Mathematical

1
Mathematical
Models
Constructing Functions And Optimisation
2
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.
Total fencing needed would be the perimeter
(adding up all sides)
Draw a picture
x
w
w
x
Area of rectangle is length x times width w
This is the area as a function of x and w. We
want area as a function of x.
3
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.
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.
If we solve for w in this equation, we can
substitute it in for w in the area equation below.
x
w
w
x
To find maximum area, well look at the graph.
4
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.
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
The graph is a parabola that opens down. Put this
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.
w
w
x
5
(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
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Another Example

• Let P (x, y) be a point on the graph of y x2
  8
• Express the distance d from P to the point (0,
  -1) as a function of x.
• What is d if x 0?
• What is d if x -1?
• Use a graphing utility to graph d d(x).
• For what values of x is d smallest?

The first thing to do is draw a picture. Well
take each part and do it on a slide.
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• Let P (x, y) be a point on the graph of y x2
  8
• Express the distance d from P to the point (0,
  -1) as a function of x.

• Let P (x, y) be a point on the graph of y x2
  8
• Express the distance d from P to the point (0,
  -1) as a function of x.

This is a parabola vertically translated down 8.
Lets use the distance formula to express the
distance from (x, y) to (0, -1)
This is a formula for the distance from P to (0,
-1) as a function of x and y. We only want it as
a function of x so we need another equation
relating x and y to solve and substitute for y.
(0, -1)
(x, y)
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• Let P (x, y) be a point on the graph of y x2
  8
• 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
We can then substitute for y in the distance
equation above. y x2 8
(0, -1)
(x, y)
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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?
c) What is d if x -1?
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d) Use a graphing utility to graph d d(x). e)
For what values of x is d smallest?
This is an even function so will also be smallest
d at x - 2.55
11
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 x axis so its
position at any time is changing but can be
written as (x, 0)
Express the distance d between the cars as a
function of time.
Using the distance formula, we can find the
distance between (x, 0) and (0, y) to find the
distance between the two cars
Lets draw a picture putting the cars on a
coordinate system letting the origin be the
intersection.
(3, 0)
(0, -2)
The first car is moving along the y axis so its
position at any time is changing but can be
written as (0, y)
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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 distance d between the cars as a
function of time.
We need to find equations for x and y in terms of
t
We now have the distance as a function of time
The first car is moving along the y axis. Using
d rt we have d 30t. It started at -2 on the
y axis so its y axis position is y -2 30t
Similarly the second car is moving along the x
axis. Using d rt we have d 40t. It started
at 3 on the x axis but is moving in the negative
x direction so its x axis position is x 3 40t
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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.
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 lets zoom in and see
if it ever is (meaning the cars did crash).
They dont crash and the closest they get is
about ¼ km apart.
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Acknowledgement I wish to thank Shawna Haider
from Salt Lake Community College, Utah USA for
her hard work in creating this PowerPoint. www.sl
cc.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
Stephens School Carramar www.ststephens.wa.edu.
au