C A S i n th e C l a s s room :
W h a t a re S o m e G ood W a y s to Ma ke U s e of
C om p u te r A l g eb ra S y s tem s i n Y ou r
M a th em a ti c s C l a s s ?
C o m b i n i n g C A S a n d D y n a m i c G e o m e try C a p a b i li ti e s
59th Annual Meeting of the ICTM
Peoria, IL
Session 131
October 18, 2008
Dr. Donald Porzio
Mathematics Faculty
Illinois Mathematics and Science Academy
1500 W. Sullivan Rd., Aurora, IL 60506
630-907-5966
dporzio@imsa.edu
Activity 1: Modeling Fixed Perimeter, Maximal Area Problems
A standard problem students are asked to solve involves maximizing the area of a rectangular region
given that the region is to have a fixed perimeter. The following is an example of such a problem.
Jessica has 80 meters of wire-mesh fence to make a rectangular pen for her rabbits.
What dimensions will produce the maximal area for such a pen?
One difficulty students often have on this type of problem is formulating an algebraic equation to
represent the problem situation. Another difficulty may be that students do not understand the dynamic
nature of this problem situation; the area of the pen changes as the dimensions of the pen change, even
though its perimeter is fixed. As we will see, the capabilities of Cabri Geometry on the TI-89 allow
students to construct a rectangular pen that dynamically models the problem situation, thus allowing them
to “see” what happens to its area as its length and width change, but its perimeter does not!
APPS (FlashApps), then select Cabri
To open a new Cabri Geometry “session”, press
Geometry followed by New. You must then enter a variable name to open the new Cabri Geometry
window. Now press ENTER twice and we’re all set!
First, we construct a horizontal ray - select 6:Ray from the F2 menu - at the top of the screen. Now, we
will place a point on the ray 40 mm away from its endpoint (since the semi-perimeter of the rectangular
pen is 40 m). To do this, first press
| to access the Geometry Format menu, and set
Length & Area to MM. Now select 6:Numerical Edit from the F7 menu, press ENTER
and type the number 40. Then press 2ND
3 (Units), select 2:Length, and then press ESC ,
(and remember to press ESC each time you complete a construction). Finally, select
9:Measurement Transfer from the F4 menu, select the number 40 mm, and then select the ray.
A point will appear on the ray exactly 40 mm from its endpoint. Next, hide ray - select 1:Hide/Show
from the F7 menu and then select the ray - and construct a segment connecting the endpoint of the ray
with the newly constructed point - select 5:Segment from the F2 menu, select one point, and then
select the other point. Now, place a point on the just-constructed segment - select 2:Point on Object
from the F2 menu, move the pointer to a place on the segment where you'd like the point, and then select
the segment. Select 4:Label from the F7 menu and then label the points A, B, and C going left to right.
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To begin the construction of the rectangle pen, select 8:Compass in the F4 menu to create a circle
with center in the lower left of the screen and with a radius equivalent in length to AB - select points A
and , then move the cursor to the lower left corner of the screen and press ENTER . Draw a horizontal
radius of this circle – select 5:Segment from the F2 menu, then select the center of the circle and a
point on the circle. This segment will be one side of the rectangular pen. Now draw perpendicular lines
at the endpoints of this segment - select 1:Perpendicular Line from the F4 menu, select one
endpoint and the segment, and then repeat this for the other endpoint.
Next, use the Compass command again to construct a circle with the same center as the first circle and
with a radius equivalent in length to BC . Construct the intersection point of this circle and the
perpendicular line through its center - select 3:Intersection Point from the F2 menu and then select
the circle and the line - before using the Perpendicular Line command again to construct a line
perpendicular to the line through the center of the circle and passing through this point. Use the
Intersection Point command again to construct the point of intersection of this line to the
perpendicular line containing the endpoint on the radius of first circle constructed.
Now it's time to eliminate some the clutter in our figure. Select 1:Hide/Show from the F7 menu and
then select all the construction lines and circles, leaving just the four vertices of a rectangle and AC at the
top of the screen. Lastly, construct the rectangle itself - select 4:Polygon from the F3 menu and then
select each vertex in order before pressing ENTER after selecting the fourth vertex.
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Now we need to display the different measurements of interest in this problem -the length of a side,
perimeter, and area of the rectangle. To display the length, select 1:Distance & Length from the F6
menu and then select two endpoints on a side of the rectangle.
