A florist offers three sizes of flower arrangements containing roses, daisies, and crysan- themums. Each small arrangement contains one rose, three daisies, and three crysan- themums. Each medium...

A florist offers three sizes of flower arrangements containing roses, daisies, and crysan- themums. Each small arrangement contains one rose, three daisies, and three crysan- themums. Each medium arrangement contains two roses, four daisies, and six crysan- themums. Each large arrangement contains four roses, eight daisies, and six crysan- themums. One day the florist noted that a total of 24 roses, 50 daisies, and 48 crysan- themums had been used in filling orders for these three types of arrangements. (a)  Write down a system of linear equations that describes this information. (b)  Use Gaussian elimination to reduce the corresponding augmented matrix to row echelon form. (c)  Use back-substitution to find the numbers of each type of arrangement that were made.

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embizze | High School Teacher | (Level 1) Educator Emeritus

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To write the system of equations, assign variables. Let x be the number of small arrangements, y the number of medium, and z the number of large arrangements.

The first equation involves roses. Each small requires 1 rose, mediums take 2, and large require 4. There are a total of 24 roses so:

x+2y+4z=24 In a similar vein we get the other equations, so the system is:

x+2y+4z=24
3x+4y+8z=50
3x+6y+6z=48

The augmented matrix is:

`[[1,2,4,24],[3,4,8,50],[3,6,6,48]]`

A matrix is row reduced if in each row the first nonzero element is 1, and the entries in the column below the one are all zeros. For a 3x4 matrix, the reduced form looks like:

`[[1,x,x,x],[0,1,x,x],[0,0,1,x]]` where x is some real number.

Use row operations to reduce the matrix:

`[[1,2,4,24],[0,-2,-4,-22],[0,0,-6,-24]]` Add R2-3R1 for row 2 and R3-3R1 for row 3.

`[[1,2,4,24],[0,1,2,11],[0,0,1,4]]`  Multiply row 2 by -1/2 and row 3 by -1/6

This is in row reduced form.

z=4 from the last row.

y+8=11 ==> y=3 (Substitute z=4 into y+2z=11)

x+6+16=24 ==> x=2

So there were 2 small, 3 medium, and 4 large.

** Note that instead of back substitution, we could have put the matrix in reduced row echelon form -- entries above the leading 1 are converted to zero:

`[[1,2,4,24],[0,1,2,11],[0,0,1,4]] -> [[1,0,0,2],[0,1,2,11],[0,0,1,4]] -> [[1,0,0,2],[0,1,0,3],[0,0,1,4]]`

Now you can read the solution -- x=2,y=3,z=4 as above.

Sources:

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