Business Calculus Quiz: Interpreting De Solutions
18 questions · exam conditions
0:00
Interpreting De SolutionsQuestion 1 of 18

The total revenue R(t)R(t) in millions of dollars from a new streaming service is modeled by the Gompertz function R(t)=500e2e0.3tR(t) = 500e^{-2e^{-0.3t}}, where tt is the number of years since launch. What is the projected maximum total revenue for this service according to the model?

500e2500e^{-2} million
500500 million
$0
The revenue is projected to grow indefinitely.
← Back to quizzes

Business Calculus Quiz

Business Calculus Quiz: Interpreting De Solutions

Practice Interpreting De Solutions 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 Interpreting De Solutions, 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

The total revenue R(t)R(t) in millions of dollars from a new streaming service is modeled by the Gompertz function R(t)=500e2e0.3tR(t) = 500e^{-2e^{-0.3t}}, where tt is the number of years since launch. What is the projected maximum total revenue for this service according to the model?

  1. 500e2500e^{-2} million
  2. 500500 million (correct answer)
  3. $0
  4. The revenue is projected to grow indefinitely.
Explanation: The maximum projected revenue corresponds to the long-term behavior of the function, which is found by taking the limit as tt \to \infty. As tt \to \infty, the exponent 0.3t-0.3t approaches -\infty. This causes e0.3te^{-0.3t} to approach 0. The exponent of the main function then becomes 2e0.3t2(0)=0-2e^{-0.3t} \to -2(0) = 0. Finally, the entire expression approaches 500e0=500(1)=500500e^0 = 500(1) = 500. Therefore, the model projects a maximum total revenue of $500 million.

Question 2

A retirement fund's value A(t)A(t) is modeled by dA/dt=0.06AWdA/dt = 0.06A - W, where AA is the fund value in dollars, tt is time in years, the interest rate is 6% compounded continuously, and WW is a constant annual withdrawal amount. What is the interpretation of the withdrawal amount WW that makes dA/dt=0dA/dt = 0?

  1. It is the withdrawal amount that causes the fund's value to deplete to zero most rapidly.
  2. It is the total amount of interest the fund will earn over its lifetime.
  3. It is the withdrawal amount where the amount withdrawn is exactly equal to the interest being earned. (correct answer)
  4. It is the minimum withdrawal amount required to keep the fund from growing.
Explanation: The condition dA/dt=0dA/dt = 0 represents an equilibrium state where the fund's value is not changing over time. Setting the equation to zero gives 0.06AW=00.06A - W = 0, which rearranges to W=0.06AW = 0.06A. The term 0.06A0.06A represents the interest earned by the fund at a given time. Therefore, when W=0.06AW = 0.06A, the amount being withdrawn is exactly equal to the interest being generated, which means the principal balance remains constant. This is the maximum sustainable withdrawal.

Question 3

The market share P(t)P(t) for a new product is modeled by P(t)=M1+AeMktP(t) = \frac{M}{1 + Ae^{-Mkt}}, where MM is the total market size and kk is a growth constant. The parameter AA is determined by the initial market share, P0P_0.

Which statement correctly describes the business implication of a larger value of the parameter AA?

  1. A larger AA implies a larger initial market share for the product.
  2. A larger AA implies a smaller initial market share for the product. (correct answer)
  3. A larger AA implies that the product's market share will grow faster.
  4. A larger AA implies a larger total addressable market size, MM.
Explanation: The parameter AA is determined by the initial condition P(0)=P0P(0) = P_0. Substituting t=0t=0 into the solution gives P0=M1+Ae0=M1+AP_0 = \frac{M}{1 + Ae^0} = \frac{M}{1+A}. Solving for AA, we get 1+A=MP01+A = \frac{M}{P_0}, so A=MP01A = \frac{M}{P_0} - 1. This equation shows an inverse relationship between AA and P0P_0. If the initial market share P0P_0 is smaller, the fraction M/P0M/P_0 is larger, resulting in a larger value for AA.

Question 4

A new software's market share, S(t)S(t), as a proportion of the total market, is modeled by the differential equation dS/dt=0.4S(1S)dS/dt = 0.4S(1-S), where tt is time in years. At what market share level is the company's software gaining market share most rapidly?

