Question

An automobile company uses three types of steel S₁, S₂ and S₃ for producing three types of cars C₁, C₂and C₃. Steel requirements (in tons) for each type of cars are given below : 

	Cars C1	C2	C3
Steel S1	2	3	4
S2	1	1	2
S3	3	2	1

Using Cramer's rule, find the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of three types respectively.

Answer 1

Let x, y and z denote the number of cars that can be produced of each type . Then, 
\[2x + 3y + 4z = 29\] 
\[x + y + 2z = 13\] 
\[3x + 2y + z = 16\] 
Using Cramer's rule, we get
\[D = \begin{vmatrix}2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1\end{vmatrix}\] 
\[ = 2(1 - 4) - 3(1 - 6) + 4(2 - 3)\] 
\[ = - 6 + 15 - 4\] 
\[ = 5\] 
\[ D_1 = \begin{vmatrix}29 & 3 & 4 \\ 13 & 1 & 2 \\ 16 & 2 & 1\end{vmatrix}\] 
\[ = 29(1 - 4) - 3(13 - 32) + 4(26 - 16)\] 
\[ = - 87 + 57 + 40\] 
\[ = 10\] 
\[ D_2 = \begin{vmatrix}2 & 29 & 4 \\ 1 & 13 & 2 \\ 3 & 16 & 1\end{vmatrix}\] 
\[ = 2(13 - 32) - 29(1 - 6) + 4(16 - 39)\] 
\[ = - 38 + 145 - 92\] 
\[ = 15\] 
\[ D_3 = \begin{vmatrix}2 & 3 & 29 \\ 1 & 1 & 13 \\ 3 & 2 & 16\end{vmatrix}\] 
\[ = 2(16 - 26) - 3(16 - 39) + 29(2 - 3)\] 
\[ = - 20 + 69 - 29\] 
\[ = 20\] 
Thus,
\[x = \frac{D_1}{D} = \frac{10}{5} = 2\] 
\[y = \frac{D_2}{D} = \frac{15}{5} = 3\] 
\[z = \frac{D_3}{D} = \frac{20}{5} = 4\]
Therefore, 2 C₁ cars, 3 C₂ cars and 4 C_​3 cars can be produced using the three types of steel.