Math Problem Statement

Solve the following problems by using the inverse of the matrix involved. (a) An automobile factory produces two models, A and B. Model A requires 1/2 labor hour to paint and 1 labor hour to polish; model B requires 1 labor hour for each process. During each hour that the assembly line is operating, there are 65 labor hours available for painting and 80 labor hours for polishing. How many of each model can be produced each hour if all the labor hours available are to be utilized? (b) Suppose each model A requires 10 widgets and 14 shims and each model B requires 7 widgets and 10 shims. The factory can obtain 830 widgets and 1170 shims each hour. How many cars of each model can it produce while using all the parts available?

Solution

To solve these problems, we need to set up a system of linear equations for each scenario and use matrix inversion to find the solutions.

Part (a)

The given information is:

  • Model A requires 12\frac{1}{2} labor hour for painting and 1 labor hour for polishing.
  • Model B requires 1 labor hour for each process.
  • There are 65 labor hours available for painting and 80 labor hours for polishing.

Let:

  • xx = number of Model A produced each hour
  • yy = number of Model B produced each hour

We can form the following equations:

  1. Painting hours: 12x+1y=65\frac{1}{2}x + 1y = 65
  2. Polishing hours: 1x+1y=801x + 1y = 80

This gives us the system:

\frac{1}{2}x + y = 65 \\ x + y = 80 \end{cases}$$ We can rewrite this system in matrix form: $$\begin{bmatrix} \frac{1}{2} & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 65 \\ 80 \end{bmatrix}$$ #### Solution To solve this, we find the inverse of the matrix and multiply it by the constants vector: $$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 1 \\ 1 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 65 \\ 80 \end{bmatrix}$$ --- ### Part (b) The given information is: - Model A requires 10 widgets and 14 shims. - Model B requires 7 widgets and 10 shims. - The factory has 830 widgets and 1170 shims available each hour. Let: - $$x$$ = number of Model A produced each hour - $$y$$ = number of Model B produced each hour We form the following equations: 1. **Widgets**: $$10x + 7y = 830$$ 2. **Shims**: $$14x + 10y = 1170$$ This gives us the system: $$\begin{cases} 10x + 7y = 830 \\ 14x + 10y = 1170 \end{cases}$$ We can rewrite this system in matrix form: $$\begin{bmatrix} 10 & 7 \\ 14 & 10 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 830 \\ 1170 \end{bmatrix}$$ #### Solution Similarly, we solve by finding the inverse of the matrix and multiplying it by the constants vector: $$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 14 & 10 \end{bmatrix}^{-1} \begin{bmatrix} 830 \\ 1170 \end{bmatrix}$$ --- I’ll proceed with the calculations for both parts.The solutions for the problems are as follows: ### Part (a) - **Model A**: 30 units per hour - **Model B**: 50 units per hour ### Part (b) - **Model A**: 55 units per hour - **Model B**: 40 units per hour This ensures all available labor hours and parts are utilized fully. Would you like further details on the calculations or have any questions? --- Here are some related questions for further practice: 1. How would the solution change if Model A's painting time increased to 1 hour? 2. How can you verify if the solutions for each part maximize the usage of resources? 3. What would happen if the total hours or parts available decreased slightly? 4. Can the inverse matrix method solve any system of linear equations? 5. How would the solution be affected if the factory's availability changes dynamically? **Tip**: Always check if a matrix is invertible before attempting to use its inverse to solve equations.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Inversion
System of Linear Equations

Formulas

Matrix form: AX = B
Inverse of matrix A: A⁻¹
Solution using inverse: X = A⁻¹B

Theorems

Matrix Inversion Theorem

Suitable Grade Level

College Algebra or High School Advanced Math

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