College Algebra, Chapter 7, 7.2, Section 7.2, Problem 46

A specialty-car manufacturer has plants in Auburn, Biloxi and Chattanooga. Three models are produced, with daily production given in the following matrix.







Because of a wage increase, February profits are lower than January profits. The profit per car is tabulated by model in the following matrix.







a.) Calculate $AB$.

b.) Assuming that all cars produced were sold, what was the daily profit in January from the Biloxi plant?

c.) What was the total daily profit (from all three plants) in February?



a.) $\displaystyle AB = \left[ \begin{array}{ccc}
12 & 10 & 0 \\
4 & 4 & 20 \\
8 & 9 & 12
\end{array} \right] \left[ \begin{array}{cc}
1000 & 500 \\
2000 & 1200 \\
1500 & 1000
\end{array} \right] =
\left[ \begin{array}{cc}
12(1000) + 10(2000) + 0 (1500) & 12(500) + 10(1200) + 0 (1000) \\
4(1000) + 4 (2000) + 20(1500) & 4(500) + 4(1200) + 20(1000) \\
8(1000) + 9 (2000) + 12(5000) & 8(500) + 9(1200) + 12(1000)
\end{array} \right] =
\left[ \begin{array}{cc}
32000 & 18000 \\
42000 & 26800 \\
44000 & 26800
\end{array} \right]

$

b.) When we take the inner product of a row in $A$ with a column in $B$, we are multiplying the number of units sold together with its unit price per each type of car, then the result are added to produce the total profit from a specific plant. So the profit in January from the Biloxi Plant is

profit = $4(1000) + 4 (2000) + 20(1500) = \$ 42,000$

c.) Then the profit in February from all three plants is


$
\begin{equation}
\begin{aligned}

\text{profit } =& 12(500) + 10 (1200) + 0(1000) + 4(500) + 4(1200) + 20(1000) + 8(500) + 9(1200) + 12(1000)
\\
=& \$ 71,600

\end{aligned}
\end{equation}
$