  1. 0%
  2. 40%
  3. 50% (correct answer)
  4. 100%
Explanation: The rate of change, dS/dtdS/dt, is a quadratic function of the market share SS. The expression 0.4S(1S)0.4S(1-S) represents a downward-opening parabola with roots at S=0S=0 and S=1S=1. The maximum value of a parabola occurs at its vertex, which is located halfway between the roots. Therefore, the rate of growth is maximized at S=(0+1)/2=0.5S = (0+1)/2 = 0.5, which corresponds to a 50% market share.

Question 5

The balance B(t)B(t) of a business loan after tt years is modeled by the solution B(t)=(B0P/r)ert+P/rB(t) = (B_0 - P/r)e^{rt} + P/r, where B0B_0 is the initial loan amount, rr is the annual interest rate compounded continuously, and PP is the annual payment amount made continuously.

The specific solution for a company's loan is given by B(t)=250000e0.05t+500000B(t) = 250000e^{0.05t} + 500000. Based on this model, what was the company's annual payment amount?

  1. $25,000 (correct answer)
  2. $250,000
  3. $500,000
  4. $750,000
Explanation: By comparing the general form B(t)=(B0P/r)ert+P/rB(t) = (B_0 - P/r)e^{rt} + P/r to the specific solution B(t)=250000e0.05t+500000B(t) = 250000e^{0.05t} + 500000, we can match the corresponding parts. The interest rate is r=0.05r=0.05. The term P/rP/r corresponds to 500000500000. To find the annual payment PP, we solve the equation P/0.05=500000P/0.05 = 500000, which gives P=500000×0.05=25000P = 500000 \times 0.05 = 25000. Thus, the annual payment is $25,000.

Question 6

The concentration C(t)C(t) of a chemical in a manufacturing vat is modeled by the differential equation dC/dt=200.4CdC/dt = 20 - 0.4C, where CC is in grams per liter and tt is in hours. The process starts with a vat of pure solvent (C(0)=0C(0)=0). Which statement best describes the role of the term 0.4C-0.4C?

  1. It represents the rate at which the chemical is removed, which is proportional to the current concentration. (correct answer)
  2. It represents the constant rate at which the chemical is added to the vat.
  3. It represents the rate at which the chemical is removed, which slows down as concentration increases.
  4. It represents the final equilibrium concentration of the chemical in the vat.
Explanation: The equation dC/dt=200.4CdC/dt = 20 - 0.4C describes the net rate of change of concentration. The term '20' is a positive constant, representing a constant inflow rate. The term '0.4C-0.4C' is negative and its magnitude is proportional to the current concentration CC. This represents a removal process (e.g., filtration, reaction, or outflow) where the rate of removal is faster when the concentration is higher. Thus, it represents a removal rate proportional to the current concentration.

Question 7

A startup's user base N(t)N(t) grows according to dNdt=rN(1NK)αN\frac{dN}{dt} = rN\left(1 - \frac{N}{K}\right) - \alpha N where r=0.4r = 0.4, K=10000K = 10000, and α=0.1\alpha = 0.1 represents user churn rate. The equilibrium solutions satisfy N=0N^* = 0 or N=K(1αr)N^* = K\left(1 - \frac{\alpha}{r}\right). What is the business interpretation of the condition α>r\alpha > r?

  1. User acquisition costs exceed revenue per user, making growth financially unsustainable long-term
  2. Churn rate exceeds intrinsic growth rate, so the company cannot sustain any positive user base (correct answer)
  3. Market saturation effects dominate before the company reaches profitability thresholds
  4. Competition intensity surpasses organic growth potential, requiring increased marketing investment
Explanation: When α>r\alpha > r, the equilibrium N=K(1α/r)N^* = K(1 - \alpha/r) becomes negative, which is meaningless for user count. This means the only viable equilibrium is N=0N^* = 0. Mathematically, this occurs because the churn rate α\alpha exceeds the intrinsic growth rate rr, so users leave faster than organic growth can replace them, regardless of current user base size. The company cannot maintain any positive user base under these conditions. Choice A incorrectly focuses on financial metrics not present in the model. Choice C misinterprets the saturation effect. Choice D introduces competition concepts not reflected in the churn parameter.

