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Aug 13 2017

Real World Examples of Quadratic Equations #math, #maths, #mathematics, #school, #homework, #education

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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





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