A construction company is working on two building projects, Project X and Project Y. The q… — ACT

A construction company is working on two building projects, Project X and Project Y. The quantities of concrete (in tons) and steel (in tons) required for these projects are represented by the matrix $Q$ :

Q = \begin{bmatrix} 80 & 25 \\ 120 & 40 \end{bmatrix}

where the first row represents Project X, the second row represents Project Y, the first column represents concrete, and the second column represents steel.

The purchase cost per ton and the transportation cost per ton for these materials are represented by the matrix $C$ :

C = \begin{bmatrix} 110 & 15 \\ 650 & 45 \end{bmatrix}

where the first row represents concrete, the second row represents steel, the first column represents the purchase cost (in dollars per ton), and the second column represents the transportation cost (in dollars per ton).

If the product matrix $P = QC$ represents the total cost details for the two projects, what is the total transportation cost for Project Y, in dollars?

Answer: 3600 dollars

Answer

The total transportation cost for Project Y is 3,600 dollars.

The correct answer is 3,600. To find the total transportation cost for Project Y, we calculate the entry in the second row (Project Y) and second column (transportation cost) of the product matrix

P = QC

. This is calculated as

120 \text{ tons of concrete} \times 15 \text{ dollars per ton} + 40 \text{ tons of steel} \times 45 \text{ dollars per ton} = 1,800 + 1,800 = 3,600

dollars.

Step-by-Step Solution

Identify the row representing Project Y and the column representing transportation cost

Row 2 of

Q

[120, 40]

and Column 2 of

C

[15, 45]^T

To find the total transportation cost for Project Y, we must compute the entry in the second row (Project Y) and second column (transportation cost) of the product matrix

P = QC

Multiply the row elements by the corresponding column elements

120 \times 15 = 1800

and

40 \times 45 = 1800

This calculates the individual transportation costs for the concrete and steel required for Project Y.

Sum the products to find the total transportation cost

1800 + 1800 = 3600

Adding these individual costs gives the total transportation cost for Project Y.

Key Concept

Matrix multiplication involves multiplying the elements of each row of the first matrix by the corresponding elements of each column of the second matrix and summing the products.

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