Business Calculus Quiz: Units And Dimensional Consistency

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Units And Dimensional ConsistencyQuestion 1 of 20

A population model gives $P(t) = 5000e^{0.03t}$ people, where $t$ is years since 2020. The expression $\frac{1}{P(t)} \frac{dP}{dt}$ represents the relative growth rate. If we multiply this by 100 to express it as a percentage, what are the final units?

dimensionless percentage growth coefficient

percent per person per year

percent per year (or %/year)

years per percent per population change

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Business Calculus Quiz

Business Calculus Quiz: Units And Dimensional Consistency

Practice Units And Dimensional Consistency in Business Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Units And Dimensional Consistency, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

dimensionless percentage growth coefficient
percent per person per year
percent per year (or %/year) (correct answer)
years per percent per population change

Explanation: When you encounter questions about relative growth rates, you're dealing with a fundamental concept in business calculus that measures how fast something grows compared to its current size. Let's work through the units systematically. We start with

P(t) = 5000e^{0.03t}

people, so

P(t)

has units of people. The derivative

\frac{dP}{dt}

represents the rate of change of population, giving us units of people per year. Now for the relative growth rate:

\frac{1}{P(t)} \frac{dP}{dt}

. This equals

\frac{\text{people/year}}{\text{people}} = \frac{1}{\text{year}}

. The "people" units cancel out, leaving us with units of "per year" or

\text{year}^{-1}

. When we multiply by 100 to convert to percentage, we get "percent per year." Looking at the wrong answers: (A) suggests the result is dimensionless, but we still have the time component from the derivative. (B) incorrectly keeps "per person" in the units, missing that the division by

P(t)

eliminates the population dimension. (D) completely reverses the relationship and introduces unnecessary complexity. The correct answer is (C) because relative growth rate fundamentally measures the percentage change per unit of time. This is why in finance and economics, we express growth rates as "5% per year" rather than just "5%." Study tip: Remember that relative growth rates always have units of "per time" because you're measuring how the percentage changes over time. The population units always cancel out in the ratio.

Question 2

The rate of heat transfer in a manufacturing process is $H(t) = 50 + 10\sin(2t)$ BTU/hour, where $t$ is time in hours. The total heat transferred over an 8-hour period requires evaluating $\int_0^8 H(t) \, dt$ . If we then divide this result by the time period (8 hours) to find the average rate, what are the final units?

BTU (representing total heat transferred over the period)
BTU per hour (representing average heat transfer rate) (correct answer)
BTU per hour squared (representing heat transfer acceleration)
hours per BTU (representing thermal time efficiency)

Explanation: When you encounter problems involving rates and averages, focus carefully on how units behave through mathematical operations. Here, you're working with a rate function

H(t) = 50 + 10\sin(2t)

measured in BTU/hour. The integral

\int_0^8 H(t) \, dt

calculates the total heat transferred. Since you're integrating a rate (BTU/hour) over time (hours), the units become BTU/hour × hours = BTU. This gives you the cumulative amount of heat transferred over the 8-hour period. When you divide this total by 8 hours to find the average rate, you're performing:

\frac{\text{Total heat}}{\text{Time period}} = \frac{\text{BTU}}{8 \text{ hours}} = \text{BTU/hour}

This returns you to a rate unit, which makes sense because you're finding an average rate of heat transfer. Answer A is incorrect because it describes the units before division by time. The total heat transferred would indeed be in BTU, but that's not what the question asks for after dividing by the time period. Answer C represents an impossible unit for this context. BTU per hour squared would suggest some kind of acceleration of heat transfer, which isn't what averaging produces. Answer D flips the correct units upside down, giving a measure that would represent time per unit of heat rather than heat per unit of time. Remember this pattern: when you calculate an average rate by dividing a total quantity by time, you always end up with the same units as the original rate function.

