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# Real World Examples of

Quadratic Equations

A **Quadratic Equation** looks like this:

Quadratic equations pop up in many real world situations!

Here we have collected some examples for you, and solve each using different methods:

Each example follows three general stages:

- Take the real world description and make some equations
- Solve!
- Use your common sense to interpret the results

## Balls, Arrows, Missiles and Stones

When you throw a ball (or shoot an arrow, fire a missile or throw a stone) it goes up into the air, slowing as it travels, then comes down again faster and faster.

and a Quadratic Equation tells you its position at all times!

## Example: Throwing a Ball

### A ball is thrown straight up, from 3 m above the ground, with a velocity of 14 m/s. When does it hit the ground?

Ignoring air resistance, we can work out its height by adding up these three things:

(Note: **t** is time in seconds)

The height starts at 3 m:

The t = 0.2 is a negative time, impossible in our case.

The t = 3 is the answer we want:

The ball hits the ground after 3 seconds!

Here is the graph of the Parabola h = 5t 2 + 14t + 3

It shows you the **height** of the ball vs **time**

Some interesting points:

(0,3) When t=0 (at the start) the ball is at 3 m

( 0.2,0) says that 0.2 seconds BEFORE we threw the ball it was at ground level. This never happened! So our common sense says to ignore it.

(3,0) says that at 3 seconds the ball is at ground level.

Also notice that the ball goes **nearly 13 meters** high.

Note for the enthusiastic: You can find exactly where the top point is!

The method is explained in Graphing Quadratic Equations. and has two steps:

Find where (along the horizontal axis) the top occurs using ** b/2a** :

- t = b/2a = ( 14)/(2 5) = 14/10 =
**1.4 seconds**

Then find the height using that value (1.4)

- h = 5t 2 + 14t + 3 = 5(1.4) 2 + 14 1.4 + 3 =
**12.8 meters**

So the ball reaches the highest point of 12.8 meters after 1.4 seconds.

## Example: New Sports Bike

You have designed a new style of sports bicycle!

Now you want to make lots of them and sell them for profit.

Your **costs** are going to be:

- $700,000 for manufacturing set-up costs, advertising, etc
- $110 to make each bike

Based on similar bikes, you can expect **sales** to follow this Demand Curve :

- Unit Sales = 70,000 200P

Where P is the price.

For example, if you set the price:

- at $0, you just give away 70,000 bikes
- at $350, you won’t sell any bikes at all
- at $300 you might sell
**70,000 200 300 = 10,000**bikes

So. what is the best price? And how many should you make?

**Let us make some equations!**

How many you sell depends on price, so use P for Price as the variable

- Unit Sales = 70,000 200P
- Sales in Dollars = Units Price = (70,000 200P) P = 70,000P 200P 2
- Costs = 700,000 + 110 x (70,000 200P) = 700,000 + 7,700,000 22,000P = 8,400,000 22,000P
- Profit = Sales-Costs = 70,000P 200P 2 (8,400,000 22,000P) = 200P 2 + 92,000P 8,400,000

Profit = 200P 2 + 92,000P 8,400,000

Yes, a Quadratic Equation. Let us solve this one by Completing the Square .

### Solve: 200P 2 + 92,000P 8,400,000 = 0

**Step 1** Divide all terms by -200

P 2 – 460P + 42000 = 0

**Step 2** Move the number term to the right side of the equation:

P 2 – 460P = -42000

**Step 3** Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:

(b/2) 2 = ( 460/2) 2 = ( 230) 2 = 52900

P 2 – 460P + 52900 = 42000 + 52900

(P – 230) 2 = 10900

**Step 4** Take the square root on both sides of the equation:

P – 230 = 10900 = 104 (to nearest whole number)

**Step 5** Subtract (-230) from both sides (in other words, add 230):

P = 230 104 = 126 or 334

What does that tell us? It says that the profit is ZERO when the Price is $126 or $334

But we want to know the maximum profit, don’t we?

**It is exactly half way in-between!** At $230

And here is the graph:

Profit = 200P 2 + 92,000P 8,400,000

The best sale price is **$230**. and you can expect:

- Unit Sales = 70,000 200 x 230 = 24,000
- Sales in Dollars = $230 x 24,000 = $5,520,000
- Costs = 700,000 + $110 x 24,000 = $3,340,000
- Profit = $5,520,000 $3,340,000 =
**$2,180,000**

A very profitable venture.

## Example: Small Steel Frame

Your company is going to make frames as part of a new product they are launching.

The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be **28 cm 2**

The inside of the frame has to be **11 cm by 6 cm**

What should the width **x** of the metal be?

Area of steel before cutting:

Area = (11 + 2x) (6 + 2x) cm 2

Area = 66 + 22x + 12x + 4x 2

Area = 4x 2 + 34x + 66

Area of steel after cutting out the 11 6 middle:

Area = 4x 2 + 34x + 66 66