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Real World Quadratic Problem

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seungyeon ha

on 4 December 2014

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Transcript of Real World Quadratic Problem

Building Your Own Roller Coasters
"The speed of roller coasters"
To design a roller coaster, you would also want to know how long does roller coaster take.
Real World Quadratic Problem: Rollercoasters
By Anand, Seungyeon, Seungyoon
Real World Problem
You are the owner of a popular theme park and have decided to build a new roller coaster
Your competitor's roller coaster
Building your roller coaster
Promotional Material for New Rollercoaster
In addition to creating the new rollercoasters, you want to sell models of your rollercoasters in the theme park’s gift shop.

It will cost 350,000$ for manufacturing of the models and advertisement of the rollercoasters, plus an additional 15$ for each model made.


Now it's your responsibility to build a thrilling roller coaster using a quadratic equation.
In order to make roller coaster thrilling, it's necessary to make its slope steeper. Therefore you decide to build a roller coaster with steeper and higher heel compared to your competitor's roller coaster.
Near your theme park, there is a rival park that always competes with yours for the profit.
Therefore you should make much thriller coaster than your competitor's.
The maximum height of its heel is 80m.
Horizontal distance between the starting point and the ending point is about 160m.
1) Considering the shape of roller coaster rail as a parabola, 'a' should be negative since parabola opens downward.
y= ax^2 + bx + c
2) As parabola opens downward, vertex will be its maximum point.
3) Therefore, as the maximum height of its heel is 80m, y value of vertex will be 80.
1) Assuming the starting point of the rail as origin (0,0) x intercept should be 0 and 160 since its horizontal distance is 160m.

5) Considering the fact that axis of symmetry always passes through the vertex, x value of vertex is the same as the equation of axis of symmetry. Thus, its x coordinate will be 80.
80meter
(x,80)
Axis of Symmetry
2) Considering that axis of symmetry is a half-way between the x intercepts, it will be x=80
(midpoint between (0,0) and (160,0) is (80,0))
Reason: Using midpoint formula,
x= (0+160)/2 = 80
y= (0+0)/2 = 0
>> (80,0)
160m
origin (0,0)
(160,0)
(80,0)
axis of symmetry
x=80
(80,80)
6) As a result, the vertex for this roller coaster parabola is (80,80)
7) Apply vertex form of a parabola's equation to get the equation for this roller coaster parabola
(80,80)
1) Vertex form is usually written in (y = a(x-h)^2+k)
h= x coordinate of vertex
k= y coordinate of vertex
2) The vertex of this parabola = (80,80)
y=a(x-80)^2+80)


3) Knowing that one of the points on the parabola is (160,0), substitute this coordinate on the equation to get the a value
0=a(160-80)^2+80
0=a(80)^2+80
0=6400a+80
-80 = 6400a
-80/6400 = a
a= -1/80
The equation is y=-1/80(x-80)^2+80
Your competitor's roller coaster heel
y=-1/80(x-80)^2+80
By using vertical stretch, you can make a steeper and higher parabola.
Applying Vertical Stretch on the equation, y=-1/80(x-80)^2+80
* vertical stretch is method to stretch away from the x axis
y=2(-1/80(x-80)^2+80)
y=-1/40(x-80)^2+160
Vertical Stretch by a factor of 2
Because of safety issue, you decide to make two times steeper and higher heel in your new roller coaster
Comparison
Your new roller coaster will be two times higher.
Graph A
Graph B
Maximum point (=y value of vertex) of Graph A = 80
Maximum point (=y value of vertex) of Graph B = 160
Your new roller coaster will be two times steeper
Value of a is -1/80 in Graph A
Value of a is -1/40 in Graph B
Your roller coaster will be MUCH MORE FUN than your rival's!!
To calculate the speed of the roller coaster, you can start with a basic formula:

Distance = (initial velocity)(time) + ½ (acceleration)(time)^2

The equation can help you to find the speed of how far can the roller coaster travel when it comes down from the first hill.

Considering the sales of previous models sold in the gift shop, you can assume the sales will follow a Demand Curve of 10,000 - 100P, where 10,000 is the total number of units manufactured and P is the price of the unit.

The Demand Curve function gives us a prediction as to how many units will be sold.
You want to find out for which value of P ( which price) yields the highest profit; to do this, you first need to find the profit function.
Since profit is equivalent to the Sale in Dollars minus the cost, we will first find those two terms.
To find the Sale in Dollars, multiply the demand curve function by P, giving us 10,000P - 100P^2.
To find the cost, multiply the cost to
manufacture each model (15) with our
demand curve function. (10,000-100P)
and add it to our overall cost .(350,000)
That gives us 150,000 - 1500P +
350,000 which can be simplified into
500,000 -1500P.
Now to find profit, we take Sales in Dollars and subtract our cost function from it; that gives us -100P^2 + 11500P - 500,000. We can simplify this by dividing it by a common factor of 100 to get -P^2 +115P - 5000.
If we use -b/2a, we get the vertex of 57.5;putting this into the function gives us -1693.75.
Since our vertex is below the xaxis (57.5, -1693.75) we can see that using this model, you would incur only losses from using this sale model to sell model rollercoasters.


Source
http://mathforum.org/mathimages/index.php/Parabola
http://www.regentsprep.org/regents/math/algtrig/atp9/funclesson1.htm
https://prezi.com/zg-fackgdyks/quadratic-function-project-algebra-2-by-amy-aviles/
https://www.geogebratube.org/student/m22881
http://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.185580.html
http://ffden-2.phys.uaf.edu/211_fall2002.web.dir/shawna_sastamoinen/Roller_Coasters.htm
http://www.ck12.org/algebra/Use-Graphs-to-Solve-Quadratic-Equations/rwa/Roller-Coaster-Relationships/
http://en.wikipedia.org/wiki/List_of_roller_coaster_rankings#Tallest_steel_roller_coasters

However, by removing the advertising costs, we can reduce the main cost from 350,000 to 100,000. This would change the final equation to 0= -P^2 +115P-2500. This would put the vertex above the x-axis, making the maximum profit possible 806.25 dollars.
Finally, if we halve the number of units manufactured, making it 5000,
the equation would be further changed to 0= -P^2 +115P-1750. This would change to maximum profit to 1556.25 dollars - not enough to justify manufacturing and selling the models.

THANK YOU
additionally, your new roller coaster will become third tallest roller coaster in the world
Full transcript