catches the ball exactly 6.0 seconds after it is thrown.
    complete the following tasks:
    1-Create a quadratic function describing the height of the ball above street level with respect to time t.
    2- After how many seconds will the ball reach its maximum height above street level?
    3- Find the maximum height of the ball above street level .Round to the nearest foot.
    4- Stephanie uses a machine to launch the ball back up to William ,who catches it exactly 6.0 seconds after it is launched. What is the upward velocity in feet per second of the ball as it leaves the launcher? thank you so much.

  2. The path of a rock thrown vertically upward from the surface of the moon can be modeled by the equation: h(t)=-0.8t^2 +24t. H(t) represents the height in feet and t represents the time in seconds,

    1.how long did it take the rock to reach its highest point?
    2.how high did the rock go?
    When did the rock reach half its maximum height?
    4. When did the rock hit the ground?

    • Thanks for writing! For 1, I put the equation in the calculator, and got the maximum is at 15 seconds (the x) (or you can use -b/2a for x). For 2, the maximum at 15 seconds is at y = 180, so the rock went up to 180 feet. It reached half of its maximum height when y = 90; this is when x = 4.39 seconds, 4 – the rock hit the ground when y = 0, or at x = 30 seconds.
      Hope this helps! Lisa

    • Hi!
      Thanks for writing! Yes, feel free to use examples, but as long as you state where it came from (shelovesmath.com) and no one presents it as their own work.
      Let me know if you have any suggestions to make it better 🙂

  3. A suspended bridge has two cables secured at either end of the span by two supporting towers. The cables are attached to the tops of the towers. In the section between the two towers, the cables form a parablic curve. At their lowest point, the cables are 15 feet from the surface of the bridge. The towers are 400 feet apart, and the vertical distance from the surface of the bridge to the top of each tower is 415 feet. What is the quadratic erquation that describes the curve of the cables between the towers? Use the lowest point as the y-intercept.

  4. My teacher said that we need to copy some problem solving for quadratic inequalities … I need 2 .. pls. help me

  5. This problem is tough for me….

    The revenue (in dollars) for the sale of x units of a new product is given by the function R(x)=200x-0.2x^. The cost of producing x units of the new product is given by C(x)=24x+21900. Assume the company can produce between 0 and 1000 units of the new product. Find the number of units that need to be sold to produce a profit of at least $15,200

    • Thanks for writing! Since profit = revenues – cost, we have profit = 200x-.2x^2 – (24x+21900). I put that in the graphing calculator and found the maximum point (vertex) for x between 0 and 1000 units, and y over 15200. I get 440 units with a profit of $16820. Does that make sense? Lisa

      • Kind of, but my professor doesn’t allow calculators, do you think you could do a step by step? My professor has this on every test but skips steps while explaining it on the board and I keep getting it wrong. Thanks for all of your help!

        • Sure! The quadratic is -.2x^2 + 176x – 21900 after you do the subtraction. Then use -b/2a to find the vertex “x” part: -176/-.4 = 440. Then plug this in to get the “y” part. Does that make sense? Lisa

          • YES!!! 🙂

            Thank you so much for all your help!!! This has become one of my favorite help sites, I recommend it to everyone!!


  6. Annual box office revenue presented by
    B (x)=-0.19x^2+1.2x+7.6

    Where x is years after 2000
    0 <(or)=x <(or)=7
    What year was revenue at its highest?

    I used the opposite of b over 2a=x and get 12.63
    Then put 12.6 (the vertex) back into the equation and get -7.444
    The book tells me the answer is 2003 but I don't see where to go to get 2003.

    Thank you in advance.

    • Thanks for writing! For the maximum x value, I get -b/2a = -1.2(2*.19) = about 3.158. So it will be a little after 3 years after 2000, which would be 2003. Does that make sense? Lisa

  7. The profits of Mr. Unlucky’s company can be represented by the equation y= -3x^2+18x-4 the variable y is the amount of profit in millions of dollars and x is the number of years of operation. He realizes his company is on the downturn and wishes to sell before he ends up in debt.

