A manufacturer produces three products x, y, z which he sells in two markets.

Solution:

(a) 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 isRs 46000 and the same in market II isRs 53000.

(b) The unit cost prices of x, y, and z are respectively given as Rs 2.00, Rs 1.00, and 50 paise.

Consequently, the total cost prices of all the products 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}2.00 \\ 1.00 \\ 0.50\end{array}\right]$

$=10000 \times 2.00+2000 \times 1.00+18000 \times 0.50$

$=20000+2000+9000$

$=31000$

Since the total revenue in market I isRs 46000, the gross profit in this marketis (Rs 46000 − Rs 31000) Rs 15000.

The total cost prices of all the products in market II can be represented in the form of a matrix as:

$\left[\begin{array}{lll}6000 & 20000 & 8000\end{array}\right]\left[\begin{array}{l}2.00 \\ 1.00 \\ 0.50\end{array}\right]$

$=6000 \times 2.00+20000 \times 1.00+8000 \times 0.50$

$=12000+20000+4000$

$=$ Rs 36000

Since the total revenue in market II isRs 53000, the gross profit in this market is (Rs 53000 − Rs 36000) Rs 17000.