Section 3.3
Quadratic Functions
Application Problems
Quadratic Functions
Find the dimensions of a rectangular field that
can be enclosed with 3000 feet of fence and
that has the largest possible area.
Quadratic Functions
• Step 1: Make a drawing and label sides
x
y
y
x
• Step 2: Write some formulas (perimeter and
area)
P  2x  2 y
A  xy
Quadratic Functions
• 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.
y  1500  x
P  2x  2 y
3000  2 x  2 y
3000  2 x  2 y
3000
2

2x

2
1500  x  y
2y
2
Quadratic Functions
• Step 3: We solved for y, now plug y into the formula
for area.
y  1500  x
Substitute into
area.
A( x)  x  y
A ( x )  x (1500  x )
Distributive
property
A ( x )  1500 x  x
This is our area function. It’s a quadratic
function that opens down because a is negative.
2
Quadratic Functions
• 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:
A ( x )  1500 x  x
2
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.
A
x
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.
Quadratic Functions
• 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.
A ( x )  1500 x  x
a  1
2
A
b  1500
X(max)
x
How do I find x(max)?
X(max) is also the xcoordinate of my vertex. I
know how to do that.
b
2a

 1500
2(  1)
 750 ft
Quadratic Functions
• So far, this is what I have:
x= 750ft
y
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.
Quadratic Functions
• 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.
x= 750ft
y
y
x= 750ft
Using the formula for perimeter: y  1500  x
y  1500  750
y  750 ft .
Quadratic Functions
Find the dimensions of a rectangular field that
can be enclosed with 3000 feet of fence and
that has the largest possible area.
x  750 ft .
y  750 ft .
750ft
750ft.
750ft.
750ft
Quadratic Functions
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)
If a souvenir costs
$1.50 and I sell it
for $2.00, then my
profit is $0.50 per
souvenir.
P=
Careful!!! The profit
per item depends
on the amount of
10 cent increases I
make.
It says that I sell 275
souvenirs per day.
(0.50) (275)
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
Quadratic Functions
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
p  (0.50  0.10 x )(275  25 x )
We need this equation to look like the
x-intercept form.
Factor out the
coefficient of x in both parentheses.
 0.50

  275

p  (0.10) 
 x  (  25) 
 x
 0.10

 25

p  (0 .1 0 )  5  x  (  2 5)   1 1  x 
Multiply the constants together.
p  (0.10)(  25)  5  x    11  x 
p   2.5( x  5)( x  11)
Now it looks like x-intercept form of a
quadratic function!!!!
Quadratic Functions
• 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:
P ( x )   2.5( x  5)( x  11)
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.
P
x
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.
Quadratic Functions
• 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.
P
P ( x )   2.5( x  5)( x  11)
s  5
t  11
x
X(max)
How do I find x(max)?
X(max) is also the xcoordinate of my vertex. I
know how to do that.
st
2

 5  11
2
3
Quadratic Functions
• So far, this is what I have:
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) = $2.30
And that’s your final answer!!!
Quadratic Functions
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:
s   16 t  v 0 t  s 0
2
where s 0 is the initial height (in ft.) and v 0 is
the initial velocity (ft/sec). The velocity is
positive if it is traveling upward, and negative
if it is traveling downward.
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?
s   16 t  v 0 t  s 0
2
Step 1: Fill in the equation with the information
you have available.
s   16 t  1600 t  0
2
Quadratic Functions
s ( t )   16 t  1600 t
Step 2: find the x coordinate of the vertex to find
when the rocket reaches its maximum height.
2
b
2a

 1600
2(  16)
s ( t )   16 t  1600 t
2
s
 50
The rocket
reaches its
maximum height
at 50 seconds.
(50, 40,000)
s (50)   16(50)  1600(50)
2
s (50)  40, 000 ft
t
The rocket
reaches a
maximum height
of 40,000ft.
Quadratic Functions
You try: Page 172 #51