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A manufacturer produces three products $x,\, y,\, z$ which he sells in two markets. Annual sales are indicated below:
| Market | $x$ | $y$ | $z$ |
| $I$ | $10,000$ | $2,000$ | $18,000$ |
| $II$ | $6,000$ | $20,000$ | $8,000$ |
If unit sale prices of $x, \,y$ and $z$ are Rs. $2.50$, Rs. $1.50$ and Rs. $1.00,$ respectively, find the total revenue in each market with the help of matrix algebra.
$46000$ and $ 53000$
$46000$ and $ 53000$
$46000$ and $ 53000$
$46000$ and $ 53000$
Solution
The unit sale prices of $x, \,y$ and $z$ are respectively given as Rs $2.50,$ Rs $1.50,$ and Rs $1.00$
Consequently, the total revenue in market $I$ can be represented in the form of a matrix as
$\left[\begin{array}{lll}10000 & 2000 & 18000\end{array}\right]\left[\begin{array}{l}2.50 \\ 1.50 \\ 1.00\end{array}\right]$
$=10000 \times 2.50+2000 \times 1.50+18000 \times 1.00$
$=25000+3000+18000$
$=46000$
The total revenue in market $II$ can be represented in the form of a matrix as:
$\left[\begin{array}{lll}6000 & 20000 & 8000\end{array}\right]\left[\begin{array}{l}2.50 \\ 1.50 \\ 1.00\end{array}\right]$
$=6000 \times 2.50+20000 \times 1.50+8000 \times 1.00$
$=15000+30000+8000$
$=53000$
Therefore, the total revenue in market $I$ is Rs. $46000$ and the same in market $II$ is Rs. $ 53000$