Question 3

The value of a piece of equipment is $V(t)$ in dollars, where $t$ is the number of years since its purchase. Its rate of depreciation is given by $r(t) = -V'(t)$ , measured in dollars per year. Which expression represents the value of the equipment after 5 years, $V(5)$ ?

$\int_{0}^{5} r(t) \,dt$
$V(0) - \int_{0}^{5} r(t) \,dt$ (correct answer)
$V(0) + \int_{0}^{5} r(t) \,dt$
$r(5) - r(0)$

Explanation: By the Fundamental Theorem of Calculus, the integral of a rate of change gives the net change. So,

\int_{0}^{5} V'(t) \,dt = V(5) - V(0)

. Since

r(t) = -V'(t)

, we have

\int_{0}^{5} -r(t) \,dt = V(5) - V(0)

, which means

-\int_{0}^{5} r(t) \,dt = V(5) - V(0)

. Rearranging for

V(5)

, we get

V(5) = V(0) - \int_{0}^{5} r(t) \,dt

. In terms of units,

V(0)

is the initial value in dollars. The integral

\int_{0}^{5} r(t) \,dt

represents the total depreciation in dollars over 5 years. The initial value minus the total depreciation gives the final value.

Question 4

Let $C(t)$ be the total number of customers who have entered a store $t$ hours after it opens. The store manager observes that at 11 a.m. ( $t=2$ ), the rate at which customers are entering is decreasing. The statement $C''(2) = -50$ is proposed to model this situation. What are the units of the value $-50$ ?

customers per hour
customers per hour squared (correct answer)
customers
hours per customer

Explanation: The function

C(t)

has units of customers. The first derivative,

C'(t)

, represents the rate of change of customers with respect to time, so its units are customers per hour. The second derivative,

C''(t)

, represents the rate of change of the first derivative,

C'(t)

, with respect to time. Therefore, its units are (customers per hour) per hour, which is written as customers per hour squared. This represents the acceleration or deceleration of customer entry.

Question 5

The number of active users on a new social media platform, $N$ , is modeled as a function of time $t$ in days since its launch. Which of the following equations cannot be a valid model for $N(t)$ because it is dimensionally inconsistent?

$N(t) = 5000t e^{-0.1t}$
$N(t) = \frac{100000}{1 + 50e^{-0.2t}}$
$N(t) = 100t + 400\sqrt{t}$
$N(t) = 5000 + \ln(t + t^2)$ (correct answer)

Explanation: For an equation to be dimensionally consistent, terms that are added or subtracted must have the same units, and arguments of transcendental functions like logarithms must be dimensionless. In choice D, the argument of the natural logarithm is

t + t^2

. The variable

t

has units of days, and

t^2

has units of days squared. It is not possible to add quantities with different units (days + days squared). Furthermore, the argument itself is not dimensionless. Therefore, this model is dimensionally inconsistent. The other models can be made consistent by assigning appropriate units to the constants.

Question 6

The operating cost for a delivery truck consists of fuel and driver salary. The fuel cost rate, $F(v)$ , is $10 + 0.01v^2$ dollars per hour, where $v$ is the speed in miles per hour. The driver is paid $25 per hour. Which of the following expressions represents the total operating cost per mile?

$35 + 0.01v^2$
$0.02v$
$\frac{25 + F(v)}{v}$ (correct answer)
$\int (25 + F(v)) \,dv$

Explanation: The total operating cost rate is the sum of the driver's salary rate and the fuel cost rate:

25 + F(v)

. The units of this sum are dollars per hour. The truck's speed

v

has units of miles per hour. To find the cost per mile, we divide the cost per hour by the miles per hour:

\frac{\text{dollars}/ ext{hour}}{ ext{miles}/ ext{hour}} = \frac{\text{dollars}}{ ext{mile}}

. Thus, the expression is

\frac{25 + F(v)}{v}

Question 7

The rate of data flow into a server is $R(t)$ gigabytes (GB) per day, where $t$ is days from the start of the month. What is the correct interpretation of the expression $\frac{1}{10} \int_{0}^{10} R(t) \,dt$ ?