Next, with the Distance & Length command still
active, select any side of the rectangle to display its
perimeter. Finally, select 2:Area from the F6 menu
and again select any side to display the area. After
obtaining these measurements, press ESC , select
each measurement - press the ENTER key twice type in what each measurement represents, and then
drag each to an open space of the screen.
Now, we’ll put the statistical capabilities of the TI-89 to work. Open the F6 menu and select 7:Collect
Data followed by 2:Define Entry. Select the measurements for length, area, and perimeter in that
order. As each is selected, a flashing, dashed box should appear around the number. Next, press
, ,
or select 7:Collect Data from the F6 menu, followed by 1:Store Data. This will load all three of
these measurements into the Data/Matrix editor under the variable sysdata (as the message at the
bottom the screen indicates). Actually, you really need to check that this variable is empty or nonexistent
before loading any data.
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Now, move the pointer over point B, press ALPHA ,
and then use the arrow keys,
,
, to move point
B back and forth along the segment. Note what happens
to the rectangle and the measurements. You’ve created
a “dynamic” model of Jessica’s rectangular pen! As you
drag point B to different positions, press
, so as
to collect more data.
After collecting 10-15 data values, press APPS and then select 6:Data/Matrix Editor followed
by 2:Open. If sysdata appears next to Variable: in the dialog box, press ENTER . Otherwise,
cursor down twice and right once to show a list of all the Data variables, scroll down to select
sysdata, and then press ENTER twice to open the variable sysdata.
As you can see, the perimeter did not change when we
selected different lengths for the rectangle (which is
good). Now let’s look at the relationship between the
length and area by generating a scatter plot.
To begin, select the F2: Plot Setup menu and then
select F1: Define. Next select Scatter for the Plot
Type submenu and then designate column c1 for the
x-values and column c2 for the y-values.
To draw the scatter plot, press
F1 (Y=) to access the Y = Editor, turn off any extra functions
or plots that are on, and select 9:ZoomData from the F2:Zoom menu. Graph looks mighty
quadratic! Now let’s try regression to see if this is actually a quadratic relationship.
Again press APPS and then select 6:Data/Matrix Editor followed by 1:Current to re-enter
sysdata. To calculate a quadratic regression, select the F5:Calc menu, set Calculation Type to
QuadReg, designate column c1 for the x-values and column c2 for the y-values, and then store the
RegEq to y1(x), or some other variables in the Y= Editor if you wish. The regression model for this
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data (remember yours may be different) produced the function y = –x2 + 40x within machine accuracy.
(Note: You’ll probably want to discuss why the regression equation was not “exactly” y = –x2 + 40x.)
Press
F3 (GRAPH) to graph the regression equation and scatter plot together. As expected, the
graph of the regression equation passes through all the data points (see Figure 30). Now to solve the
original problem! (Note: At this point, it’s probably best to store –x2 + 40x in y1(x) to avoid difficulties
that might occur if the calculator-generated regression equation is used.)
To determine graphically the dimensions of the rectangle of maximal area, we can locate the coordinates
of the absolute maximum of the graph by first selecting 4:Maximum from the F5:Math menu and
then following the instructions provided. The resulting point, (20, 400) corresponds to a length of 20 m
and a maximal area of 400 m2. We might also determine these dimensions algebraically by first recalling
40
b
that the x-coordinate of the vertex of a parabola is
. For this parabola, we get
so that x = 20.
2a
2(1)
Finally, we can examine how the equation for the area of the rectangle, y = –x2 + 40x might have been
obtained algebraically. The right-hand side of the equation y = –x2 + 40x can be factored (using the CAS
capabilities of the TI-89 if the factorization wasn't so obvious), producing the equivalent equation y = x(60
– x). Then students can be asked to relate this factored form of the equation to the information from the
original problem that was used in the construction of the dynamic model of the rectangle. Students should
come to realize that if x represents the length of the rectangle, then the quantity 60 – x represents the width
of that rectangle (since the perimeter of the rectangle is always 80 meters), and the product x(60 – x)
represents the area of Jessica’s rectangular pen. Here again, this opportunity for analysis “after-the-fact”
should help make it easier for students to understand from where the algebraic equation that models the
problem was obtained, as compared to the more typical scenario where the teacher helps students derive
the equation, or derives the equation for students, before the problem-solving process is even undertaken.
