This section covers:

**Quadratic Projectile Problem****Quadratic Trajectory (Path) Problem****Optimization of Area Problem****Maximum Profit and Revenue Problems****Population Problem**,**Linear Increase/Decrease Problem****Pythagorean Theorem Quadratic Application****Quadratic Inequality Problem****Finding the Quadratic Equation – Finding the “a” Problem**

**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 , where ** t** is the time in seconds, and

**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.**

*h***Solution:**

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:

(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

**axis, and the shapes are similar. Height versus distance would be the**

*y***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?

**Solution:**

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:

# 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?

**Solution:**

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*Now, we want to maximize the area, and (from Geometry), we know that the area (the ** y**) is

**width times length**, or

**. We want to maximize the area, and when we learned about Quadratics, we learned that the vertex is the**

*w*(120 – 2*w*)**maximum**, given an

*y*point**point, (in our case, a**

*x***point). Area depends on length and width – which makes sense.**

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

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 – 2*w”)*, and this also must be positive. Since **120 – 2 w > 0**,

**. So a reasonable domain for the width is 0 to 60 feet or, in interval notation, (0, 60).**

*w*< 60In 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.

**Solution:**

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: .

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**?

**Solution:**

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:

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 , 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 ** t **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?

**Solution:**

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 “

**” here is “**

*x***”, which is time. Part (b), the maximum population, will be the**

*t***part of the vertex, since the “**

*y***” here is “**

*y***”, which is population.**

*p*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:

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!)

**Solution:**

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 – 2

*t*) and the width is (15 + 3

*t*)? 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:

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:

# 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.

**Solution:**

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:

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** **.

# 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 has three possibilities:

- means there is only
**one real solution**. - means there are
**two real solutions.** - means there are
**no real solutions**, only imaginary solutions.

Here is the problem:

Given , find the values of **b** so there are **no real solutions** to the quadratic.

**Solution:**

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

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 form**,

**standard 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!