# Thread: Systems of Equation application problem, regarding student/professor/sections per cla

1. ## Systems of Equation application problem, regarding student/professor/sections per cla

Enormous state university history department offers three courses, Ancient , medieval, and modern history, and the chairperson is trying to decide how many sections of each to offer this semester. the department is allowed to offer 45 sections total, there are 5000 students who would like to take a course, and there are 60 professor to teach them. Sections of ancient history have 100 students each, medieval hold 50 students, and modern history has 200 students each. Modern history is thought by 2 professor, while ancient and medieval history need only 1 professor per section.
How many sections of each course should the chair schedule in order to offer all the sections that they are allowed to, accommodate all students, and give one teaching assignment to each professor?

i set up my table as
_AH_.MeH_MoH__Total
Student per section| 100 | 50 | 200 | 5000
Professor..............| 1 | 1 | 2 | 60
Sections...............|0.....|0 | 0 | 45

my matrices looked like

$\left[\begin{matrix}100&50&200\\1&1&2\\0&0&0\end{matrix} \left|\begin{matrix}5000\\60\\45\end{matrix}\right]$

as you can see, i likely messed up at the "sections" part.

I ended up with

$\left[\begin{matrix}1&0&3\\0&1&-2\\0&0&0\end{matrix}\left|\begin{matrix}40\\20\\45 \end{matrix}\right]$

alternatively:
x = (400-300z)/100
y= (1000 + 100z)/50
z=z

but anyways my point is nether of the previous 2 gets to the proper answer, most likely because of "section" consisting of 3 zeros.

2. How did you get those "0"s? If I read your set up correctly, your columns correspond to the number of classes of Ancient History, Midieval history, and modern history and the last row is just say they must total to 45: AH+ MiH+ MoH= 45 so your coefficients are 1, 1, 1.

You "augmented matrix" should be $\left[\begin{matrix}100&50&200\\1&1&2\\1&1&1\end{matrix} \left|\begin{matrix}5000\\60\\45\end{matrix}\right]$

3. well unfortunatel i ended up with

$\left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}\left |\begin{matrix}-25\\20\\15\end{matrix}\right]$

4. Your solution is incorrect, as you have already noted, and is not what I got. Can you show us your row reductions?

5. i did
100R2-R1
100R3-R1
then
R1-R2
R3-R2
then
.5R1+R3
then
.5R1
1/50*R2
-1/100*R3

edit: i think i may see where i committed the problem (missed a zero for 5000 and used a negative instead of a positive).
anyway revised it is

$\left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}\left |\begin{matrix}40\\20\\15\end{matrix}\right]$

and still not getting the correct answer

6. Your first two ERO's are good. I would question the third. The fourth looks good. Incidentally, I would be explicit about where you're putting the result of an ERO. For example: 100 R2 - R1 gets stored in R2, I would write as follows: 100 R2 - R1 -> R2. Once you've done the first, second, and fourth ERO's, you can stop there and use back substitution to solve for the variables, if you want. You don't necessarily have to go all the way to the identity matrix.