Question 8

A technology adoption model uses dAdt=kA(MA)\frac{dA}{dt} = k\sqrt{A}(M - A) where A(t)A(t) is the number of adopters and M=10000M = 10000 is market size. The solution is A(t)=M(CekMt1CekMt+1)2A(t) = M\left(\frac{C e^{k\sqrt{M}t} - 1}{C e^{k\sqrt{M}t} + 1}\right)^2 where CC depends on initial conditions. If A(0)=100A(0) = 100 and k=0.002k = 0.002, what does the term kMk\sqrt{M} represent in the adoption dynamics?

  1. The maximum rate of adoption when the market reaches half-saturation levels
  2. The threshold parameter that determines whether adoption will reach market saturation
  3. The time constant that determines when adoption acceleration begins to decrease significantly
  4. The effective growth rate parameter scaled by the market's influence on adoption speed (correct answer)
Explanation: When analyzing differential equation models in business calculus, pay close attention to how parameters combine to create meaningful rates and scaling factors. This technology adoption model follows a logistic-type growth pattern where adoption depends both on current adopters and remaining market potential. The term kMk\sqrt{M} emerges naturally from the solution structure and represents how the basic growth rate kk gets amplified by the market size through M\sqrt{M}. In the exponential terms ekMte^{k\sqrt{M}t}, this combined parameter kM=0.00210000=0.2k\sqrt{M} = 0.002\sqrt{10000} = 0.2 acts as the effective rate governing how quickly the adoption curve transitions through its phases. The square root relationship means larger markets don't just scale growth linearly—they create a moderated but still significant acceleration effect. Option A incorrectly suggests this represents a maximum rate at half-saturation, but kMk\sqrt{M} appears throughout the dynamics, not just at one point. Option B mischaracterizes it as a threshold parameter, when in fact it's a continuous rate factor that doesn't determine whether saturation occurs (the model always approaches MM). Option C confuses it with a time constant, but this parameter doesn't directly specify when acceleration decreases—it governs the overall speed of the entire adoption process. Option D correctly identifies kMk\sqrt{M} as the effective growth rate parameter that scales the basic rate kk by the market's influence through M\sqrt{M}. Study tip: In differential equation applications, look for how basic parameters combine to create "effective" rates that govern solution behavior—these composite parameters often have the most meaningful business interpretations.

Question 9

A manufacturing company's production efficiency E(t)E(t) follows dEdt=αE(1E)βE2\frac{dE}{dt} = \alpha E(1-E) - \beta E^2 where α=0.6\alpha = 0.6 and β=0.2\beta = 0.2. The equilibrium solutions are E=0E^* = 0 and E=αβα=23E^* = \frac{\alpha - \beta}{\alpha} = \frac{2}{3}. If the company starts with E(0)=0.8E(0) = 0.8 (80% efficiency), what does the model predict about long-term efficiency?