The total GB of data that flowed into the server during the first 10 days.
The rate of data flow on day 10.
The average rate of data flow in GB per day during the first 10 days. (correct answer)
The increase in the rate of data flow from day 0 to day 10.

Explanation: The expression

\frac{1}{b-a} \int_{a}^{b} f(x) \,dx

calculates the average value of the function

f(x)

on the interval

[a, b]

. Here, the expression calculates the average value of the rate function

R(t)

on the interval

[0, 10]

. The units of

R(t)

are GB/day. The units of the average value of a function are the same as the units of the function itself. Therefore, the expression represents the average rate of data flow in GB per day over the first 10 days.

Question 8

A company's production output, $Q$ , is given by a function $Q(K, L) = 100K^{0.3}L^{0.7}$ , where $K$ is the capital investment in units of $1,000 and $L $is the labor input in units of 1000 worker-hours. What are the units of the partial derivative$ \frac{\partial Q}{\partial L}$?

Units of output per $1,000.
Units of output per 1000 worker-hours. (correct answer)
Units of output per worker-hour.
Worker-hours per unit of output.

Explanation: The partial derivative

\frac{\partial Q}{\partial L}

measures the rate of change of the output

Q

with respect to the labor input

L

, holding capital

K

constant. The units of a derivative are the units of the dependent variable divided by the units of the independent variable. Here, the units of

Q

are 'units of output' and the units of

L

are '1000 worker-hours'. Therefore, the units of

\frac{\partial Q}{\partial L}

are 'units of output per 1000 worker-hours'.

Question 9

A manufacturing process is subject to a budget constraint given by the equation $150K + 40L = 20000$ , where $K$ is the number of machine-hours used and $L$ is the number of worker-hours used. If this equation is used to determine the marginal rate of technical substitution, represented by the absolute value of the derivative $\frac{dK}{dL}$ , what are the units of this derivative?

worker-hours per machine-hour
dollars per worker-hour
It is a dimensionless quantity.
machine-hours per worker-hour (correct answer)

Explanation: When you encounter a budget constraint equation like this, you're looking at the relationship between two inputs - machine-hours and worker-hours. The marginal rate of technical substitution tells you how much of one input you can trade for another while staying on the same budget. To find the units of

\frac{dK}{dL}

, start by solving the constraint equation for K:

150K + 40L = 20000

, so

K = \frac{20000 - 40L}{150}

. Taking the derivative:

\frac{dK}{dL} = -\frac{40}{150} = -\frac{4}{15}

. Since we want the absolute value, the marginal rate of technical substitution is

\frac{4}{15}

. The key insight is understanding what this fraction represents. The numerator comes from the coefficient of L (which has units of dollars per worker-hour), and the denominator comes from the coefficient of K (dollars per machine-hour). When you divide these coefficients, the dollar units cancel out, leaving you with machine-hours per worker-hour. Choice A reverses the units - it would be correct if we calculated

\frac{dL}{dK}

instead. Choice B incorrectly suggests the derivative retains monetary units, but the dollars cancel when dividing coefficients. Choice C claims it's dimensionless, missing that we're comparing two different types of hours, not the same units. Study tip: For marginal rates of substitution, always think "how much of the first variable per unit of the second variable." The units of

\frac{dK}{dL}

will always be "K-units per L-unit" - just follow the fraction format.

Question 10

A factory operates for an 8-hour shift ( $0 \le t \le 8$ ). The rate at which it incurs costs is $C(t)$ in dollars per hour, and the rate at which it produces goods is $P(t)$ in items per hour. Both rates vary over time. Which expression represents the average cost per item produced during the entire 8-hour shift?

