A florist plans to make 5 Identical bridesmaid bouquets for a weeding. She budgeted £ 610 and she want to 24 flowers for each bouquet. She want to use three types of flowers namely; roses, tulips and lilies . A survey of prices shows roses cost £6, tulips £4 and lilies £3 respectively. She want to have twice as many roses as the other two flowers combined in each bouquet. Use matrix algebra to find the number of roses, tulips and lilies she as buy.
R, T, L are the flowers in each bouquet
total budget per bouqet=cost of flowers/per bouquet
610/5=R*6 + T*4 + L*3
R=2(T+L)
24=R+T+L
that are three unknowns, three equations.
R....T.....L.....=
1... 1....1.....24
1...-2...-2.....0
6....4....3....122
I put these into my calculator, http://www.wolframalpha.com/widgets/view.jsp?id=db5bdc8ad46ab6087d9cdfd8a8662ddf
and the answers you get are for each bunch, for totals, you have to multiply by five.
Let's use matrix algebra to solve this problem.
Let's represent the number of roses, tulips, and lilies she needs to buy for each bouquet respectively as variables x, y, and z.
We have three equations based on the given conditions:
1. The total cost of each bouquet should not exceed £610. Using the given prices, this can be represented as:
6x + 4y + 3z ≤ 610
2. She wants twice as many roses as the other two flowers combined in each bouquet. This can be represented as:
x = 2(y + z)
3. She wants to make 5 identical bouquets, so the total number of flowers of each type should be 24 * 5 = 120. Therefore, we have:
x + y + z = 120
Now, let's use matrix algebra to solve these equations.
Step 1: Write the augmented matrix using the coefficients of the variables:
[ 6 4 3 | 610 ]
[ 1 -2 -2 | 0 ]
[ 1 1 1 | 120 ]
Step 2: Perform row operations to get row-echelon form or reduced row-echelon form.
Using matrix row operations, we perform the following operations:
R2 = R2 - R1 (Multiply R1 by -1 and add it to R2)
R3 = R3 - R1 (Multiply R1 by -1 and add it to R3)
R2 = R2/(-2) (Divide R2 by -2 to make the coefficient of x equal to 1)
The augmented matrix after these operations becomes:
[ 6 4 3 | 610 ]
[ 0 -10 -10 | -610 ]
[ 0 -3 -2 | -490 ]
Step 3: Continue the row operations to simplify the matrix further.
R3 = R3 - (-(3/10)) * R2 (Multiply R2 by -(3/10) and add it to R3)
R2 = R2/(-10) (Divide R2 by -10)
R1 = R1 - (4/6) * R2 (Multiply R2 by (4/6) and add it to R1)
The augmented matrix after these operations becomes:
[ 6 0 0 | 200 ]
[ 0 1 1 | 61 ]
[ 0 0 1 | 40 ]
Step 4: Convert the matrix back into equations:
6x = 200 -> x = 33.33 (approx.)
y + z = 40
y + z = 40
Step 5: Solve for y and z:
y + z = 40
Since there are no further constraints specified, y and z can take any values as long as their sum is 40. For example, y can be 20 and z can be 20.
Therefore, the florist needs to buy approximately 33 roses, 20 tulips, and 20 lilies in each bouquet.
To solve this problem using matrix algebra, we can set up a system of equations based on the given information.
Let's denote the number of roses, tulips, and lilies she needs to buy as x, y, and z respectively. From the information given, we have the following equations:
Equation 1: 6x + 4y + 3z = 610 (the total cost should be equal to £610)
Equation 2: x = 2(y + z) (she wants twice as many roses as the other two flowers combined)
We can rewrite Equation 2 as:
x - 2y - 2z = 0
Now, we can express these equations in matrix form. Let A be the coefficient matrix, X the variable matrix, and B the solutions matrix. We have:
A = | 6 4 3 |
| 1 -2 -2 |
X = | x |
| y |
| z |
B = | 610 |
| 0 |
To find the variable matrix X, we can solve the matrix equation AX = B by multiplying both matrices on the left and then finding the inverse of A:
AX = B
X = A^(-1) * B
First, let's find the inverse of matrix A:
A^(-1) = 1 / (6 * -2 + 4 * 1) * | -2 -4 |
| -1 -3 |
Multiplying A^(-1) and B, we get:
X = | -2 -4 | * | 610 |
| -1 -3 | | 0 |
Simplifying this multiplication, we find:
X = | (-2 * 610) + (-4 * 0) | | -1220 |
| (-1 * 610) + (-3 * 0) | | -610 |
Therefore, x = -1220 and y = -610.
Since the number of flowers cannot be negative, we can conclude that there is no solution to this system of equations. The florist's plan is not possible given the budget and constraints mentioned.