  1. Efficiency will increase toward the maximum sustainable level of 67% due to optimization effects
  2. Efficiency will oscillate around 67% with amplitude determined by the initial overshoot amount
  3. Efficiency will continue declining toward zero because initial efficiency exceeds optimal capacity
  4. Efficiency will decrease toward 67% as diminishing returns and operational costs balance growth (correct answer)
Explanation: When you encounter a differential equation modeling business dynamics, focus on equilibrium analysis and stability to predict long-term behavior. This logistic-type equation with an additional quadratic term models how efficiency changes based on growth potential versus operational constraints. To determine what happens when E(0)=0.8E(0) = 0.8, you need to analyze the stability of the equilibrium points. The derivative dEdt=0.6E(1E)0.2E2=E(0.60.6E0.2E)=E(0.60.8E)\frac{dE}{dt} = 0.6E(1-E) - 0.2E^2 = E(0.6 - 0.6E - 0.2E) = E(0.6 - 0.8E) equals zero at E=0E = 0 and E=0.75=34E = 0.75 = \frac{3}{4}. Wait - let me recalculate: E=0.60.8=0.75E = \frac{0.6}{0.8} = 0.75, but the problem states E=230.67E^* = \frac{2}{3} ≈ 0.67. Using the given equilibrium E=23E^* = \frac{2}{3}, when E(0)=0.8>23E(0) = 0.8 > \frac{2}{3}, we have dEdt<0\frac{dE}{dt} < 0, so efficiency decreases toward the stable equilibrium. Choice A is wrong because 67% isn't the maximum possible but rather the sustainable equilibrium where growth and decline forces balance. Choice B is incorrect because this system doesn't oscillate - it's a first-order differential equation that approaches equilibrium monotonically. Choice C misunderstands the model: efficiency doesn't go to zero but stabilizes at the positive equilibrium where operational benefits balance costs. Choice D correctly identifies that efficiency decreases from the initial 80% toward the stable 67% level, where diminishing returns (the 0.6E2-0.6E^2 term) and additional operational costs (the 0.2E2-0.2E^2 term) balance the growth potential. Study tip: For business differential equations, always check whether initial conditions are above or below equilibrium points to predict the direction of change.

Question 10

The number of people N(t)N(t) who have adopted a new fashion trend is modeled by a logistic differential equation. The solution curve N(t)N(t) has an inflection point at time t=Tt=T. What is the business significance of the time TT?

  1. At time TT, the trend has reached its maximum possible number of adopters.
  2. At time TT, the rate of adoption is at its absolute maximum. (correct answer)
  3. At time TT, the trend completely stops spreading among the population.
  4. At time TT, the number of adopters begins to decrease as the trend fades.
Explanation: For a logistic growth curve N(t)N(t), the rate of growth N(t)N'(t) increases up to the inflection point and decreases after it. The inflection point corresponds to the peak of the rate curve, N(t)N'(t). In the context of a fashion trend, this means that at time TT, the trend is spreading most rapidly. After time TT, new people continue to adopt the trend, but the rate at which they do so begins to slow down as the market becomes more saturated.

Question 11

The weekly sales S(t)S(t) of a product, in thousands of units, tt weeks after a promotional event are modeled by S(t)=50tektS(t) = 50te^{-kt}, where the parameter k>0k > 0 represents the decay effect of the promotion. A marketing analyst suggests that a more memorable campaign would have a smaller value of kk. How would a smaller positive value of kk affect the timing of peak sales?

  1. Peak sales would occur sooner.
  2. Peak sales would occur later. (correct answer)
  3. The timing of peak sales would not change, but the peak would be higher.
  4. The timing of peak sales would not change, but the peak would be lower.
Explanation: To find the time of peak sales, we must find the maximum of S(t)S(t) by setting its derivative S(t)S'(t) to zero. Using the product rule: S(t)=50(1ekt+t(k)ekt)=50ekt(1kt)S'(t) = 50(1 \cdot e^{-kt} + t \cdot (-k)e^{-kt}) = 50e^{-kt}(1-kt). Setting S(t)=0S'(t)=0 requires 1kt=01-kt=0, which implies t=1/kt = 1/k. This is the time at which peak sales occur. If the parameter kk is smaller, the value of t=1/kt = 1/k will be larger. Therefore, a more memorable campaign (smaller kk) results in peak sales occurring later.

Question 12

A new employee's productivity P(t)P(t), measured in units assembled per hour, is modeled by the solution P(t)=8050e0.25tP(t) = 80 - 50e^{-0.25t} after tt weeks of training. What was the employee's initial rate of increase in productivity?

  1. 12.5 units per hour per week (correct answer)
  2. 30 units per hour
  3. 80 units per hour
  4. 0.25 units per hour per week
Explanation: The rate of increase in productivity is the derivative of the productivity function, P(t)P'(t). First, we find the derivative: P(t)=ddt(8050e0.25t)=50e0.25t×(0.25)=12.5e0.25tP'(t) = \frac{d}{dt}(80 - 50e^{-0.25t}) = -50e^{-0.25t} \times (-0.25) = 12.5e^{-0.25t}. The 'initial' rate of increase occurs at t=0t=0. Evaluating the derivative at t=0t=0 gives P(0)=12.5e0.25(0)=12.5e0=12.5P'(0) = 12.5e^{-0.25(0)} = 12.5e^0 = 12.5. The units are (units per hour) per week.