$\frac{\int_{0}^{8} C(t) \,dt}{\int_{0}^{8} P(t) \,dt}$ (correct answer)
$\frac{1}{8} \int_{0}^{8} \frac{C(t)}{P(t)} \,dt$
$\frac{C(8) - C(0)}{P(8) - P(0)}$
$\int_{0}^{8} (C(t) - P(t)) \,dt$

Explanation: When you encounter a problem asking for "average cost per item," think about what this means in practical terms: total costs divided by total items produced. This requires understanding how to extract totals from rate functions using integration. Since

C(t)

represents the cost rate (dollars per hour) and

P(t)

represents the production rate (items per hour), you need to find the total cost and total production over the 8-hour shift. The integral

\int_{0}^{8} C(t) \,dt

gives you total costs in dollars, while

\int_{0}^{8} P(t) \,dt

gives you total items produced. Therefore, the average cost per item is

\frac{\int_{0}^{8} C(t) \,dt}{\int_{0}^{8} P(t) \,dt}

, making choice A correct. Choice B represents the average of the instantaneous cost-per-item ratios

\frac{C(t)}{P(t)}

over time, which is fundamentally different from the overall average cost per item. This would weight each moment equally rather than accounting for varying production levels. Choice C attempts to use endpoint values

C(8) - C(0)

and

P(8) - P(0)

, but these represent changes in rates, not totals. Since

C(t)

and

P(t)

are already rates, their differences have no meaningful interpretation for this problem. Choice D subtracts production rate from cost rate and integrates, giving you a meaningless quantity that mixes different units (dollars per hour minus items per hour). Study tip: When working with rate functions, remember that integration converts rates to totals. Always ask yourself: "Do I need the total amount or just the rate?" This distinction is crucial for setting up business calculus problems correctly.

Question 11

A factory's pollutant emission rate is modeled by $E(t) = \frac{100}{t+2}$ kilograms per hour (kg/h), where $t$ is the number of hours after the start of a shift. Which expression calculates the average emission rate in kilograms per minute over the first 5 hours of the shift?

$\frac{1}{300} \int_{0}^{5} E(t) \,dt$ (correct answer)
$\frac{1}{5} \int_{0}^{5} E(t) \,dt$
$12 \int_{0}^{5} E(t) \,dt$
$\frac{1}{60} \int_{0}^{5} E(t) \,dt$

Explanation: When you encounter average rate problems in calculus, you need to carefully track both the mathematical average formula and any unit conversions required. The average value of a function

f(t)

over interval

[a,b]

\frac{1}{b-a}\int_a^b f(t)\,dt

. Here, you want the average of

E(t)

over the first 5 hours, so that's

\frac{1}{5-0}\int_0^5 E(t)\,dt = \frac{1}{5}\int_0^5 E(t)\,dt

. However, this gives you the average in kg/hour, but the question asks for kg/minute. Since there are 60 minutes in an hour, you must divide by 60 to convert from kg/hour to kg/minute. This gives you

\frac{1}{5 \times 60}\int_0^5 E(t)\,dt = \frac{1}{300}\int_0^5 E(t)\,dt

, which is answer A. Answer B gives the average emission rate in kg/hour, not kg/minute—it's missing the unit conversion. Answer C multiplies by 12 instead of dividing by 60, which would give you an average rate 720 times too large and in the wrong units. Answer D divides by 60 but forgets to account for the 5-hour time interval in the average formula, giving you one-fifth of the correct answer. Remember that average value problems often include unit conversion traps. Always check whether your final answer has the units requested in the question, and don't forget that the average value formula requires dividing by the length of the time interval.

Question 12

The total cost, $C$ , of production for a company depends on the amount of raw material, $s$ (in kilograms). The amount of material needed depends on the number of units produced, $n$ . The number of units produced depends on the hours of labor, $h$ . The relationships are given by the functions $C(s)$ , $s(n)$ , and $n(h)$ . What are the units of the derivative $\frac{dC}{dh}$ ?

dollars per kilogram
kilograms per unit
dollars per hour (correct answer)
dollars per unit

