My Math Forum need help 2

 Algebra Pre-Algebra and Basic Algebra Math Forum

 May 28th, 2014, 03:28 PM #1 Newbie   Joined: May 2014 From: east freedom pennsylvania Posts: 5 Thanks: 0 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.
 May 28th, 2014, 06:43 PM #2 Newbie   Joined: May 2014 From: China Posts: 3 Thanks: 0 I want to know whether it has some order for example we must finish A so we can do B later
 May 28th, 2014, 08:24 PM #3 Newbie   Joined: May 2014 From: east freedom pennsylvania Posts: 5 Thanks: 0 No Order Was just given instructions of how many copies of each software
 May 29th, 2014, 06:38 AM #4 Senior Member   Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 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.
 May 29th, 2014, 07:20 AM #5 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,150 Thanks: 730 Math Focus: Physics, mathematical modelling, numerical and computational solutions 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! Thanks from perfect_world
May 29th, 2014, 07:26 AM   #6
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Quote:
 Originally Posted by Benit13 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.

 May 29th, 2014, 07:40 PM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,932 Thanks: 1127 Math Focus: Elementary mathematics and beyond 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.
 May 29th, 2014, 09:05 PM #8 Newbie   Joined: May 2014 From: east freedom pennsylvania Posts: 5 Thanks: 0 Thanks Awesome cd!
 May 30th, 2014, 02:04 AM #9 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,150 Thanks: 730 Math Focus: Physics, mathematical modelling, numerical and computational solutions $\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 $\displaystyle \left( \begin{array}{ccc} A \\ B \\ C \end{array} \right) =\frac{1}{40}\left( \begin{array}{ccc} -4 & 14 & -8 \\ -2 & -3 & 36 \\ 8 & -8 & -24 \end{array} \right) \left( \begin{array}{ccc} 398 \\ 164 \\ 58 \end{array} \right) = \left( \begin{array}{ccc} 6 \\ 20 \\ 12 \end{array} \right)$.
 May 30th, 2014, 02:53 AM #10 Senior Member   Joined: Apr 2014 From: UK Posts: 903 Thanks: 331 If they are making 3 add-ons, then the time to make 1 of each is all that's required, then just copy it Thanks from perfect_world

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