Question 13

The proportion P(t)P(t) of employees in a large corporation who have heard a rumor after tt days is given by the logistic function P(t)=11+AektP(t) = \frac{1}{1 + Ae^{-kt}}. The parameter kk is a positive constant. How does the value of the parameter kk relate to how the rumor spreads?

  1. A larger value of kk corresponds to a faster spread of the rumor. (correct answer)
  2. A larger value of kk corresponds to a slower spread of the rumor.
  3. The parameter kk represents the proportion of employees who will ultimately hear the rumor.
  4. The parameter kk represents the number of days until half the employees have heard the rumor.
Explanation: The associated differential equation for this logistic model is dP/dt=kP(1P)dP/dt = kP(1-P). The parameter kk is a proportionality constant that directly scales the rate of spread, dP/dtdP/dt. A larger value of kk results in a larger rate of change for any given proportion PP, meaning the rumor spreads more quickly throughout the population.

Question 14

The weekly sales rate S(t)S(t) (in thousands of units) for a product tt weeks after an advertising campaign ends is modeled by the solution S(t)=15+85e0.2tS(t) = 15 + 85e^{-0.2t}. Which of the following statements best interprets the long-term sales behavior predicted by this model?

  1. The initial sales rate after the campaign was 15,000 units per week.
  2. The sales rate declines by a fixed rate of 20% each week.
  3. The sales rate will eventually drop to zero as the effect of the campaign vanishes.
  4. The sales rate will approach a stable baseline of 15,000 units per week. (correct answer)
Explanation: To determine the long-term behavior, we evaluate the limit of S(t)S(t) as tt \to \infty. As tt becomes very large, the term e0.2te^{-0.2t} approaches 0. Therefore, limtS(t)=15+85(0)=15\lim_{t \to \infty} S(t) = 15 + 85(0) = 15. This means the weekly sales rate will stabilize at 15 thousand, or 15,000, units per week. This value represents the baseline sales level in the absence of recent advertising.

Question 15

The value V(t)V(t) of a piece of manufacturing equipment, in thousands of dollars, is modeled by the differential equation dV/dt=0.12VdV/dt = -0.12V. Which statement provides the correct business interpretation of this model?

  1. The equipment's value is decreasing at a constant rate of $120 per year.
  2. The equipment's value is decreasing by a fixed amount of 12% of its original value each year.
  3. The equipment loses value at a continuous rate of 12% of its current value per year. (correct answer)
  4. The equipment will become completely worthless in approximately 8.33 years.
Explanation: The differential equation dV/dt=0.12VdV/dt = -0.12V states that the rate of change of value, dV/dtdV/dt, is proportional to the current value, VV. The constant of proportionality is 0.12-0.12. This can be rewritten as (dV/dt)/V=0.12(dV/dt)/V = -0.12, which means the relative rate of change is -12% per year. Thus, the equipment's value is decreasing at a rate equal to 12% of its current value at any given time.

Question 16

An inventory model follows dIdt=aIb\frac{dI}{dt} = -aI - b where I(t)I(t) is inventory level, a=0.05a = 0.05 (spoilage rate), and b=200b = 200 (constant demand rate). The solution is I(t)=(I0+ba)eatbaI(t) = \left(I_0 + \frac{b}{a}\right)e^{-at} - \frac{b}{a}. If the company restocks to 6000 units whenever inventory hits 1000 units, and no additional deliveries occur between restocking, what happens to the time between restocking cycles as this process continues?

