Business Calculus Quiz: Limits In Context
20 questions · exam conditions
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Limits In ContextQuestion 1 of 20

A company's profit function is P(x)=−2x3+15x2+36x−50P(x) = -2x^3 + 15x^2 + 36x - 50 thousand dollars, where xx is the number of units produced (in hundreds). If lim⁡h→0P(4+h)−P(4)h=12\lim_{h \to 0} \frac{P(4+h) - P(4)}{h} = 12, what is the most accurate interpretation of this limit in the business context?

When producing 400 units, profit is increasing at a rate of $12,000 per hundred additional units produced
When producing 400 units, profit is increasing at a rate of $1,200 per additional unit produced
The company's profit will increase by exactly $12,000 if production increases from 400 to 500 units
The marginal profit at 400 units is $12 per unit, indicating optimal production level has been reached
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Business Calculus Quiz

Business Calculus Quiz: Limits In Context

Practice Limits In Context 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 Limits In Context, 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

A company's profit function is P(x)=−2x3+15x2+36x−50P(x) = -2x^3 + 15x^2 + 36x - 50 thousand dollars, where xx is the number of units produced (in hundreds). If lim⁡h→0P(4+h)−P(4)h=12\lim_{h \to 0} \frac{P(4+h) - P(4)}{h} = 12, what is the most accurate interpretation of this limit in the business context?

  1. When producing 400 units, profit is increasing at a rate of $12,000 per hundred additional units produced (correct answer)
  2. When producing 400 units, profit is increasing at a rate of $1,200 per additional unit produced
  3. The company's profit will increase by exactly $12,000 if production increases from 400 to 500 units
  4. The marginal profit at 400 units is $12 per unit, indicating optimal production level has been reached
Explanation: The limit represents the derivative P'(4) = 12, which is the instantaneous rate of change of profit with respect to x at x = 4. Since x is measured in hundreds of units and P(x) is in thousands of dollars, the rate is $12,000 per hundred additional units. Choice B has wrong units (should be per hundred units, not per unit). Choice C confuses instantaneous rate with total change over an interval. Choice D incorrectly suggests this rate indicates optimality.

Question 2

The total cost, in dollars, for a company to produce qq units of a product is given by the function C(q)C(q). The following limit is known: lim⁡h→0C(500+h)−C(500)h=32\lim_{h \to 0} \frac{C(500+h) - C(500)}{h} = 32 What is the best interpretation of this mathematical statement in a business context?

  1. The total cost to produce the first 500 units is divided equally, with each unit costing $32 on average.
  2. The total cost to produce 501 units will be exactly $32 more than the total cost to produce 500 units.
  3. When 500 units are being produced, the cost is increasing at a rate of $32 per unit. (correct answer)
  4. The price of each unit should be set to $32 in order to break even on the 500th unit produced.
Explanation: The expression lim⁡h→0C(500+h)−C(500)h\lim_{h \to 0} \frac{C(500+h) - C(500)}{h} is the definition of the derivative of the cost function, C′(q)C'(q), evaluated at q=500q=500. This represents the instantaneous rate of change of cost with respect to the number of units produced. Therefore, C′(500)=32C'(500) = 32 means that when production is at 500 units, the cost is increasing at a rate of $32 per unit. This value is also known as the marginal cost, which approximates the cost of producing the 501st unit.

Question 3

Let Cˉ(q)=C(q)q\bar{C}(q) = \frac{C(q)}{q} be the average cost function for manufacturing qq electronic components. A financial analyst determines that lim⁡q→∞Cˉ(q)=15\lim_{q \to \infty} \bar{C}(q) = 15 What is the most accurate business implication of this finding?

  1. The marginal cost to produce any additional component will eventually be $15.
  2. As the number of components produced becomes very large, the average cost per component approaches $15. (correct answer)
  3. The total cost of production will eventually stabilize and approach a maximum of $15.
  4. The fixed costs associated with setting up production are $15.
Explanation: The limit lim⁡q→∞Cˉ(q)=15\lim_{q \to \infty} \bar{C}(q) = 15 describes the long-run behavior of the average cost function. It means that as the production quantity qq increases without bound, the average cost per unit gets closer and closer to $15. This is often due to fixed costs being spread over a very large number of units.