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Activity 2: Another Minimization Problem
Here’s an optimization problem involving minimizing distance that has interesting geometric overtones.
Pipelines for carrying water need to be run from a pumping station on a river to two
different cities on the same side of the river. The pipelines are to be run directly from
the pumping station to each city. One city is 5 miles from the river, and the other city
3 miles. The distance between the points on the river closes to each city is 6 miles (or
we can get real contrived and say that the distance between the cities is miles). Where
should the pumping station be built on the river, relative to the two cities, to minimize
the amount of pipeline used?
| to access the Geometry Format menu,
Open a new Cabri Geometry “session” then press
and set Length & Area to MM. Length & Area to Pixel. This unit was chosen because
centimeters would have produced too large a figure and millimeters too small a figure. In fact, in order to
get a reasonable figure, we not only need to measure in pixels, but we also need to multiple all lengths by
a factor of 10. Thus, to begin, construct a segment having length 60 px, then construct perpendicular lines
to this segment passing through each endpoint. Place a point on each perpendicular (label them A and B)
and adjust their position so that they are 50 px and 30 px from the endpoints of the segment
Place a point on the segment (label it R) and construct segments connecting R to the city at point A and to
the city at point B. While you are at it, label the endpoints of the segment X and Y, then hide the
extraneous numbers and the two perpendiculars to help unclutter the screen.
Now, select 1:Distance & Length from the F6 menu and measure the distances between R and A,
R and B, and R and X. Use 6:Calculate in the F6 menu to determine the sum of the distance from R
to A and R to B, which corresponds to the amount of pipeline that will need to be used.
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Once again, it’s statistics time. After making sure the variable sysdata has been deleted or emptied,
open the F6 menu, select 7:Collect Data followed by 2:Define Entry, and highlight the
measurements for the distance between R and X and the sum of the distances from R to the cities. Now
start collecting data by dragging point R and pressing
, , or selecting 7:Collect Data from the
F6 menu followed by 1:Store Data. After collecting a sufficient number of data points (say 10 to
15), enter the variable sysdata in the Data/Matrix Editor and generate a scatter plot.
Looking at this scatter plot, it’s hard to tell what regression model to use. So, let’s try our favorite three:
quadratic, cubic and quartic. A quick check of the results shows that, once again, none of the models
produces a perfect fit, though all are quite good. Using the quartic regression model (since it produced the
best fit) and rounding the results, we get the ordered pair (37.5, 100) as a solution.
Based on the results of the quartic regression, we can conclude that the minimum amount of pipeline - 10
miles worth - is used if the pumping station is built 3.75 miles from point X, the closest point on the river
to the city at point A.
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This solution can be verified using the TI-89’s
CAS capabilities. If we let x be the distance (in
miles) from X to R, then the distance from A to
R is given by
25  x 2 and the distance from R
to B is given by (6  x) 2  9 . Finding the
derivative of the sum of these two distances,
setting it equal to 0, and solving for x produces
an identical result as that found using the model.
Now, what about those geometric overtones I mentioned? Go back to our figure and construct a point at
the correct spot on the river. To do this, recognize that 37.5 px is 5/8th of the way from point X to point Y
(and what does that 5/8th have to do with the positions of the two cities relative to the river). Use
6:Calculate in the F6 menu to place the decimal equivalent to 5/8th on the screen. Next, select
3:Dilation from the F5 menu, select the decimal value for 5/8th, select point Y, and finally, select
point X. This newly constructed point (label it R) represents where the pumping station should be placed
to minimize the amount of pipeline used. Now, select 4:Reflection from the F5 menu, select point B,
and select point Y. Label this point B. (Note: you may need to move the river toward the top of the
screen to be able to see the reflected point.) Looks an awful lot like X, R, B are collinear. To verify this,
open the F6 menu, select 8:Check Property followed by 1:Collinear.
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So what was that all about? Basically, the geometric model provided us with an opportunity to “discover”
(or verify) one of the more important theorems in transformational geometry. We can even take this a step
further using the TI-89’s CAS capabilities and show that, no matter how far apart (along the river) the two
cities are (say y miles), if the pumping station is placed 5/8ths of the way from point X to point Y, the
points on the river closest to the two cities.
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