Explanation: Using the chain rule,

\frac{dC}{dh} = \frac{dC}{ds} \cdot \frac{ds}{dn} \cdot \frac{dn}{dh}

. Let's analyze the units of each part: The units of

\frac{dC}{ds}

are dollars per kilogram. The units of

\frac{ds}{dn}

are kilograms per unit. The units of

\frac{dn}{dh}

are units per hour. Multiplying the units gives:

(\frac{\text{dollars}}{\text{kg}}) \cdot (\frac{\text{kg}}{\text{unit}}) \cdot (\frac{\text{units}}{\text{hour}}) = \frac{\text{dollars}}{\text{hour}}

Question 13

The population density surrounding a new retail location is given by $\rho(x)$ in people per square mile, where $x$ is the distance in miles from the location. Assuming the population is distributed symmetrically around the location, which integral correctly represents the total population within a 3-mile radius of the location?

$2\pi \int_{0}^{3} x \rho(x) \,dx$ (correct answer)
$\pi \int_{0}^{3} (\rho(x))^2 \,dx$
$\int_{0}^{3} \rho(x) \,dx$
$2\pi \int_{0}^{3} \rho(x) \,dx$

Explanation: When you encounter population density problems involving circular regions, you're working with polar integration concepts. The key insight is that population density gives you people per unit area, so you need to account for how much area exists at each distance from the center. The correct approach uses the fact that at distance

x

from the center, you have a thin circular ring with radius

x

and thickness

dx

. The area of this ring is

2\pi x \, dx

(circumference times thickness). Since

\rho(x)

gives population density at distance

x

, multiplying density by area gives you the population in that ring:

\rho(x) \cdot 2\pi x \, dx

. Integrating from 0 to 3 sums up all these rings, giving you

2\pi \int_{0}^{3} x \rho(x) \,dx

, which is answer A. Answer B incorrectly squares the density function and uses

\pi

, which would relate to finding area rather than population and makes no physical sense for this problem. Answer C simply integrates the density function without accounting for the circular geometry—this would work for a one-dimensional problem but ignores that we need area elements in two dimensions. Answer D includes the

2\pi

factor but omits the crucial

x

term, failing to account for how the circumference (and thus area) of each ring increases with distance from the center. Remember: in circular population problems, always include both the

2\pi

factor for circular symmetry and the radius term

x

to account for increasing ring areas as you move outward from the center.

Question 14

A manufacturer's cost function is $C(x) = 0.5x^2 + 10x + 200$ where $C$ is in dollars and $x$ is the number of items produced. The expression $\frac{C(x)}{x}$ represents the average cost per item. What happens to the units when we compute $\frac{d}{dx}\left(\frac{C(x)}{x}\right)$ ?

The units become dimensionless since we differentiate a ratio
The units remain dollars per item since derivatives preserve ratios
The units become items per dollar per unit change
The units become dollars per item per item produced (correct answer)

Explanation: When you encounter derivatives of economic functions, always track how units change through each mathematical operation. This question tests your understanding of how differentiation affects units in business contexts. Let's trace the units step by step. The cost function

C(x) = 0.5x^2 + 10x + 200

has units of dollars, and

x

represents items produced. So

\frac{C(x)}{x}

has units of

\frac{\text{dollars}}{\text{items}}

, which is dollars per item (average cost). When you differentiate

\frac{C(x)}{x}

with respect to

x

, you're finding how the average cost changes as production increases by one item. Since differentiation measures "change in output per unit change in input," you get:

\frac{d}{dx}\left(\frac{C(x)}{x}\right)

has units of

\frac{\text{dollars per item}}{\text{items}} = \frac{\text{dollars}}{\text{items}^2}

This translates to "dollars per item per item produced" – showing how average cost changes per additional item produced. Choice A is wrong because derivatives don't automatically make ratios dimensionless. Choice B incorrectly assumes derivatives preserve the original units rather than adding a "per unit change" dimension. Choice C reverses the relationship and creates nonsensical units. Choice D correctly captures that we're measuring the rate of change of average cost (dollars per item) with respect to production level (items), giving us dollars per item per item produced. Study tip: When differentiating any business function, always add "per unit change in the variable" to the original units. This systematic approach prevents unit confusion in optimization problems.