Question 4

The cost C(x)C(x), in thousands of dollars, to remove xx percent of a toxic substance from a water supply is given by a function, where 0≤x<1000 \le x < 100. It is determined that lim⁡x→100−C(x)=+∞\lim_{x \to 100^-} C(x) = +\infty What does this limit imply about the cleanup operation?

  1. The cost of removing the substance increases at a constant rate as the percentage removed increases.
  2. Removing 100% of the substance is not just technologically difficult, but the cost to do so becomes prohibitively large. (correct answer)
  3. The total cost to remove almost all of the substance will be a very large, but finite, amount of money.
  4. The benefit of removing the last few percent of the substance is not economically justifiable.
Explanation: The limit indicates that as the percentage of the substance removed, xx, approaches 100% from the left side (since you can't remove more than 100%), the cost C(x)C(x) increases without bound. This represents a vertical asymptote at x=100x=100. In a business or economic context, this means the cost would become infinitely large, making it practically impossible to achieve 100% removal due to prohibitive costs.

Question 5

The profit, in thousands of dollars, from manufacturing xx hundred specialized microchips is given by the function P(x)P(x). A calculation shows that lim⁡h→0P(20+h)−P(20)h=4.5\lim_{h \to 0} \frac{P(20+h) - P(20)}{h} = 4.5 Based on this result, which conclusion is the most appropriate?

  1. The total profit from manufacturing 2,000 microchips is $4,500.
  2. When producing 2,000 microchips, the profit is increasing at a rate of $4.50 per microchip.
  3. The profit from selling the next 100 microchips after the first 2,000 will be exactly $4,500.
  4. At a production level of 2,000 microchips, the marginal profit is $4,500 per hundred microchips. (correct answer)
Explanation: The limit represents the derivative P′(20)P'(20). Since xx is in hundreds of units, x=20x=20 corresponds to 20×100=2,00020 \times 100 = 2,000 microchips. P(x)P(x) is in thousands of dollars. The derivative P′(x)P'(x) is therefore in units of (thousands of dollars) per (hundred units). So, P′(20)=4.5P'(20) = 4.5 means the marginal profit is $4.5 thousand dollars per hundred microchips, or $4,500 per hundred microchips, at the production level of 2,000 units.

Question 6

A company's profit is P(q)P(q) and its revenue is R(q)R(q) for selling qq items. An analyst finds that lim⁡q→∞P(q)R(q)=0.12\lim_{q \to \infty} \frac{P(q)}{R(q)} = 0.12 Assuming that revenue continues to grow as q→∞q \to \infty, what is the best interpretation of this limit?

  1. In the long run, the company's profit approaches 12% of its total revenue. (correct answer)
  2. For every additional item sold, the profit increases by a flat rate of 12 cents.
  3. The company's total profit will eventually stabilize at a maximum value of $0.12.
  4. The profit margin is increasing, and its rate of increase is 12% per unit.
Explanation: The expression P(q)R(q)\frac{P(q)}{R(q)} represents the profit margin, which is the ratio of profit to revenue. The limit describes the long-run behavior of this ratio. As the quantity produced and sold (qq) becomes very large, this ratio approaches 0.12, or 12%. This means that for very large scales of operation, the company's profit tends to be about 12% of its total revenue.

Question 7

A company is comparing two advertising campaigns. The number of new customers acquired from Campaign A is NA(t)N_A(t) and from Campaign B is NB(t)N_B(t), where tt is the number of days the campaign has run. At the two-week mark (t=14t=14), their respective rates of acquisition are: NA′(14)=lim⁡h→0NA(14+h)−NA(14)h=100N'_A(14) = \lim_{h \to 0} \frac{N_A(14+h) - N_A(14)}{h} = 100 NB′(14)=lim⁡h→0NB(14+h)−NB(14)h=120N'_B(14) = \lim_{h \to 0} \frac{N_B(14+h) - N_B(14)}{h} = 120 What is the most valid conclusion that can be drawn from this data?

