Section 3.3 Quadratic Functions

Application Problems

Quadratic Functions

Find the dimensions of a rectangular field thatcan be enclosed with 3000 feet of fence and

that has the largest possible area.

• Step 1: Make a drawing and label sides

• Step 2: Write some formulas (perimeter and area)

Quadratic Functions

y

x

x

y

2 2P x yA x y

• Step 3: Substitute one equation inside the other. To do this, solve the perimeter equation for x or y and plug into the Area formula. We plug into the Area formula because we are trying to maximize area.

Quadratic Functions

2 23000 2 23000 2 23000 2 22 2 2

1500

P x yx yx yx y

x y

1500y x

• Step 3: We solved for y, now plug y into the formula for area.

Quadratic Functions

1500y x ( )A x x y

Substitute into area.

( )A x x (1500 )x2( ) 1500A x x x

Distributive property

This is our area function. It’s a quadratic function that opens down because a is negative.

• This part you don’t have to do, but just so that you understand what’s going on: If I were to plot my area function, this is what it would look like:

Quadratic Functions

2( ) 1500A x x x

This graph represents how the area changes with different values for x. Remember that x in your area equation represents the length dimension. In this graph, length dimensions are on the x axis, and the corresponding area is on the y axis.

x

A

If you want to know what the largest possible area is, then you can easily see from the graph, that the maximum area occurs at the vertex.

• Notice that if I look for the x in the graph that corresponds to this vertex, this will be the x that gives me maximum area. This x(max) also represents the length of my enclosure.

Quadratic Functions

2( ) 15001 1500

A x x xa b

x

A

X(max)

How do I find x(max)?

X(max) is also the x-coordinate of my vertex. I know how to do that. 1500 750

2 2( 1)b fta

• So far, this is what I have:

Quadratic Functions

y

x= 750ft

y

x= 750ft

We have just proven that if I make my fencing with 750ft of length, I will get the maximum area possible.

How do I find my width size, knowing that I have a total of 3000ft of fencing?

You can use common sense or use the formula for perimeter. By the way, using common sense is also using the perimeter formula, but in your head.

• Common sense tells me that if I have a total of 3000 ft. of fencing, and I have used 1500 ft, then I have 1500ft left for my two widths. This means that y has to be 1500 divided by 2, y = 750ft.

Using the formula for perimeter:

Quadratic Functions

y

x= 750ft

y

x= 750ft

15001500 750750 .

y xyy ft

Quadratic Functions

Find the dimensions of a rectangular field thatcan be enclosed with 3000 feet of fence and

that has the largest possible area.750 .750 .

x fty ft

750ft.

750ft

750ft.

750ft

You try: Page 171 #45

Quadratic Functions

A vendor can sell 275 souvenirs per day at a price of $2.00 each. The cost to the vendor is $1.50 per souvenir. Each $0.10 profit increase decreases sales by 25 per day. What price should be charged to maximize profit?

Quadratic Functions

First of all, let’s understand how to choose my variable: Each 10 cent increase in price will affect, obviously, the price of the souvenir and also the amount of souvenirs I sell.

A $0.10 increase in price = 1 (10 cent increase)A $0.20 increase in price = 2 (10 cent increases)A $0.40 increase in price = 4 (10 cent increases)

x will represent the amount of ten cent increases

Quadratic Functions

Let’s write an equation for profit:P = (amount of profit per item)(amount of items sold)

P = (0.50) (275)

Quadratic Functions

If a souvenir costs $1.50 and I sell it for $2.00, then my profit is $0.50 per souvenir.

It says that I sell 275 souvenirs per day.

Careful!!! The profit per item depends on the amount of 10 cent increases I make.

Careful!!! The amount of items I sell depends on the amount of 10 cent increases I make.

Each $0.10 profit increase decreases sales by 25 per day

x will represent the amount of ten cent increases

P = (amount of profit per item)(amount of items sold)

P = (0.50 + 0.10 x) (275 – 25x)P = (0.50+0.10x)(275-25x)

We have our equation for profit and it looks like the x-intercept form of a quadratic function

Quadratic Functions

Quadratic Functions

We need this equation to look like the x-intercept form. Factor out the coefficient of x in both parentheses.

(0.50 0.10 )(275 25 )p x x

0.50 275(0.10) ( 25)0.10 25

p x x

(0.10) 5 ( 25) 11p x x Multiply the constants together.

(0.10)( 25) 5 112.5( 5)( 11)

p x xp x x

Now it looks like x-intercept form of a

quadratic function!!!!

• This part you don’t have to do, but just so that you understand what’s going on: If I were to plot my profit function, this is what it would look like:

Quadratic Functions

( ) 2.5( 5)( 11)P x x x

This graph represents how the profit changes with different values for x. Remember that x in your profit equation represents the amount of 10 cent increases. In this graph, the amount of 10 cent increases are on the x axis, and the corresponding profit is on the y axis.

x

P

If you want to know what the largest possible profit is, then you can easily see from the graph, that the maximum profit occurs at the vertex.

• Notice that if I look for the x in the graph that corresponds to this vertex, this will be the x that gives me maximum profit. This x(max) also represents the amount of 10 cent increases.

Quadratic Functions

x

P

X(max)

How do I find x(max)?

X(max) is also the x-coordinate of my vertex. I know how to do that. 5 11 3

2 2s t

( ) 2.5( 5)( 11)5 11

P x x xs t

• So far, this is what I have:

Quadratic Functions

We have just proven that if I make 3 ten cent increases ($0.30), that’s where I’ll have the most profit.

The original problem asks: What price should be charged to maximize profit?

Price that should be charged is:$2.00 + 3(0.10) = $2.30And that’s your final answer!!!

You try: Page 171 #46

Quadratic Functions

A rocket is fired upward from ground level with an initial velocity of 1600 ft/sec. When does the rocket reach its maximum height and how high is it at that time?

Quadratic Functions

Use the following equation for the height (in feet) of an object moving along a vertical line after t seconds:

where is the initial height (in ft.) and is the initial velocity (ft/sec). The velocity is positive if it is traveling upward, and negative if it is traveling downward.

Quadratic Functions

20 016s t v t s

0s 0v

A rocket is fired upward from ground level with an initial velocity of 1600 ft/sec. When does the rocket reach its maximum height and how high is it at that time?

Step 1: Fill in the equation with the information you have available.

Quadratic Functions

20 016s t v t s

216 1600 0s t t

Quadratic Functions2( ) 16 1600s t t t

Step 2: find the x coordinate of the vertex to find when the rocket reaches its maximum height.

t

s1600 502 2( 16)ba

The rocket reaches its

maximum height at 50 seconds.

2

2

( ) 16 1600

(50) 16(50) 1600(50)(50) 40,000

s t t t

ss ft

(50, 40,000)

The rocket reaches a

maximum height of 40,000ft.

You try: Page 172 #51

Quadratic Functions