Miscellaneous · Q7: Manufacturer's revenue, cost and profit

Basics you need

Part (a): Revenue in each market

1. Build the sales matrix and the price matrix

   Sales matrix S: rows are markets (I, II), columns are products ($x, y, z$⁠), entries are units sold per year.

   $$S = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}$$

   Price matrix P: one row per product ($x, y, z$⁠), entries in rupees per unit.

   $$P = \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix}$$

   S is $2\times3$ and P is $3\times1$⁠.

   The table already has markets down the side and products across the top, so S copies it exactly. The prices are written as a column whose rows follow the same product order $x, y, z$ as the columns of S. That way, when a row of S meets the column P, units of $x$ are paired with the price of $x$⁠, and so on.
   A matrix records a table: entry $s_{ij}$ is in row $i$ (market) and column $j$ (product). Basics: rows, columns and the address $a_{ij}$

2. Why the product is SP (and not PS)

   $SP$ is $(2\times3)(3\times1)$⁠: the inner numbers agree (3 products), so it is defined, of order $2\times1$⁠, one entry per market. $PS$ would be $(3\times1)(2\times3)$⁠, and the inner numbers 1 and 2 disagree, so it is not defined.

   The order of the product is decided by meaning, and the orders confirm it. Revenue needs each market's sales of a product paired with that product's price, so the "products" index must be the one that is summed over. In SP the shared index is the 3 products, and the result is indexed by market, which is what we want.
   Section 3.4.5: AB is defined when the number of columns of A equals the number of rows of B. Lesson: matrix multiplication

3. Compute the revenue SP

   Each entry is one market's row of sales times the price column:

   $$\begin{aligned} \text{I} &: 10000(2.50) + 2000(1.50) \\ &\qquad + 18000(1.00) \\ &= 25000 + 3000 + 18000 \\ &= 46000 \\ \text{II} &: 6000(2.50) + 20000(1.50) \\ &\qquad + 8000(1.00) \\ &= 15000 + 30000 + 8000 \\ &= 53000 \end{aligned}$$ $$SP = \begin{bmatrix} 46000 \\ 53000 \end{bmatrix}$$

   So the total revenue is ₹46,000 in market I and ₹53,000 in market II.

   Each term is (units) $\times$ (rupees per unit) $=$ rupees, so each entry of SP is a sum of rupee amounts. For example, in market I product $x$ brings in $10000\times2.50 = 25000$ rupees.
   Row times column. Basics: pair up, multiply, add

Part (b): Cost and gross profit

1. Build the cost matrix and compute the cost SC

   50 paise is ₹0.50 (100 paise make a rupee). Cost matrix C, one row per product ($x, y, z$⁠), in rupees per unit, is $3\times1$⁠:

   $$C = \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix}$$

   As in Part (a), Step 2, $SC$ is $(2\times3)(3\times1)$⁠, so it is defined and $2\times1$⁠:

   $$\begin{aligned} \text{I} &: 10000(2.00) + 2000(1.00) \\ &\qquad + 18000(0.50) \\ &= 20000 + 2000 + 9000 \\ &= 31000 \\ \text{II} &: 6000(2.00) + 20000(1.00) \\ &\qquad + 8000(0.50) \\ &= 12000 + 20000 + 4000 \\ &= 36000 \end{aligned}$$ $$SC = \begin{bmatrix} 31000 \\ 36000 \end{bmatrix}$$
   All money must be in the same unit before we add or subtract, so 50 paise is converted to ₹0.50 first. The cost of producing the goods sold in a market is computed exactly like the revenue, with costs in place of prices.
   ₹1 = 100 paise, so 50 paise = ₹0.50.

2. Gross profit = revenue − cost

   Both SP and SC are $2\times1$⁠, so their difference is defined, entry by entry:

   $$\begin{aligned} \text{I} &: 46000 - 31000 = 15000 \\ \text{II} &: 53000 - 36000 = 17000 \end{aligned}$$ $$SP - SC = \begin{bmatrix} 15000 \\ 17000 \end{bmatrix}$$

   So the gross profit is ₹15,000 in market I and ₹17,000 in market II, a total of $15000 + 17000 = 32000$⁠, that is ₹32,000.

   "Gross profit" here means sales revenue minus the cost of the goods sold, before any other expenses. The question does not say whether it wants it per market or in total; the answer key gives the two market figures, and we give the total as well.
   Subtraction of matrices of the same order, Section 3.4.1. Basics: adding only like with like

3. A shortcut that checks the answer: $S(P - C)$

   By the distributive law, $SP - SC = S(P - C)$⁠. The profit per unit on each product is

   $$P - C = \begin{bmatrix} 2.50 - 2.00 \\ 1.50 - 1.00 \\ 1.00 - 0.50 \end{bmatrix} = \begin{bmatrix} 0.50 \\ 0.50 \\ 0.50 \end{bmatrix}$$

   so each unit sold earns ₹0.50, whatever the product. Then

   $$\begin{aligned} \text{I} &: 0.50(10000 + 2000 + 18000) \\ &= 0.50(30000) = 15000 \\ \text{II} &: 0.50(6000 + 20000 + 8000) \\ &= 0.50(34000) = 17000 \end{aligned}$$

   the same profits as in Part (b), Step 2 ✓.

   This is an independent route to the same numbers, so it confirms both the revenue and the cost calculations at once. It also explains the result in words: market II sells 34,000 units against 30,000 in market I, and at 50 paise profit per unit that is ₹2,000 more profit.
   Distributive law $A(B - C) = AB - AC$⁠, Section 3.4.6. Lesson: properties of the operations

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