To solve this problem, we need to set up a system of equations based on the information provided about the hours required for each operation and the total available hours each day.
Let's denote:
- [tex]\( x \)[/tex] as the number of Basic sets produced each day,
- [tex]\( y \)[/tex] as the number of Homeowner sets produced each day,
- [tex]\( z \)[/tex] as the number of Pro sets produced each day.
The given table provides the following information on the hours required for each type of set:
[tex]\[
\begin{array}{|c|c|c|c|}
\hline
& \text{Basic} & \text{Homeowner} & \text{Pro} \\
\hline
\text{Painting} & 1.2 & 2 & 2.9 \\
\text{Assembly} & 0.8 & 1.4 & 1.7 \\
\text{Packaging} & 0.6 & 0.9 & 1.6 \\
\hline
\end{array}
\][/tex]
Also given are the total hours available for each operation per day:
- 91.4 hours for painting,
- 57.8 hours for assembly,
- 46.8 hours for packaging.
Using this information, we can set up the following system of linear equations:
1. For painting:
[tex]\[ 1.2x + 2y + 2.9z = 91.4 \][/tex]
2. For assembly:
[tex]\[ 0.8x + 1.4y + 1.7z = 57.8 \][/tex]
3. For packaging:
[tex]\[ 0.6x + 0.9y + 1.6z = 46.8 \][/tex]
To find [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex], we can represent this linear system in matrix form [tex]\( A \mathbf{x} = \mathbf{b} \)[/tex], where [tex]\( A \)[/tex] is the coefficient matrix, [tex]\( \mathbf{x} \)[/tex] is the vector of unknowns, and [tex]\( \mathbf{b} \)[/tex] is the vector of the right-hand side constants:
[tex]\[
A =
\begin{bmatrix}
1.2 & 2.0 & 2.9 \\
0.8 & 1.4 & 1.7 \\
0.6 & 0.9 & 1.6
\end{bmatrix}
,
\mathbf{x} =
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
,
\mathbf{b} =
\begin{bmatrix}
91.4 \\
57.8 \\
46.8
\end{bmatrix}
\][/tex]
To solve for [tex]\( \mathbf{x} \)[/tex], we can use the inverse of the matrix [tex]\( A \)[/tex], denoted [tex]\( A^{-1} \)[/tex]. The solution is given by:
[tex]\[
\mathbf{x} = A^{-1} \mathbf{b}
\][/tex]
The computed solution is:
[tex]\[
\mathbf{x} = \begin{bmatrix}
x \\
y \\
z
\end{bmatrix} = \begin{bmatrix}
5.999999999999552 \\
16.00000000000018 \\
18.000000000000064
\end{bmatrix}
\][/tex]
Rounding to the nearest whole number, we find:
[tex]\[
x \approx 6, \quad y \approx 16, \quad z \approx 18
\][/tex]
Therefore, the manufacturer can produce:
- [tex]\( 6 \)[/tex] Basic sets,
- [tex]\( 16 \)[/tex] Homeowner sets,
- [tex]\( 18 \)[/tex] Pro sets per day.