  1. After 14 days, Campaign B has acquired more total customers than Campaign A.
  2. In the long run, Campaign B will be the more effective campaign for acquiring customers.
  3. Campaign B will generate exactly 20 more customers than Campaign A on the 15th day.
  4. On the 14th day, Campaign B is generating new customers at a higher rate than Campaign A. (correct answer)
Explanation: When you encounter derivatives in a business context, remember that the derivative represents the instantaneous rate of change at a specific point. The limit definition f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} tells you how fast something is changing at exactly time t=at = a. Here, NA′(14)=100N'_A(14) = 100 means Campaign A is acquiring new customers at a rate of 100 customers per day at the 14-day mark, while NB′(14)=120N'_B(14) = 120 means Campaign B is acquiring customers at 120 customers per day at that same moment. Since 120 > 100, Campaign B has a higher rate of customer acquisition on day 14, making D correct. Let's examine why the other options fail: A is wrong because derivatives tell us about rates, not cumulative totals. Campaign B could have a higher rate at day 14 but still have fewer total customers if it started poorly. B is incorrect because knowing the rate at one point doesn't predict long-term performance – rates can change over time. C makes a common error by confusing instantaneous rates with actual quantities. The derivative of 120 means the rate at day 14, not the number of customers gained on day 15. Study tip: Don't confuse derivatives (rates of change) with the original function values (actual quantities). When you see f′(a)f'(a), think "How fast is this changing at point aa?" not "What's the total value?" This distinction appears frequently in business calculus applications.

Question 8

The demand function for a product is q(p)q(p), where qq is the quantity demanded at price pp. The elasticity of demand is given by E(p)=−p⋅q′(p)q(p)E(p) = \frac{-p \cdot q'(p)}{q(p)}. If we know that for a certain price p0p_0, lim⁡p→p0q(p)=500andlim⁡h→0q(p0+h)−q(p0)h=−25\lim_{p \to p_0} q(p) = 500 \quad \text{and} \quad \lim_{h \to 0} \frac{q(p_0+h) - q(p_0)}{h} = -25 What can be concluded about the demand at price p0p_0?

  1. The demand is increasing at a rate of 25 units for every dollar increase in price.
  2. The total number of units that can ever be sold is limited to 500, with a maximum price of $25.
  3. The total revenue at price p0p_0 is decreasing because the quantity demanded is 500.
  4. The rate of change of demand with respect to price is −25-25, and the quantity demanded is approaching 500. (correct answer)
Explanation: When you encounter limits and derivatives together in a business calculus problem, you're being asked to interpret the fundamental definitions of these concepts in a real-world context. The given information tells us two key things through limit notation. First, lim⁡p→p0q(p)=500\lim_{p \to p_0} q(p) = 500 means that as the price approaches p0p_0, the quantity demanded approaches 500 units. Second, lim⁡h→0q(p0+h)−q(p0)h=−25\lim_{h \to 0} \frac{q(p_0+h) - q(p_0)}{h} = -25 is the definition of the derivative q′(p0)q'(p_0), which represents the instantaneous rate of change of quantity with respect to price at p0p_0. Option D correctly interprets both pieces of information: the rate of change of demand with respect to price is -25 (meaning demand decreases by 25 units per dollar price increase), and quantity demanded approaches 500. Option A misses the negative sign, incorrectly stating demand is increasing when it's actually decreasing. Option B misinterprets the limit as a maximum constraint on total units and confuses the derivative value with a price, which makes no mathematical sense. Option C incorrectly assumes we can determine revenue behavior from quantity alone—revenue depends on both price and quantity, and we'd need the actual price value p0p_0 to calculate revenue. Remember that limits describe approaching values (not necessarily achieved values), and the derivative limit formula lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} always represents instantaneous rate of change. Pay close attention to signs—they indicate direction of change.

Question 9

The number of active users on a new social media platform, in millions, is given by U(t)U(t), where tt is the number of months since its launch. The expression U(6)−U(3)3\frac{U(6) - U(3)}{3} represents the average rate of user growth over a specific period. Which limit precisely defines the instantaneous rate of user growth at the end of the 3rd month?

