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July 2nd, 2017, 03:34 PM  #1 
Newbie Joined: Jun 2017 From: Wisconsin Posts: 15 Thanks: 0  Stuck on a GaussJordan Word Problem
I'm stuck on one of the problems. A similar question from the book asks: An electronics company produces three models of stereo speakers, models A, B and C, and can deliver them by truck, van or SUV. A truck holds 2 boxes of model A, 2 boxes of model B and 3 boxes of model C. A van holds 3 boxes of model A, 4 boxes of model B and 2 boxes of model C. An SUV holds 5 boxes of model A, 7 boxes of model B, and 1 box of model C. There are 2 parts to this question and I don't understand part b. b) Model C has been discontinued. If 25 boxes of model A and 33 boxes of model B, how many vehicles of each type should be used to operate at full capacity? I understand the math behind the problem but, am having trouble wrapping my head around how they arrived at the solution. Solving the matrix gives: 1 0 0 2 0 1 0 2 0 0 1 3 Since model C is discontinued, you'll have 2 equations. Solving this matrix gives: 1 0 1/2 1/2 0 1 2 8 So x = 1/2z + 1/2 and y = 8  2z It says "This equation shows that z must be an odd number for x to be a whole number and not a fraction." Why must z be an odd number? Solving the inequality: 8  2z > 0 z < 4 So z must be 1 or 3. Plug z into the equation when z = 1 to solve for x and y, gives you: x = 1 and y = 6 and when z = 3: x = 2 and y = 2 Therefore, the solution using the most trucks is to use 2 trucks, 2 vans, and 3 SUVs. The other solution is to use 1 trucks, 2 vans, and 3 SUVs. I don't understand where these values are coming from. Last edited by skipjack; July 2nd, 2017 at 04:50 PM. 
July 2nd, 2017, 06:01 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,142 Thanks: 1417 
I assume there are x trucks, y vans and z SUVs. Boxes of model A delivered = 2x + 3y + 5z = 25 Boxes of model B delivered = 2x + 4y + 7z = 33 Those equations can easily be solved to give x = (1/2)z + 1/2 and y = 8  2z. The equation x = 1/2z + 1/2 is equivalent to z = 2x  1. Hence if x is an integer, z is odd. If z is odd, y = 8  2z cannot be zero (as that would require z = 4). Hence if y is nonnegative, y > 0, and so 8  2z > 0, which implies z < 4. Hence if z is nonnegative (as well as being odd), it is 1 or 3. If z = 1, x = 1/2 + 1/2 = 1 and y = 8  2z = 6. If z = 3, x = 3/2 + 1/2 = 2 and y = 8  2z = 2. In summary (x, y, z) = (1, 6, 1) or (2, 2, 3). Although the question didn't ask for the information, the solution using the most trucks is (x, y, z) = (2, 2, 3). The other solution, (x, y, z) = (1, 6, 1) uses 1 truck, 6 vans, and 1 SUV, contrary to what you posted. 
July 2nd, 2017, 07:54 PM  #3 
Newbie Joined: Jun 2017 From: Wisconsin Posts: 15 Thanks: 0 
Thank you for you answer! This problem is much more clear to me. I just have a few additional questions: 1. How do we know z is odd? I plugged in some values into z = 2x  1 and got odd numbers. Is this how we would know or is there a more obvious way? 2. Also I'm a little confused as to how you knew z couldn't equal 4 before solving the inequality. Last edited by skipjack; July 3rd, 2017 at 11:26 AM. 
July 3rd, 2017, 11:19 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,142 Thanks: 1417 
If x is an integer, 2x is even, and so 2x  1 is odd. As z is odd (assuming that x is an integer) and 4 is even, z cannot equal 4. 
July 3rd, 2017, 12:17 PM  #5 
Newbie Joined: Jun 2017 From: Wisconsin Posts: 15 Thanks: 0 
Thank you sooo much!


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