Question

An automobile company uses three types of steel S₁, S₂ and S₃ for producing three types of cars C₁, C₂ and 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

$$\begin{gathered} \mathrm{Let}x,y\mathrm{and}z\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{cars}\mathrm{that}\mathrm{can}\mathrm{be}\mathrm{produced}\mathrm{of}\mathrm{each}\mathrm{type}.\mathrm{Then}, \\ 2x+3y+4z=29 \\ x+y+2z=13 \\ 3x+2y+z=16 \\ \mathrm{Using}\mathrm{Cramer}\text{'}\mathrm{s}\mathrm{rule},\mathrm{we}\mathrm{get} \\ D=\left|\begin{array}{ccc}2& 3& 4\\ 1& 1& 2\\ 3& 2& 1\end{array}\right| \\ =2(1-4)-3(1-6)+4(2-3) \\ =-6+15-4 \\ =5 \\ {D}_1=\left|\begin{array}{ccc}29& 3& 4\\ 13& 1& 2\\ 16& 2& 1\end{array}\right| \\ =29(1-4)-3(13-32)+4(26-16) \\ =-87+57+40 \\ =10 \\ {D}_2=\left|\begin{array}{ccc}2& 29& 4\\ 1& 13& 2\\ 3& 16& 1\end{array}\right| \\ =2(13-32)-29(1-6)+4(16-39) \\ =-38+145-92 \\ =15 \\ {D}_3=\left|\begin{array}{ccc}2& 3& 29\\ 1& 1& 13\\ 3& 2& 16\end{array}\right| \\ =2(16-26)-3(16-39)+29(2-3) \\ =-20+69-29 \\ =20 \\ \mathrm{Thus}, \\ x=\frac{D_1}{D}=\frac{10}{5}=2 \\ y=\frac{D_2}{D}=\frac{15}{5}=3 \\ z=\frac{D_3}{D}=\frac{20}{5}=4 \end{gathered}$$

Therefore, 2 C₁ cars, 3 C₂ cars and 4 C_​3 cars can be produced using the three types of steel.