Quadratics and Polynomials

Quadratic (polynomial) Application

Modelling

Applications

(Contextual Problems)

• In this section we want to look at the applications that

quadratic equations and functions have in the real

world.

• There are several standard types: problems where the

formula is given, falling object problems, problems

involving geometric shapes. Just to name a few. There

are many other types of application problems that use

quadratic equations, however, we will concentrate on

these types to simplify the matter

Optimisation Problems

The number of bacteria in a refrigerated food

is given by N(T) = 20T2 – 20T + 120,

for -2≤T ≤14 and where T is the temperature

of the food in Celsius.

Q: At what temperature will the number of

bacteria be minimal?

• First we can see that we have a quadratic

function given to us. Moreover, the

parabola would open up. So that means that

the vertex of the parabola represents a

minimum value. This means the first thing

we need to do is determine the vertex, since

we want the minimal number of bacteria.

Optimisation cont’d

• Use the formula for the vertex:

• to find the vertex since it is much more

efficient than the completing the square

method for finding the vertex in word

problems. Since our function is in terms of

T, the formula we really use is:

2v

bx

a

20 1

2 * 20 2T

Optimisation cont’d

• So the T value of the vertex is ½. Now we

must look back at the question to see what

we really wanted. Since we want the

temperature at which the number is

minimum and T is the temperature in

Celsius, this is the value we want.

Therefore, the minimum number of bacteria

are present when the temperature is ½0

Celsius.

Area Problems: triangle

Calculate length of a

triangle if we know

that height is 6 cm

shorter than a base and

area is 40cm2.

b (base)

h

(height)

Area of a Triangle

1.) Define variables:

h – height of a triangle

b – base of a triangle

A – area of a triangle (A = )

2.) What else is given:

h is 6cm shorter than base: h = b – 6

A = 40cm2

1

2bh

24

2

b b ac

a

Area of a Triangle (cont’d)

2

2

2

1

2

1* * ( 6)

2

2 6

40 402 6

3 40 0

A bh

A b b

b bA

b bA

b b

Area of a Triangle (cont’d)

We know from quadratic formula:

In our example, we are using ‘b’ as the x

variable.

DO NOT CONFUSE YOURSELF!!!

24

2

b b acx

a

Area of a Triangle (cont’d)

2

1,2

1

2

( 1) ( 1) 4 * 3 * ( 40)

2 * (3)

1481

6

1481

6

481 21.335

1481 6

6

b

b

b

h

Therefore b2 < 0 and we can’t

use this solution. Only b1

Optimisation Problem

My best friend recently had twins. They are ready for a play

pan and she has bought online from China a ‘Playpen Set’.

This is just load of PlayPen Border Sheet with joints so

that she can make whatever shape she wants. She has got a

100m length of playpen. She wants to make equal playpan

areas for her twins. What’s the best way to maximise the

areas? Two separate play-pens or one with dividing wall?

If my friend wants the total area

to be maximum, what dimensions

should be used to make the pens?

• Since we are looking to maximize the area,

we need to generate a function for the area

so that we will only need to find the

maximum of this function.

The first thing we should do is decide on some labels

and variables. Lets call the longer side x and the shorter

side y.

Y

X

Then we need to express y in terms of x, that way we will only have to deal

with one variable:

Now that we have no other information, let’s see what do we need to find:

Area.

Area of a rectangle = length * height

A = x * y = x * (100 – 3x) / 2

= 50x – 1.5* x2

Then let’s look at what else we were given?

•She has got in total 100m of the fence. So, the maximum perimeter we can

get is 100m. Perimeter is simply a sum of all walls:

P = Y + X + Y + X + X

P = 100 therefore:

100 = 3x + 2y

y = (100 − 3x) / 2

A = – 1.5x2 + 50x

We know that this will have a max vertex at:

xv = - b/2a = -50/2*(-1.5) = -50/(-3) = 50/3 = 16.667

This means at the length of a side x of 16.667m we

will have maximum area . But we don’t need area,

we need the length of Y. (substitute back):

y = (100 − 3x) / 2 = 24.995

We need to be very careful to find exactly

what the situation is asking for. Sometimes

the problem is asking for a value of the

vertex, sometimes the problem is asking for

the solutions to the quadratic and sometimes

the problem is merely asking to evaluate a

quadratic function. We must carefully read

each question to determine exactly what is

being asked.