Question #216780

A company that manufactures cars used three different types of steels S1, S2, and S3 for producing three cars C1, C2, and C3. The tons of steel requirement in each type of car is given in below table Cars Steel C1 C2 C3 S1 1 1 1 S2 1 2 4 S3 4 2 3 Determine the number of cars of each type which can be produced using 45, 120 and 130 tones of steel of the three types respectively. (Hint: use gaussian method)


1
Expert's answer
2021-07-13T16:09:30-0400
C1C2C3S1111S2124S3423\def\arraystretch{1.5} \begin{array}{c:c:c:c} & C_1 & C_2 & C_3 \\ \hline S_1 & 1 & 1 & 1 \\ \hdashline S_2 & 1 & 2 & 4 \\ \hdashline S_3 & 4 & 2 & 3 \end{array}

x+y+z=45x+2y+4z=1204x+2y+3z=130\begin{matrix} x+y+z=45 \\ x+2y+4z=120 \\ 4x+2y+3z=130 \end{matrix}

(11145124120423130)\begin{pmatrix} 1 & 1 & 1 & & 45 \\ 1 & 2 & 4 & & 120 \\ 4 & 2 & 3 & & 130\\ \end{pmatrix}

R2=R2R1R_2=R_2-R_1


(1114501375423130)\begin{pmatrix} 1 & 1 & 1 & & 45 \\ 0 & 1 & 3 & & 75 \\ 4 & 2 & 3 & & 130\\ \end{pmatrix}

R3=R34R1R_3=R_3-4R_1


(111450137502150)\begin{pmatrix} 1 & 1 & 1 & & 45 \\ 0 & 1 & 3 & & 75 \\ 0 & -2 & -1 & & -50\\ \end{pmatrix}

R1=R1R2R_1=R_1-R_2


(102300137502150)\begin{pmatrix} 1 & 0 & -2 & & -30 \\ 0 & 1 & 3 & & 75 \\ 0 & -2 & -1 & & -50\\ \end{pmatrix}

R3=R3+2R2R_3=R_3+2R_2


(1023001375005100)\begin{pmatrix} 1 & 0 & -2 & & -30 \\ 0 & 1 & 3 & & 75 \\ 0 & 0 & 5 & & 100\\ \end{pmatrix}

R3=R3/5R_3=R_3/5


(102300137500120)\begin{pmatrix} 1 & 0 & -2 & & -30 \\ 0 & 1 & 3 & & 75 \\ 0 & 0 & 1 & & 20\\ \end{pmatrix}

R1=R1+2R3R_1=R_1+2R_3


(100100137500120)\begin{pmatrix} 1 & 0 &0 & & 10 \\ 0 & 1 & 3 & & 75 \\ 0 & 0 & 1 & & 20\\ \end{pmatrix}

R2=R23R3R_2=R_2-3R_3


(100100101500120)\begin{pmatrix} 1 & 0 &0 & & 10 \\ 0 & 1 & 0 & & 15 \\ 0 & 0 & 1 & & 20\\ \end{pmatrix}


x=10,y=15,z=20x=10, y=15, z=20

Number of cars produced by type C1,C2C_1,C_2 and C3C_3 are 10,1510,15 and 2020 respectively.



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