# From Chapter 2 Engineering Costs and Cost Estimating ```From Chapter 2 Engineering Costs and Cost Estimating
1. A manufacturing firm’s specialty circuit board division has annual fixed costs of
\$100,000 and variable costs of \$20.00 per board. If they charge \$100 per circuit
board, how many circuit boards must they produce and sell in order to break
even?
(a) As many boards as possible
(b) 5
(c) 1,000
(d) 1,250
Solution: To breakeven, total costs = total revenue, where total costs = total fixed costs
+ total variable costs. \$100,000 + \$20X = \$100X. X = \$100,000/\$80 = 1250 circuit
boards.
2. A large utilities contractor bought five used trucks for \$100,000 from a smaller
bankrupt utilities contractor. Six years ago, the vehicles were purchased for
\$300,000. It is estimated that the annual maintenance costs for the vehicles will be
\$1,500 per year and each vehicle will have a salvage value of \$5,000 in four
years. Which of the following would the large utilities contractor not include in
her economic analysis?
(a) Original purchase price of \$300,000
(b) The used purchase price of \$100,000
(c) The annual maintenance costs of \$1,500
(d) The salvage value of the vehicles
Solution: The original purchase price of \$300,000 (answer (a)) is considered a sunk
cost and is irrelevant to the economic analysis for the large utilities contractor. She
should only be concerned with the costs that directly impact her business now such as
the salvage value of the vehicles, her purchase price of \$100,000, and the annual
maintenance costs.
3. In breakeven analysis, the profit at the breakeven point is equal to
(a) The total cost
(b) Zero
(c) The total revenue
(d) The variable cost multiplied by the number of items sold
Solution: In economic analysis, the breakeven point is the point at which the total
revenue is equal to the total cost and neither a profit nor a loss is incurred. Thus,
since total revenue – total cost = total profit and both revenue and cost are equal,
the total profit must be zero, so (b) is the correct answer.
4. A company manufactures electrical metering devices that monitor power quality. The
company’s fixed cost is \$68,000 per month. The variable cost is \$80 per metering device.
The selling price per device can be modeled by S = 170 – 0.05 Q where S is the selling
price and Q is the number of metering devices sold. How many metering devices must
the company sell per month in order to realize a maximum profit?
A. 900 metering devices
B. 1800 metering devices
C. 3400 metering devices
D. As many metering devices as it can
Solution:
Profit = total revenue − total cost
Total revenue = S  Q
Total cost = fixed cost + variable cost = 68,000 + 80Q
Profit = (170 – 0.05Q) Q – (68,000 + 80Q)
Profit = 170Q − 0.05Q 2 − 68,000 − 80Q
Profit = − 0.05Q 2 + 90Q – 68,000
To find the number of devices the company needs to sell per month, we take the first
derivative of the profit equation and set it equal to 0. After taking the first derivative,
double-check that your answer is a maximum by taking the second derivative. Recall
from calculus that maximums will have a negative second derivative.
Letting P = profit and taking the first derivative of the profit equation:
P
= − 0.10Q + 90
Q
Setting the first derivative equal to 0 and solving for Q:
0 = − 0.10Q + 90
0.10Q = 90
Q = 900 metering devices, so (a) is the correct answer.
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