Solve the following problem by using the inverse of the matrix involved.An automobile factory produces two models, A and B. Model A requires 1 labour hour to paintand 30 minutes of labour to polish; model B requires 1 labour hour for each process. Duringeach hour that the assembly line is operating, there are 100 labour hours available for paintingand 80 labour hours for polishing. How many of each model can be produced each hour if allthe labour hours available are to be utilised? Round your answers to the nearest whole number.
Question
Solve the following problem by using the inverse of the matrix involved.An automobile factory produces two models, A and B. Model A requires 1 labour hour to paintand 30 minutes of labour to polish; model B requires 1 labour hour for each process. Duringeach hour that the assembly line is operating, there are 100 labour hours available for paintingand 80 labour hours for polishing. How many of each model can be produced each hour if allthe labour hours available are to be utilised? Round your answers to the nearest whole number.
Solution
To solve this problem, we will use the concept of matrix inversion.
First, we need to represent the problem as a system of linear equations.
Let's denote the number of Model A cars produced as x and the number of Model B cars as y.
From the problem, we know that:
1x + 1y = 100 (This represents the labour hours for painting)
0.5x + 1y = 80 (This represents the labour hours for polishing)
We can represent this system as a matrix equation AX = B, where A is the matrix of coefficients, X is the column vector (x, y), and B is the column vector on the right side of the equation.
So, A = [[1, 1], [0.5, 1]], X = [[x], [y]], and B = [[100], [80]].
To find X, we need to multiply both sides of the equation by the inverse of A. The inverse of a 2x2 matrix [[a, b], [c, d]] is (1/(ad-bc))*[[d, -b], [-c, a]].
The determinant of A (ad-bc) is (11 - 10.5) = 0.5. So, the inverse of A is (1/0.5)*[[1, -1], [-0.5, 1]] = [[2, -2], [-1, 2]].
Multiplying the inverse of A by B gives us X = [[2, -2], [-1, 2]] * [[100], [80]] = [[0], [60]].
So, the factory can produce 0 of Model A cars and 60 of Model B cars per hour to fully utilise the available labour hours. Since we can't produce a fraction of a car, and we need to round to the nearest whole number, the factory can't produce any Model A cars and can produce 60 Model B cars per hour.
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