Question 15

The concentration of a drug in the bloodstream is modeled by $C(t) = \frac{10t}{t^2 + 4}$ mg/L, where $t$ is time in hours. When computing the related rate $\frac{dC}{dt} \cdot \frac{dt}{dV}$ where $V$ is blood volume in liters, what units result from this chain rule application?

mg per liter per liter (or mg/L²) (correct answer)
mg per liter per hour per liter
mg per hour (representing total drug metabolism rate)
liters per mg per hour (representing clearance efficiency)

Explanation: Using the chain rule: dC/dt has units (mg/L)/hour, and dt/dV has units hour/L. Their product is [(mg/L)/hour] × [hour/L] = mg/L². Choice B incorrectly keeps both hour terms. Choice C incorrectly suggests the hours cancel to give total metabolism. Choice D completely reverses the units and adds an incorrect interpretation.

Question 16

The demand function for a product is $D(p) = \frac{1000}{p^2}$ units, where $p$ is price in dollars. Consumer surplus is calculated as $CS = \int_{p_1}^{p_2} D(p) \, dp$ between two price levels. What units does this consumer surplus integral have?

dollars (representing total monetary consumer benefit)
units per dollar (representing purchasing efficiency)
dollar-units (a monetary measure of consumer benefit) (correct answer)
units squared per dollar (representing demand intensity)

Explanation: When you encounter questions about consumer surplus integrals, focus on dimensional analysis—multiply the units of what you're integrating by the units of integration to find the result's units. The demand function

D(p) = \frac{1000}{p^2}

gives you quantity demanded in units for any price

p

in dollars. When you integrate

\int_{p_1}^{p_2} D(p) \, dp

, you're multiplying the demand (units) by an infinitesimal change in price (

dp

, measured in dollars). This gives you units × dollars = dollar-units. Think of this geometrically: you're finding the area under the demand curve between two price points. The vertical axis shows quantity (units), the horizontal axis shows price (dollars), so the area has dimensions of units × dollars. Choice A is wrong because dollars alone would suggest pure monetary value, but we're integrating quantity over price, not price over quantity. Choice B (units per dollar) would result from integrating

\frac{1}{D(p)}

with respect to quantity, which isn't what we're doing here. Choice D (units squared per dollar) incorrectly suggests we're squaring the demand function somewhere in our calculation. Choice C correctly identifies that consumer surplus from this integral has units of dollar-units—a hybrid measure representing the monetary value of consumer benefit measured in quantity-price terms. Study tip: Always perform dimensional analysis on integrals by multiplying the integrand's units by the differential's units. This catches many common errors and helps you understand what economic quantities actually represent.

Question 17

A company's profit margin is $M(x) = \frac{R(x) - C(x)}{R(x)}$ where $R(x)$ is revenue in thousands of dollars and $C(x)$ is cost in thousands of dollars, both functions of $x$ thousand units produced. When we compute $\frac{dM}{dx}$ , what are the units of this derivative?

thousands of dollars per thousand units
per thousand units (or 1/thousand units) (correct answer)
dimensionless rate of margin change
thousand units per thousand dollars

Explanation: When analyzing derivatives in business calculus, always start by identifying the units of the original function, then determine how differentiation affects those units. The profit margin

M(x) = \frac{R(x) - C(x)}{R(x)}

is a ratio where both the numerator and denominator have units of "thousands of dollars." When you divide quantities with the same units, the result is dimensionless - profit margin is expressed as a decimal or percentage with no units. Since

M(x)

is dimensionless and

x

represents "thousand units," the derivative

\frac{dM}{dx}

has units of "dimensionless per thousand units," which simplifies to "per thousand units" or "1/thousand units." Looking at the wrong answers: Choice A suggests "thousands of dollars per thousand units," but this ignores that profit margin itself has no units since it's a ratio. Choice C states "dimensionless rate of margin change," which incorrectly treats the derivative as having no units - while the margin itself is dimensionless, its rate of change per unit of production is not. Choice D gives "thousand units per thousand dollars," which would be the reciprocal of what we need and represents the wrong type of relationship entirely. Study tip: When finding units for derivatives, use the pattern: if

f(x)

has units U and

x

has units V, then

\frac{df}{dx}

has units "U per V." Remember that ratios of quantities with identical units are dimensionless, but their derivatives with respect to other variables are not.