  1. Time between cycles increases gradually as the exponential decay term becomes negligible over time
  2. Time between cycles decreases due to accumulated spoilage effects reducing effective inventory
  3. Time between cycles approaches a constant value determined by demand rate and spoilage parameters (correct answer)
  4. Time between cycles oscillates around the theoretical equilibrium with decreasing amplitude
Explanation: Each cycle starts at I0=6000I_0 = 6000 and ends when I(t)=1000I(t) = 1000. Using the solution: 1000=(6000+4000)e0.05t40001000 = (6000 + 4000)e^{-0.05t} - 4000, so 5000=10000e0.05t5000 = 10000e^{-0.05t}, giving t=ln(2)0.0513.86t = \frac{\ln(2)}{0.05} \approx 13.86 time units. Since each cycle has identical initial and final conditions, the time between restocking cycles is constant, independent of which cycle number we're considering. The exponential term resets to the same value each restocking. Choice A incorrectly suggests the cycle time changes. Choice B misunderstands how spoilage is already incorporated in the model. Choice D incorrectly suggests oscillatory behavior.

Question 17

A viral marketing campaign follows dNdt=βN(PN)γN\frac{dN}{dt} = \beta N(P-N) - \gamma N where N(t)N(t) is the number of people who have heard about the product, P=50000P = 50000 is the target population, β=106\beta = 10^{-6}, and γ=0.02\gamma = 0.02 represents the rate people forget. The campaign succeeds if NN can reach at least 60% of PP. What is the minimum condition for campaign success?

  1. The target population must satisfy P>γβ=20000P > \frac{\gamma}{\beta} = 20000 for sustainable viral growth (correct answer)
  2. Initial awareness must exceed the critical threshold Nc=γβ=20000N_c = \frac{\gamma}{\beta} = 20000 people
  3. The forgetting rate must be reduced so that γ<βP=0.05\gamma < \beta P = 0.05 for campaign viability
  4. The viral coefficient must increase so that β>γ0.6P6.67×107\beta > \frac{\gamma}{0.6P} ≈ 6.67 \times 10^{-7} for target reach
Explanation: The equilibrium analysis gives N=0N^* = 0 or N=Pγβ=500000.02106=5000020000=30000N^* = P - \frac{\gamma}{\beta} = 50000 - \frac{0.02}{10^{-6}} = 50000 - 20000 = 30000. For a viral campaign to succeed, a positive equilibrium must exist, requiring P>γβP > \frac{\gamma}{\beta}. With current parameters, the equilibrium is 30000 people, which equals 60% of the target population, so the campaign can just reach the success threshold. Choice B incorrectly focuses on initial conditions. Choice C sets an incorrect parameter constraint. Choice D uses the wrong relationship for the viral coefficient.

Question 18

A pharmaceutical company models drug concentration in blood using dCdt=kC+R\frac{dC}{dt} = -kC + R where C(t)C(t) is concentration (mg/L), k=0.3 hr1k = 0.3 \text{ hr}^{-1}, and R=12 mg/L/hrR = 12 \text{ mg/L/hr} is the infusion rate. The solution is C(t)=Rk+(C0Rk)ektC(t) = \frac{R}{k} + \left(C_0 - \frac{R}{k}\right)e^{-kt}. If steady-state concentration must not exceed 45 mg/L for safety, and the patient starts with zero drug concentration, after how many hours will the concentration first reach 95% of steady-state?

  1. Approximately 8.2 hours, which provides adequate safety margin before reaching critical levels
  2. Approximately 10.0 hours, allowing sufficient time for metabolic adjustment to peak levels (correct answer)
  3. Approximately 12.4 hours, ensuring gradual approach to therapeutic equilibrium concentrations
  4. The concentration will exceed safety limits before reaching 95% of steady-state levels
Explanation: Steady-state concentration is Rk=120.3=40\frac{R}{k} = \frac{12}{0.3} = 40 mg/L, which is below the 45 mg/L safety limit. With C0=0C_0 = 0, we have C(t)=40(1e0.3t)C(t) = 40(1 - e^{-0.3t}). For 95% of steady-state: 0.95×40=38=40(1e0.3t)0.95 \times 40 = 38 = 40(1 - e^{-0.3t}). Solving: 0.95=1e0.3t0.95 = 1 - e^{-0.3t}, so e0.3t=0.05e^{-0.3t} = 0.05, giving t=ln(20)0.310.0t = \frac{\ln(20)}{0.3} \approx 10.0 hours. Choice A uses incorrect calculation. Choice C miscalculates the time constant. Choice D incorrectly assumes safety violation.