  1. lim⁡t→3U(t)\lim_{t \to 3} U(t)
  2. lim⁡h→0U(3+h)−U(3)h\lim_{h \to 0} \frac{U(3+h) - U(3)}{h} (correct answer)
  3. lim⁡t→6U(t)−U(3)t−3\lim_{t \to 6} \frac{U(t) - U(3)}{t-3}
  4. lim⁡h→0U(6+h)−U(3)h\lim_{h \to 0} \frac{U(6+h) - U(3)}{h}
Explanation: The instantaneous rate of change of a function f(x)f(x) at a point x=ax=a is given by the definition of the derivative, f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. To find the instantaneous rate of user growth at the end of the 3rd month (t=3t=3), we must apply this definition to the function U(t)U(t) at a=3a=3. This yields the expression lim⁡h→0U(3+h)−U(3)h\lim_{h \to 0} \frac{U(3+h) - U(3)}{h}.

Question 10

The proficiency of a worker assembling a complex device is modeled by N(t)N(t), the number of devices assembled per hour after tt days of training. The rate of improvement is the derivative, N′(t)N'(t). It is observed that lim⁡t→∞N′(t)=0\lim_{t \to \infty} N'(t) = 0 What does this imply about the worker's long-term performance?

  1. Eventually, the worker will stop being able to assemble any devices, and their output will be zero.
  2. The worker's proficiency will increase indefinitely, but the rate of increase will slow down.
  3. The worker's hourly assembly rate will level off and approach a stable, constant value. (correct answer)
  4. The total number of devices the worker can ever assemble is finite and will approach a limit.
Explanation: The limit lim⁡t→∞N′(t)=0\lim_{t \to \infty} N'(t) = 0 means that the rate of change of the worker's proficiency approaches zero in the long run. It does not mean that the proficiency itself (N(t)N(t)) approaches zero. Instead, it implies that the worker's skill improvement diminishes over time, and their assembly rate N(t)N(t) stabilizes, or approaches a plateau. The worker is no longer getting faster, but maintains a consistent peak proficiency.

Question 11

The value of an investment, V(t)V(t), in thousands of dollars, is modeled as a function of time tt in years. The function is continuous for t>0t > 0. However, due to a sudden market crash at t=3t=3, the company must re-evaluate the asset. The value is modeled such that lim⁡t→3V(t)=50\lim_{t \to 3} V(t) = 50, but the officially recorded value is V(3)=35V(3) = 35. What is the best financial interpretation of this situation?

  1. The investment was depreciating and was expected to be worth $35,000 at year 3, but a market bubble temporarily inflated its value to $50,000.
  2. The rate at which the investment was losing value changed from $50,000/year to $35,000/year at the 3-year mark.
  3. Just before and just after 3 years, the investment's value was trending toward $50,000, but an immediate event caused a sudden write-down of its value on that day. (correct answer)
  4. The investment's value was consistently $50,000 before year 3 and consistently $35,000 after year 3.
Explanation: The statement lim⁡t→3V(t)=50\lim_{t \to 3} V(t) = 50 means that as time approached 3 years from both the left and the right, the value of the investment was heading towards $50,000. However, the actual function value $V(3) = 35indicatesthatattheprecisemomentindicates that at the precise momentt=3$, the value was different. This represents a removable discontinuity, best explained by a sudden, instantaneous event (like a market crash and subsequent accounting adjustment) that changed the asset's value from its expected trend.

Question 12

Let R(p)R(p) be the daily revenue in dollars from selling a smartphone app at a price of pp dollars. An analysis of the revenue function yields: lim⁡p→9.99R(p)−R(9.99)p−9.99=−500\lim_{p \to 9.99} \frac{R(p) - R(9.99)}{p - 9.99} = -500 What is the correct interpretation of this limit?

  1. When the price is set at $9.99, the revenue is decreasing at a rate of $500 per dollar increase in price. (correct answer)
  2. If the price is increased from $9.99 to $10.99, the total revenue will decrease by exactly $500.
  3. At a price of $9.99, the company sells 500 fewer units for each dollar the price is increased.
  4. The total revenue generated from selling the app at $9.99 is $500.
Explanation: The limit expression is the definition of the derivative, R′(p)R'(p), at p=9.99p=9.99. The value of the derivative, -500, represents the instantaneous rate of change of revenue with respect to price. The negative sign indicates that revenue is decreasing. Therefore, when the price is $9.99, revenue is decreasing at a rate of $500 for each one-dollar increase in the price $p$.