    When will Unlucky’s business show the maximum profit?
    What is the maximum profit?
    At what time will it be too late to sell his business? (When will he start losing money?)

    • Thanks for writing! To find the maximum profit and when it occurs, we’ll have to find the vertex of the quadratic. We can either use a graphing calculator or use (-b/2a, f(-b/2a)) in y = ax^2 + bx + c.

      I got the vertex is (3, 23), which means the maximum profit will occur 3 years of operation and that profit will be $23M.
      He’ll start losing money when y = 0, which is in about 5.77 years.
      Does that make sense?

    • Thanks for writing! (And I assume you meant 255, not 225?) I’d do it like this: n(n+2)=255, n^2 + 2n – 255 = 0, (n – 15) (n + 17) = 0, n = 15 and n = -17. So the two numbers could be 15 and 17 , or -15 and -17. I hope this helps! Lisa

  8. Hi! I was just wondering in the “Finding a Quadratic Equation from Points or a Graph” part:
    Why in the vertex form, (0,5), when you substitute it to ‘a’ or aka ‘x’ and ‘y’; y = 0 and x = 5? 🙂

  9. I need help! We need to make a real life application problem using quadratic equations, the problem I made is:

    Alexander works at a shipping line. He had to deliver 25 boxes containing different books for a school’s library, the area of each box is 3721 cm2 and its side measures 61 cm. If the area of the delivery truck’s container that he would use is 15 m2 and its length is one more than twice the width, Will the 25 boxes fit the 15 m2 delivery truck? If not, how many times would it take for Alexander to deliver it all?

    Well, This is just a simple problem involving Quadratic equations. What changes do you think I should make? Is it fine already?

    • Thanks for writing! I don’t think you can do this problem using areas – you have to use volumes. It still might be a quadratic problem though. Have you already come up with a problem? Lisa

  10. I need help!! Figuring out your Maximum Revenue Problem (above)

    How did u get (8,28900000) for ur vertex in ur graphing calculator ?? I am very stuck ??

  11. I’m wondering if the last example you have with the dog toy is actually incorrect. I’m an algebra 2 teacher and used this problem (by the way thank you for these, they are wonderful examples). I always taught it exactly as you showed; however, today a student tried to use the projectile motion equation to figure it out. I challenged them to try it and see if they get the same equation you have in vertex form. Now I’m realizing…the points in this problem would never actually occur with gravity on earth….there’s no way it could take 2 seconds to travel only 4 inches AND also have that be the max height. It might have a really huge initial velocity and shoot up really high so that after 2 seconds it’s only 4 inches above the starting height, but (2, 9) can not be the vertex if this is happening on Earth. ….this just took 1/2 of our math department to figure out during our lunch break. Whew!

    • Oh gosh, thanks so much for noticing this! You are right – I never thought of that! I changed it – if you get a chance, can you check it out? And please let me know if you see anything else. I really need to rewrite a lot of these first sections – I’m not too happy with them. Thanks again – bless you 🙂 Lisa

  12. A Company has releases its new phone in the market with a suggested retail price of Php 35, 500. Many consumers are waiting for this model because of its unique and more advanced features. You are working as a marketing manager of the store, one of its dealer stores. As the marketing manager it is your responsibility to maximize the sales and profit for four months, making sound recommendations regarding the pricing of the product. It has been observed in the 3rd week of the first month that for every Php 300 increase in the suggested retail price, 3 fewer customers will not buy the product. The proposal must be accurate in computations, and represented using a model.

    Please help me find the mathematical model. It’s hard form without the given average sales of the phone 🙁

    • Thanks for writing! I have a problem similar to this one here (check out the maximum revenue problem), but I’m not sure how to do yours, given they don’t give the average number of customers, like you say. If you had the number of customers , you could maximize (35,500 – 300x)(customers – 3x), where x is the number of 300 increases of the price. Does that make sense? Lisa

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