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May 28th, 2014, 03:28 PM   #1
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need help 2

You own a business that makes gaming software. Your company has decided to create 3 add-on software options. To create these add-ons, it takes a team that consists of a computer programmer, graphic artist, and mathematician.
• Add-on software A takes the programmer 9 hours, the graphic artist 6 hours, and the mathematician 1 hour to complete.
• Add-on software B takes the programmer 10 hours, the graphic artist 4 hours, and the mathematician 2 hours.
• Add-on software C takes the programmer 12 hours, the graphic artist 4 hours, and the mathematician 1 hour.
If there are 398 programming hours available, 164 graphic artist hours available, and 58 mathematician hours available, how many copies of each software can be produced?
Use the following guidelines for your answer:
• Set up the systems of equations.
• Solve the system of equations, using any preferred method for solving.
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May 28th, 2014, 06:43 PM   #2
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I want to know whether it has some order for example we must finish A so we can do B later
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May 28th, 2014, 08:24 PM   #3
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No Order

Was just given instructions of how many copies of each software
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May 29th, 2014, 06:38 AM   #4
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You'd have to be more specific. What if I was to decide not to produce add on software B or C? That would surely allow me to make more copies of add on software A, wouldn't it?

I'm not a fan of these badly written maths problems. If a teacher has given you this question, then you should ask that teacher to re-write the question. That would be a more promising mathematical solution.

Last edited by perfect_world; May 29th, 2014 at 06:50 AM.
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May 29th, 2014, 07:20 AM   #5
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The key word you are missing is 'maximum' number of copies of software by making best use of the time for each staff. Otherwise, this question is very clear. Equations can be formed that are based on the premise that every possible minute is spent by each team member performing work.

So, assuming that no one spends time pissing about doing nothing, what is number of copies of A, B and C made that allow this?

I have an answer... I'll post it in a few hours!
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May 29th, 2014, 07:26 AM   #6
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Quote:
Originally Posted by Benit13 View Post
The key word you are missing is 'maximum' number of copies of software by making best use of the time for each staff. Otherwise, this question is very clear. Equations can be formed that are based on the premise that every possible minute is spent by each team member performing work.

So, assuming that no one spends time pissing about doing nothing, what is number of copies of A, B and C made that allow this?

I have an answer... I'll post it in a few hours!
Stated explicitly. Lol.
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Last edited by perfect_world; May 29th, 2014 at 07:31 AM.
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May 29th, 2014, 07:40 PM   #7
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9A + 10B + 12C = 398
6A + 4B + 4C = 164
A + 2B + C = 58

Using elimination of variables it can readily be determined that A = 6, B = 20 and C = 12.
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May 29th, 2014, 09:05 PM   #8
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Thanks

Awesome cd!
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May 30th, 2014, 02:04 AM   #9
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$\displaystyle \left( \begin{array}{ccc}
9 & 10 & 12 \\
6 & 4 & 4 \\
1 & 2 & 1 \end{array} \right)
$$\displaystyle \left( \begin{array}{ccc}
A \\
B \\
C \end{array} \right)
= \left( \begin{array}{ccc}
398 \\
164 \\
58 \end{array} \right)
$

Left-multiplying both sides by the inverse gives

$\displaystyle
\left( \begin{array}{ccc}
A \\
B \\
C \end{array} \right)
= \left( \begin{array}{ccc}
9 & 10 & 12 \\
6 & 4 & 4 \\
1 & 2 & 1 \end{array} \right)^{-1}
\left( \begin{array}{ccc}
398 \\
164 \\
58 \end{array} \right)
$.

Using whichever method you prefer, the inverse of the matrix can be calculated to be
$\displaystyle \frac{1}{40}\left( \begin{array}{ccc}
-4 & 14 & -8 \\
-2 & -3 & 36 \\
8 & -8 & -24 \end{array} \right)$,

so $\displaystyle
\left( \begin{array}{ccc}
A \\
B \\
C \end{array} \right)
=\frac{1}{40}\left( \begin{array}{ccc}
-4 & 14 & -8 \\
-2 & -3 & 36 \\
8 & -8 & -24 \end{array} \right)
\left( \begin{array}{ccc}
398 \\
164 \\
58 \end{array} \right) =
\left( \begin{array}{ccc}
6 \\
20 \\
12 \end{array} \right)
$.
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May 30th, 2014, 02:53 AM   #10
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If they are making 3 add-ons, then the time to make 1 of each is all that's required, then just copy it
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