Question 13

The total shipping cost S(w)S(w) for a parcel of weight ww pounds is modeled by a function. The function has a discontinuity at w=50w=50, described by: lim⁡w→50−S(w)=40andlim⁡w→50+S(w)=35\lim_{w \to 50^-} S(w) = 40 \quad \text{and} \quad \lim_{w \to 50^+} S(w) = 35 Which business practice best explains this discontinuity?

  1. A fuel surcharge is applied to all parcels, which has a greater impact on lighter parcels.
  2. A bulk-shipping discount takes effect for parcels weighing 50 pounds or more. (correct answer)
  3. The average cost per pound of shipping is lower for parcels weighing more than 50 pounds.
  4. Parcels heavier than 50 pounds must be shipped by a more expensive freight method.
Explanation: The limits describe a jump discontinuity. As the weight ww approaches 50 pounds from below, the shipping cost approaches $40. As $w$ approaches 50 pounds from above, the cost approaches $35. The cost abruptly drops at the 50-pound mark. This is characteristic of a bulk discount where crossing a certain threshold (50 pounds) results in a lower price structure, possibly due to a flat rate or lower per-pound rate that makes the total cost jump down.

Question 14

The cost function for a manufacturing process is C(q)=0.1q2+5q+200C(q) = 0.1q^2 + 5q + 200 dollars, where qq is the quantity produced. An economist calculates lim⁡q→50C(q)−C(50)q−50\lim_{q \to 50} \frac{C(q) - C(50)}{q - 50} to analyze production efficiency. What does this limit represent, and what is its value?

  1. Average cost per unit at 50 units; equals $15.00 per unit
  2. Marginal cost at 50 units; equals $15.00 per additional unit produced (correct answer)
  3. Total cost savings when reducing production from 51 to 50 units; equals $15.00
  4. Fixed cost component of production; equals $15.00 regardless of quantity
Explanation: This limit is the definition of the derivative C'(50), which represents marginal cost at q = 50. Taking the derivative: C'(q) = 0.2q + 5, so C'(50) = 0.2(50) + 5 = 15. Choice A confuses marginal cost with average cost. Choice C misinterprets the limit as a discrete change rather than an instantaneous rate. Choice D incorrectly identifies this as fixed cost, which would be the constant term.

Question 15

A technology startup's user base grows according to U(t)=50000tt+10U(t) = \frac{50000t}{t + 10} users, where tt is months since launch. The growth rate at month 20 is calculated as lim⁡Δt→0U(20+Δt)−U(20)Δt\lim_{\Delta t \to 0} \frac{U(20 + \Delta t) - U(20)}{\Delta t}. If this limit equals 555.56, what strategic insight does this provide?

  1. The user base will reach exactly 50,555.56 users by month 21, indicating strong continued growth
  2. The startup should expect to gain 555.56 users on average each month for the foreseeable future
  3. At month 20, the user base is growing at approximately 556 new users per month instantaneously (correct answer)
  4. User acquisition costs are decreasing at a rate of $555.56 per month at the 20-month mark
Explanation: When you encounter a limit expression like lim⁡Δt→0U(20+Δt)−U(20)Δt\lim_{\Delta t \to 0} \frac{U(20 + \Delta t) - U(20)}{\Delta t}, you're looking at the definition of a derivative. This measures the instantaneous rate of change of the function U(t)U(t) at t=20t = 20. In business contexts, this tells you how fast something is changing at a specific moment in time. The given limit equals 555.56, which means U′(20)=555.56U'(20) = 555.56. Since U(t)U(t) represents users and tt represents months, the derivative U′(20)U'(20) has units of users per month. This is the instantaneous growth rate of the user base at month 20 — essentially a "snapshot" of how quickly users are being added at that precise moment. Answer C correctly interprets this as the instantaneous growth rate of approximately 556 new users per month at month 20. Answer A confuses the derivative with actual user count and incorrectly adds the growth rate to predict future users. Answer B misinterprets the instantaneous rate as a sustained average rate — the derivative tells you the rate at one moment, not a long-term average. Answer D completely misreads the context, introducing cost information that doesn't exist in the problem. Remember: when you see the limit definition of a derivative in business problems, you're always dealing with instantaneous rates of change. Don't confuse this with total amounts, averages over time, or unrelated business metrics. The derivative gives you the "speedometer reading" at one specific point.

