Final Assignment – Part B
ÒMultiple SolutionsÓ
By
Victor
L. BrunaudVega

1. Find as many solutions as possible
for A, B, and C that satisfy both equations: ABC
= 4 3A + 2B  C = 3 



What observations can you make about
your results? Again, we discussed some approaches to this one in class. Your
task is to prepare a writeup that explores this task. (Not
required, but you might want to consider how to explore this one with a
spreadsheet as well as with graphing tools.). 



Okay, let's start looking at these
surfaces in 3D, using the software Graphing Calculator and replacing x=A,
y=B, and z=C. I will try the
two equations separately to find their characteristics and try to analyze
them. 



As we can see in the picture at right,
the first equation, xyz=4, is a hyperboloid with the following
characteristics:










If I project in plane the intersection
between the hyperboloid and z=1 (red line, quadrants I and III) and z=1
(blue line, quadrants II and IV), I get the picture at left. 



Now I will try the second equation
using ÒGraphing CalculatorÓ and replacing A, B and C by x, y, and z,
respectively. 



As you can see at right, the result of
graphing the equation 3A + 2B + C = 3 is a plane. This plane will intersect the
z axis (x = 0, y = 0) for z = 3. Keeping z=0, the graph in the plane
shows the straight line you can see below 









If I graph both equations now, maybe I
will find a solution or more probably a set of solutions. Below there are some pictures of
graphs using both equations. 











As we could see, there are many
solutions. If z =  1 there are two intersecting
points, each in quadrants second (VI) and fourth (VIII), as you can see in
the picture at right. 





If z = 1, there is no intersection
between both surfaces. There is no solution for z=1. 








Conclusions: ¤ x and y can't both be negative.
Observing the graphs, they never intersect on quadrant III. ¤ x and y can't be negative if z is positive. 







LetÕs try something different: If ABC=4, then C=4/AB. Therefore, I can replace C in the
second equation, obtaining: 3A + 2B + 4/AB = 3. 



Now we can graph it, getting
something similar to the picture at right. This is not helping to clarify the situation. Let us try a different approach. 









We can use Excel to explore a set of
results, but these equations will not be useful. I need a new equation. Let us try saying that: If C= 2, then replacing C in the first
equation, AB= 2. Therefore, A= 2/B. Replacing A in the second equation,
we have: 3(2/B) + 2B – (2) =3. After simplification, 2B^{2} – B 6 = 0 That gives us two values for B: 3 and 2, with x = 4/6 and x=1, respectively. Can we repeat this procedure for any
value of C? 




Not really, as we can see in this
spreadsheet. When the value of C
goes between 0 and 2.5, there are no solutions. 




















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