Question 18

The marginal cost to produce a specialized microchip is given by $C'(x) = 50 - 0.02x$ dollars per chip, where $x$ is the number of chips produced. A manager calculates the value of the definite integral $\int_{1000}^{1500} C'(x) \, dx$ . What do the units and value of this integral represent?

The total cost in dollars to produce the first 1500 chips.
The average cost in dollars per chip for producing chips between the 1000th and 1500th unit.
The additional total cost in dollars incurred to increase production from 1000 chips to 1500 chips. (correct answer)
The rate of change of cost, in dollars per chip, when production is at 1500 chips.

Explanation: The units of the integral

\int C'(x) \, dx

are the product of the units of

C'(x)

and the units of

x

. Here, that is (dollars/chip) * (chips) = dollars. The definite integral

\int_{a}^{b} C'(x) \, dx

represents the total change in the cost function

C(x)

x

increases from

a

b

. Therefore, the integral represents the total increase in cost (in dollars) when production is increased from 1000 to 1500 chips.

Question 19

Let $R(t)$ be the rate at which a company's revenue is generated, and $C(t)$ be the rate at which its costs are incurred. Both are measured in thousands of dollars per month, with $t$ in months. What is the correct interpretation of the area of the region enclosed between the graphs of $R(t)$ and $C(t)$ from $t=2$ to $t=8$ ?

The maximum profit achieved between the second and eighth months, measured in thousands of dollars.
The total accumulated net profit or loss from the end of the second month to the end of the eighth month, measured in thousands of dollars. (correct answer)
The average rate of profit increase between the second and eighth months, measured in thousands of dollars per month.
The point in time, in months, where the rate of revenue equals the rate of cost, indicating a potential peak profit.

Explanation: The area between the two curves is given by the integral

\int_{2}^{8} (R(t) - C(t)) \, dt

. The integrand,

R(t) - C(t)

, represents the rate of profit in thousands of dollars per month. Integrating this rate with respect to time (months) gives the total accumulated quantity. The units are (thousands of dollars/month) * (months) = thousands of dollars. The definite integral from

t=2

t=8

calculates the total net change in profit (i.e., accumulated profit or loss) over that specific time interval.

Question 20

A firm's output $Q$ is modeled by the production function $Q(L, K) = cL^{0.6}K^{0.4}$ , where $Q$ is in thousands of units, $L$ is labor in hundreds of worker-hours, and $K$ is capital in thousands of dollars. For the equation to be dimensionally consistent, what must be the units of the constant $c$ ?

Thousands of units per (hundred worker-hours) $^{0.6}$ (thousand dollars) $^{0.4}$ (correct answer)
Thousands of units per (hundred worker-hours $\cdot$ thousand dollars)
$c$ is a dimensionless constant of proportionality.
Thousands of units per (hundred worker-hours + thousand dollars)

Explanation: The equation must be dimensionally consistent, meaning the units on both sides must be equal. The units of

Q

are 'thousands of units'. The units on the right side are (units of

c

) * (units of

L

)

^{0.6}

* (units of

K

)

^{0.4}

. Plugging in the given units: thousands of units = (units of

c

) * (hundreds of worker-hours)

^{0.6}

* (thousands of dollars)

^{0.4}

. To find the units of

c

, we isolate it by dividing: Units of

c

= (thousands of units) / [(hundreds of worker-hours)

^{0.6}

* (thousands of dollars)

^{0.4}