Final Assignment – Part B

“Multiple Solutions”


Victor L. Brunaud-Vega



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 write-up 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:


  • It doesn’t intersect any of the three axes x=0, y=0, z=0, in any point.


  • For z > 0, the product xy must be positive, therefore the curve will exist in the first and third quadrant only


  •  for z < 0, the product xy must be negative, so the curve will exist in the sixth and fourth octants.








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 “Graphing Calculator” 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.





Here is a file where you can explore this situation. 






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.










§  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. 






Let’s 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, 2B2 – 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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