Suppose a farmer has 50 meters of fencing to build
a rectangular yard. Express the rectangular area A
he can enclose as a function of the length x of a
side. Then find the dimensions to make his yard to
enclose the maximum area.
Draw a picture
x
w
w
Total fencing needed would
be the perimeter (adding up
all sides)
2x  2w  50
x
Area of rectangle
is length x times
width w
A  xw
This is the area as
a function of x and
w. We want area
as a function of x.
Suppose a farmer has 50 meters of fencing to build
a rectangular yard. Express the rectangular area A
he can enclose as a function of the length x of a
side. Then find the dimensions to make his yard to
enclose the maximum area.
x
w
If we solve for w in this
equation, we can substitute it
in for w in the area equation
w below.
2x  2w  50
x
50  2 x
w
 25  x
2
25  x
A  xw
To find maximum area,
we’ll look at the graph.
Suppose a farmer has 50 meters of fencing to build
a rectangular yard. Express the rectangular area A
he can enclose as a function of the length x of a
side. Then find the dimensions to make his yard to
enclose the maximum area.
x
w
x
A  x25  x
The graph is a parabola
that opens down. Put this
w in a graphing calculator
and trace the x where f(x)
is at its maximum.
Adjust the window until
you get a good view.
This is on the next screen.
(12.5, 156.25)
Remember x is the
side of the
rectangle and f(x)
is the area.
This would be
the x value
that would
give the
maximum area
This would be
the maximum
area.
The maximum enclosed
area would be 156.25
square meters
Another Example
Let P = (x, y) be a point on the graph of y = x2 – 8
a) Express the distance d from P to the point (0, -1) as a
function of x.
b) What is d if x = 0?
c) What is d if x = -1?
d) Use a graphing utility to graph d = d(x).
e) For what values of x is d smallest?
The first thing to do is draw a picture. We’ll take
each part and do it on a slide.
Let P = (x, y) be a point on the graph of y = x2 – 8
a) Express the distance d from P to the point (0, -1) as a
function of x.
This is a parabola vertically translated down 8.
This is a formula
for the distance
from P to (0, -1)
as a function of x
and y. We only (0, -1)
want it as a
-7 -6 -5 -4 -3 -2 -1 0
function of x so
we need another
equation relating
x and y to solve
and substitute
for y.
Let’s use the
distance formula to
express the
distance from
(x, y) to (0, -1)
2
2




d  x2  x1  y2  y1
(x, y)
2
2
d  x  0   y   1
2
2
d  x   y  1
1 2 3 4 5 6 7 8
Let P = (x, y) be a point on the graph of y = x2 – 8
a) Express the distance d from P to the point (0, -1) as a
function of x.
Since P is a point on
the graph of y = x2 – 8,
this equation will be
true about the
relationship between x
and y
d  x   y  1
2
2
We can then
substitute for y in the
distance equation
above. y = x2 – 8
(0, -1)
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
(x, y)

 x

d  x  x  8 1
2
 x
2
2
2
7

2
2
So we have our formula for the
distance from P to (0, -1) and we
are ready to answer other parts of
the question.
b) What is d if x = 0?


d  0x  0x  7
2
2
2
 49  7
c) What is d if x = -1?
d
 1
2

  1  7
2

2

d  x  x 7
 1  36  37
2
2

2
d) Use a graphing utility to graph d = d(x).
e) For what values of x is d smallest?

d  x  x 7
2
2

2
This is an even function so
will also be smallest d at
x = - 2.55
Two cars are approaching an intersection. One is
2 km south of the intersection and is moving at a
constant speed of 30 km per hour. At the same time, the
other car is 3 km east of the intersection and is moving
at a constant speed of 40 km per hour.
The second car is moving along the
Express the distance d
x axis so its position at any time is
between the cars as a
changing but can be written as (x, 0) function of time.
d
 x  0
2
 0  y 
2
(3, 0)
(0, -2)
Using
the draw
distance
formula,
Let’s
a picture
we can
find the
putting
the distance
cars on a
between
(x, 0) and
(0, y) to
coordinate
system
find the
distance
between
letting
the origin
be the
the two
cars
intersection.
The first car is moving along the y
axis so its position at any time is
changing but can be written as (0, y)
Two cars are approaching an intersection. One is
2 km south of the intersection and is moving at a
constant speed of 30 km per hour. At the same time, the
other car is 3 km east of the intersection and is moving
at a constant speed of 40 km per hour.
Express the
d x y
distance d between
2
2
the cars as a
d  3  40t   2  30t
function of time.
We need to find equations
We now have the distance
for x and y in terms of t
as a function of time
The first car is moving along
Similarly the second car is moving
the y axis. Using d = rt we
along the x axis. Using d = rt we
have d = 30t. It started at -2
have d = 40t. It started at 3 on
on the y axis so it’s y axis
the x axis but is moving in the
position is y = -2 + 30t
negative x direction so it’s x axis
position is x = 3 – 40t
2


2


By looking at the graph of the distance between the two
cars, determine if the cars crash at the intersection and
if not, find the minimum distance between them.
d
3  40t 
2
  2  30t 
2
Here is a graph showing t on
the x axis and the distance d on
the y axis. Looks like the
distance gets close to 0 so let’s
zoom in and see if it ever is
(meaning the cars did crash).
They don’t crash and the closest they get is about ¼ km
apart.
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au