Determine the number of cars of each type which can be produced using 29, 13, and 24 tons of steel of three types respectively. Use Cramer's rule.

Steps to Solve

1. Set up the system of equations

From the problem, we can define the required steel for each type of car as follows:

For car C1, the steel requirements are:

• $2x + 3y + 3z = 29$ (for S1)
• $1x + 1y + 2z = 13$ (for S2)
• $3x + 2y + 2z = z$ (for S3)

Where:

• $x$ = number of C1 cars
• $y$ = number of C2 cars
• $z$ = number of C3 cars

So we will rewrite these equations as:

[ \begin{align*} 2x + 3y + 3z &= 29 \quad \text{(1)}\ 1x + 1y + 2z &= 13 \quad \text{(2)}\ 3x + 2y + 2z &= z \quad \text{(3)} \end{align*} ]

2. Rewrite in matrix form

We can express the system in the form of $AX = B$:

[ A = \begin{pmatrix} 2 & 3 & 3 \ 1 & 1 & 2 \ 3 & 2 & 2 \end{pmatrix}, \quad X = \begin{pmatrix} x \ y \ z \end{pmatrix}, \quad B = \begin{pmatrix} 29 \ 13 \ 0 \end{pmatrix} ]

3. Calculate the determinant of matrix A

Finding the determinant of matrix (A):

[ D = \text{det}(A) = 2(1 \cdot 2 - 2 \cdot 2) - 3(1 \cdot 2 - 2 \cdot 3) + 3(1 \cdot 2 - 1 \cdot 3) ]

Calculating this gives:

[ D = 2(2 - 4) - 3(2 - 6) + 3(2 - 3) = 2(-2) - 3(-4) + 3(-1) = -4 + 12 - 3 = 5 ]

4. Calculate determinants for (D_x), (D_y), (D_z)

For (D_x): [ D_x = \begin{vmatrix} 29 & 3 & 3 \ 13 & 1 & 2 \ 0 & 2 & 2 \end{vmatrix} = 29(12 - 22) - 3(132 - 0) + 3(132 - 1*0) = 29(-2) - 39 + 78 ]

Simplifying this:

[ D_x = -58 - 39 + 78 = -19 ]

For (D_y): [ D_y = \begin{vmatrix} 2 & 29 & 3 \ 1 & 13 & 2 \ 3 & 0 & 2 \end{vmatrix} ]

For (D_z): [ D_z = \begin{vmatrix} 2 & 3 & 29 \ 1 & 1 & 13 \ 3 & 2 & 0 \end{vmatrix} ]

5. Solve for x, y, z using Cramer's rule

Using Cramer's rule, we calculate:

[ x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D} ]

Insert the values we calculated and evaluate.