SOLUTION: An electronics company produces three models of stereo speakers, models A, B, and C, and can deliver them by truck, van or station wagon. A truck holds 2 boxes of model A, 1 of mod

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Question 607392: An electronics company produces three models of stereo speakers, models A, B, and C, and can deliver them by truck, van or station wagon. A truck holds 2 boxes of model A, 1 of model B, and 3 of model C. A van holds 1 box of model A, 3 boxes of model B, and 2 boxes of model C. A station wagon holds 1 box of model A, 3 boxes of model B, and 1 box of model C. If 15 boxes of model A, 20 boxes of model B and 22 boxes of model C are to be delivered, how many vehicles of each type should be used so that all operate at full capacity?
I need help with setting up the equations for this problem. This is a matrix problem.
Is it something like this?
This is how I first translated it..
2x + 1y + 3z = truck
1x + 3y + 2z = van
1x + 3y + 1z = station wagon
Then I thought to maybe combine Boxes A, B, and C together within an equation:
Boxes A) 2x + 1y + 1z = 15
Boxes B) 1x + 3y + 3z = 20
Boxes C) 3x + 2y + 1z = 22
I'm a little lost.
Found 2 solutions by ewatrrr, stanbon:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi

Good Work on the Second Set-Up: x=#Trucks, y =#vans and z=#station wagons

2x + 1y + 1z = 15

1x + 3y + 3z = 20

3x + 2y + 1z = 22 (5, 2, 3)

x = 5, y=2 and z = 3 




 
Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables




  

  

  system%282%2Ax%2B1%2Ay%2B1%2Az=15%2C1%2Ax%2B3%2Ay%2B3%2Az=20%2C3%2Ax%2B2%2Ay%2B1%2Az=22%29

  

  

  

  First let A=%28matrix%283%2C3%2C2%2C1%2C1%2C1%2C3%2C3%2C3%2C2%2C1%29%29. This is the matrix formed by the coefficients of the given system of equations.

  

  

  Take note that the right hand values of the system are 15, 20, and 22 and they are highlighted here: 

  

  

  

  

  These values are important as they will be used to replace the columns of the matrix A.

  

  

  

  

  Now let's calculate the the determinant of the matrix A to get abs%28A%29=-5. To save space, I'm not showing the calculations for the determinant. However, if you need help with calculating the determinant of the matrix A, check out this solver.

  

  

  

  Notation note: abs%28A%29 denotes the determinant of the matrix A.

  

  

  

  ---------------------------------------------------------

  

  

  

  Now replace the first column of A (that corresponds to the variable 'x') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bx%5D (since we're replacing the 'x' column so to speak).

  

  

  

  

  

  

  Now compute the determinant of A%5Bx%5D to get abs%28A%5Bx%5D%29=-25. Again, as a space saver, I didn't include the calculations of the determinant. Check out this solver to see how to find this determinant.

  

  

  

  To find the first solution, simply divide the determinant of A%5Bx%5D by the determinant of A to get: x=%28abs%28A%5Bx%5D%29%29%2F%28abs%28A%29%29=%28-25%29%2F%28-5%29=5

  

  

  

  So the first solution is x=5

  

  

  

  

  ---------------------------------------------------------

  

  

  We'll follow the same basic idea to find the other two solutions. Let's reset by letting A=%28matrix%283%2C3%2C2%2C1%2C1%2C1%2C3%2C3%2C3%2C2%2C1%29%29 again (this is the coefficient matrix).

  

  

  

  

  Now replace the second column of A (that corresponds to the variable 'y') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5By%5D (since we're replacing the 'y' column in a way).

  

  

  

  

  

  

  Now compute the determinant of A%5By%5D to get abs%28A%5By%5D%29=-10.

  

  

  

  To find the second solution, divide the determinant of A%5By%5D by the determinant of A to get: y=%28abs%28A%5By%5D%29%29%2F%28abs%28A%29%29=%28-10%29%2F%28-5%29=2

  

  

  

  So the second solution is y=2

  

  

  

  

  ---------------------------------------------------------

  

  

  

  

  

  Let's reset again by letting A=%28matrix%283%2C3%2C2%2C1%2C1%2C1%2C3%2C3%2C3%2C2%2C1%29%29 which is the coefficient matrix.

  

  

  

  Replace the third column of A (that corresponds to the variable 'z') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bz%5D 

  

  

  

  

  

  

  Now compute the determinant of A%5Bz%5D to get abs%28A%5Bz%5D%29=-15.

  

  

  

  To find the third solution, divide the determinant of A%5Bz%5D by the determinant of A to get: z=%28abs%28A%5Bz%5D%29%29%2F%28abs%28A%29%29=%28-15%29%2F%28-5%29=3

  

  

  

  So the third solution is z=3

  

  

  

  

  ====================================================================================

  

  Final Answer:

  

  

  

  

  So the three solutions are x=5, y=2, and z=3 giving the ordered triple (5, 2, 3)

  

  

  

  

  Note: there is a lot of work that is hidden in finding the determinants. Take a look at this 3x3 Determinant Solver to see how to get each determinant.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
An electronics company produces three models of stereo speakers, models A, B, and C, and can deliver them by truck, van or station wagon.
A truck holds 2 boxes of model A, 1 of model B, and 3 of model C.
A van holds 1 box of model A, 3 boxes of model B, and 2 boxes of model C.
Astation wagon holds 1 box of model A,3 boxes of model B, and 1 box of model C.
------------------------------------------
If 15 boxes of model A, 20 boxes of model B and 22 boxes of model C are to be delivered, how many vehicles of each type should be used so that all operate at full capacity?
----
A's: 2t + 1v + 1s = 15
B's: 1t + 3v + 3s = 20
C's: 3t + 2v + 1s = 22
===================
Using a matrix method you get:
# of trucks = 5
# of vans = 2
# of station wagons = 3
====================
Cheers,
Stan H.
=============