Question 16

An investment account's value follows V(t)=5000(1.08)tV(t) = 5000(1.08)^t dollars after tt years. An investor calculates lim⁡t→10V(t)−10800t−10=864\lim_{t \to 10} \frac{V(t) - 10800}{t - 10} = 864 to analyze the investment performance. What does this limit reveal about the investment at year 10?

  1. The account value is increasing at $864 per year when the balance reaches $10,800 at year 10 (correct answer)
  2. The investor will earn exactly $864 in interest during the 10th year of the investment
  3. The average annual return over 10 years is $864, indicating consistent performance throughout the period
  4. The account will be worth $11,664 one year after reaching the $10,800 value
Explanation: The limit represents V'(10), the instantaneous rate of change of account value at t = 10 years. Since V(10) = 5000(1.08)^10 ≈ 10,800, this means when the account reaches $10,800, it's growing at $864 per year at that instant. Choice B confuses instantaneous rate with actual earnings over a year. Choice C misinterprets this as average return rather than instantaneous rate. Choice D uses the instantaneous rate incorrectly to predict future value.

Question 17

A manufacturing company's efficiency function is E(n)=100nn+25E(n) = \frac{100n}{n + 25} percent, where nn is the number of experienced workers. The production manager finds that lim⁡n→75E(n)−E(75)n−75=0.25\lim_{n \to 75} \frac{E(n) - E(75)}{n - 75} = 0.25. How should this result guide staffing decisions?

  1. Adding one more experienced worker when the team has 75 will increase efficiency to exactly 75.25%
  2. The optimal team size is 75 experienced workers because the efficiency gain rate is maximized at this point
  3. When the team has 75 experienced workers, efficiency is improving at a rate of 0.25 percentage points per additional worker (correct answer)
  4. Efficiency improvements will continue at 0.25 percentage points per worker for any team size above 75 workers
Explanation: When you encounter a limit expression like lim⁡n→75E(n)−E(75)n−75\lim_{n \to 75} \frac{E(n) - E(75)}{n - 75}, you're looking at the definition of a derivative at a specific point. This limit represents E′(75)E'(75), the instantaneous rate of change of efficiency when there are 75 experienced workers. The result of 0.25 tells us that at exactly 75 workers, efficiency is increasing at a rate of 0.25 percentage points per additional worker. This is what answer choice C correctly states - it's interpreting the derivative as a rate of change at that specific point. Let's examine why the other options miss the mark. Choice A misunderstands what the derivative represents - it gives the instantaneous rate, not the actual change from adding one worker. The derivative approximates change for small increments but isn't exact. Choice B incorrectly assumes this rate represents an optimal point. A positive derivative simply means efficiency is still increasing, but we'd need to find where E′(n)=0E'(n) = 0 to locate a maximum. Choice D makes the critical error of assuming the rate of change remains constant beyond 75 workers. Derivatives typically vary as the input changes - this 0.25 rate applies specifically at n=75n = 75. Remember that when you see the limit definition of a derivative in business contexts, it's asking about instantaneous rates of change at specific points, not overall optimization or constant rates across ranges. Focus on what the derivative means at that exact input value.

Question 18

A company's revenue function is R(x)=50x−0.5x2R(x) = 50x - 0.5x^2 thousand dollars for xx thousand units sold. The marketing department reports that lim⁡x→30−R(x)−R(30)x−30=20\lim_{x \to 30^-} \frac{R(x) - R(30)}{x - 30} = 20 and lim⁡x→30+R(x)−R(30)x−30=20\lim_{x \to 30^+} \frac{R(x) - R(30)}{x - 30} = 20. What can be concluded about the marginal revenue at 30,000 units?

  1. Marginal revenue exists and equals $20,000 per thousand additional units, indicating revenue is still increasing (correct answer)
  2. Marginal revenue is undefined at exactly 30,000 units due to a discontinuity in the revenue function
  3. Marginal revenue equals $20 per unit, and this represents the maximum possible marginal revenue
  4. The company should produce exactly 30,000 units because marginal revenue equals average revenue at this point
Explanation: Since both one-sided limits equal 20, the limit exists and equals 20, meaning the marginal revenue exists and equals $20,000 per thousand units (or $20 per unit). Since this is positive, revenue is still increasing at this production level. Choice B is wrong because both one-sided limits exist and are equal. Choice C has correct value but incorrect claim about maximum. Choice D makes an unsupported claim about optimality and incorrectly relates marginal to average revenue.

Question 19

A consulting firm tracks client satisfaction using the function S(t) = 85 + 15te^(-0.2t), where S(t) represents the satisfaction score (0-100 scale) and t is months after implementing a new service protocol.

If the rate of change of satisfaction at t = 5 months is found using lim⁡h→0S(5+h)−S(5)h=2.03\lim_{h \to 0} \frac{S(5+h) - S(5)}{h} = 2.03, what is the most appropriate interpretation for management?

  1. Client satisfaction will increase by exactly 2.03 points over the next month from the 5-month mark
  2. Client satisfaction improvement will accelerate by 2.03 points per month starting at month 5
  3. The average satisfaction increase over the first 5 months has been 2.03 points per month
  4. At the 5-month point, client satisfaction is increasing at a rate of approximately 2.03 points per month (correct answer)
Explanation: When you encounter a limit expression like lim⁡h→0S(5+h)−S(5)h\lim_{h \to 0} \frac{S(5+h) - S(5)}{h}, you're looking at the definition of a derivative—specifically, the instantaneous rate of change at a single point. This measures how fast the function is changing at exactly t = 5 months, not over an interval. The value 2.03 represents the derivative S'(5), which tells you the instantaneous rate at which satisfaction is changing at the 5-month mark. Since the units of S(t) are satisfaction points and t is in months, the derivative has units of "points per month." This means at month 5, satisfaction is increasing at a rate of approximately 2.03 points per month. Answer D correctly interprets this as the instantaneous rate of change at t = 5. Answer A incorrectly suggests this rate will continue unchanged for an entire month—but instantaneous rates can vary continuously. Answer B misinterprets the derivative as acceleration (which would be the second derivative, S''(t)), confusing rate of change with rate of change of the rate of change. Answer C incorrectly describes this as an average rate over the first 5 months, but the limit definition gives you the instantaneous rate at a single point, not an average over an interval. Remember: when you see the limit definition of a derivative, you're always dealing with instantaneous rates at a specific point, not averages over intervals or predictions about future behavior. The derivative tells you "right now" information, not "over time" information.

Question 20

A population model for a city predicts P(t)=800001+15e−0.3tP(t) = \frac{80000}{1 + 15e^{-0.3t}} people after tt years. Urban planners need to understand the growth rate when the population reaches 40,000. If lim⁡h→0P(t0+h)−P(t0)h=600\lim_{h \to 0} \frac{P(t_0 + h) - P(t_0)}{h} = 600 when P(t0)=40000P(t_0) = 40000, what is the practical significance of this limit?

  1. The population will reach exactly 40,600 people one year after reaching 40,000 people
  2. At the moment when population is 40,000, it is growing at an instantaneous rate of 600 people per year (correct answer)
  3. The average population growth rate from year 0 until population reaches 40,000 is 600 people per year
  4. The population growth will accelerate by 600 people per year squared when population exceeds 40,000 people
Explanation: The limit represents P'(t₀), the instantaneous rate of change of population at the specific time when P(t₀) = 40,000. This is the instantaneous growth rate of 600 people per year at that moment. Choice A confuses instantaneous rate with actual change over a year. Choice C misinterprets this as average rate over an interval. Choice D incorrectly describes this as acceleration (second derivative) rather than velocity (first derivative).