Business Calculus Quiz: Average Value In Context
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Average Value In ContextQuestion 1 of 13

The marginal profit for a product is given by MP(q)=−0.3q2+18q+200MP(q) = -0.3q^2 + 18q + 200 dollars per unit, where qq is the number of units produced. What is the average marginal profit as production is increased from q=20q=20 to q=50q=50 units?

$395
$440
$463
$13,200
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Business Calculus Quiz

Business Calculus Quiz: Average Value In Context

Practice Average Value 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.

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Question 1

The marginal profit for a product is given by MP(q)=−0.3q2+18q+200MP(q) = -0.3q^2 + 18q + 200 dollars per unit, where qq is the number of units produced. What is the average marginal profit as production is increased from q=20q=20 to q=50q=50 units?

  1. $395
  2. $440 (correct answer)
  3. $463
  4. $13,200
Explanation: To find the average marginal profit, we must calculate the average value of the function MP(q)MP(q) on the interval [20,50][20, 50].
  1. The formula for the average value is 1b−a∫abf(x) dx\frac{1}{b-a} \int_a^b f(x) \, dx. Here, f(q)=MP(q)f(q) = MP(q), a=20a=20, and b=50b=50. The length of the interval is b−a=30b-a = 30.
  2. Set up the integral: Average MP=130∫2050(−0.3q2+18q+200) dq\text{Average MP} = \frac{1}{30} \int_{20}^{50} (-0.3q^2 + 18q + 200) \, dq
  3. Find the antiderivative: ∫(−0.3q2+18q+200) dq=−0.1q3+9q2+200q\int (-0.3q^2 + 18q + 200) \, dq = -0.1q^3 + 9q^2 + 200q
  4. Evaluate the definite integral: [−0.1q3+9q2+200q]2050\left[ -0.1q^3 + 9q^2 + 200q \right]_{20}^{50} =(−0.1(503)+9(502)+200(50))−(−0.1(203)+9(202)+200(20))= (-0.1(50^3) + 9(50^2) + 200(50)) - (-0.1(20^3) + 9(20^2) + 200(20)) =(−12500+22500+10000)−(−800+3600+4000)= (-12500 + 22500 + 10000) - (-800 + 3600 + 4000) =(20000)−(6800)=13200= (20000) - (6800) = 13200
  5. Divide by the interval length: Average MP=1320030=440\text{Average MP} = \frac{13200}{30} = 440
The average marginal profit is $440 per unit.

Question 2

A warehouse's inventory of a seasonal product over the course of a year is modeled by the function I(t)=2000+500sin⁡(πt12)I(t) = 2000 + 500\sin(\frac{\pi t}{12}), where tt is the number of months since the beginning of the year. What is the approximate average inventory level during the first six months of the year (t=0t=0 to t=6t=6)?

  1. 2000 units
  2. 2159 units
  3. 2318 units (correct answer)
  4. 3546 units
Explanation: The average inventory level is the average value of the function I(t)I(t) on the interval [0,6][0, 6].
  1. Set up the integral for the average value: Average Inventory=16−0∫06(2000+500sin⁡(πt12)) dt\text{Average Inventory} = \frac{1}{6-0} \int_0^6 \left(2000 + 500\sin\left(\frac{\pi t}{12}\right)\right) \, dt
  2. Find the antiderivative of I(t)I(t): ∫(2000+500sin⁡(πt12)) dt=2000t−500⋅12πcos⁡(πt12)\int \left(2000 + 500\sin\left(\frac{\pi t}{12}\right)\right) \, dt = 2000t - 500 \cdot \frac{12}{\pi}\cos\left(\frac{\pi t}{12}\right)
  3. Evaluate the definite integral from 00 to 66: [2000t−6000πcos⁡(πt12)]06\left[ 2000t - \frac{6000}{\pi}\cos\left(\frac{\pi t}{12}\right) \right]_0^6 =(2000(6)−6000πcos⁡(6π12))−(2000(0)−6000πcos⁡(0))= \left(2000(6) - \frac{6000}{\pi}\cos\left(\frac{6\pi}{12}\right)\right) - \left(2000(0) - \frac{6000}{\pi}\cos(0)\right) =(12000−6000πcos⁡(π2))−(0−6000π(1))= \left(12000 - \frac{6000}{\pi}\cos\left(\frac{\pi}{2}\right)\right) - \left(0 - \frac{6000}{\pi}(1)\right) =(12000−6000π(0))−(−6000π)=12000+6000π= \left(12000 - \frac{6000}{\pi}(0)\right) - \left(-\frac{6000}{\pi}\right) = 12000 + \frac{6000}{\pi}
  4. Divide by the length of the interval (6 months): Average Inventory=16(12000+6000π)=2000+1000π\text{Average Inventory} = \frac{1}{6} \left(12000 + \frac{6000}{\pi}\right) = 2000 + \frac{1000}{\pi}
  5. Approximate the value: 2000+10003.14159...≈2000+318.31≈2318.312000 + \frac{1000}{3.14159...} \approx 2000 + 318.31 \approx 2318.31 The closest answer is 2318 units.

Question 3

The price of a commodity in dollars is modeled by the function P(t)=kt+P0P(t) = k\sqrt{t} + P_0, where tt is the number of months from the initial observation. The initial price was $45. If the average price over the first 4 months was $60, what is the price of the commodity at $t=9$ months?

  1. $33.75
  2. $67.50
  3. $75.00
  4. $78.75 (correct answer)
Explanation: This is a multi-step problem. First, we find P0P_0, then solve for kk using the average value, and finally calculate P(9)P(9).
  1. Find P0P_0: The initial price (t=0t=0) was $45. So, $P(0) = k\sqrt{0} + P_0 = 45,whichimplies, which implies P_0 = 45.Thepricefunctionis. The price function is P(t) = k\sqrt{t} + 45$.
  2. Use the average value to find kk: The average price over [0,4][0, 4] was $60. $14−0∫04(k\frac{1}{4-0} \int_0^4 (kt^{1/2} + 45) \, dt = 60 $$ $$ \frac{1}{4} \left[ k \frac{t^{3/2}}{3/2} + 45t \right]_0^4 = 60 $$ $$ \frac{1}{4} \left[ \frac{2k}{3} t^{3/2} + 45t \right]_0^4 = 60 $$ $$ \frac{1}{4} \left( (\frac{2k}{3} (4^{3/2}) + 45(4)) - (0) \right) = 60 $$ 4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$. 14(2k3(8)+180)=60\frac{1}{4} \left( \frac{2k}{3}(8) + 180 \right) = 60 Multiply by 4: 16k3+180=240\frac{16k}{3} + 180 = 240 16k3=60\frac{16k}{3} = 60 k=60⋅316=18016=454=11.25k = \frac{60 \cdot 3}{16} = \frac{180}{16} = \frac{45}{4} = 11.25
  3. Find the price at t=9t=9: Now we have the full price function P(t)=11.25t+45P(t) = 11.25\sqrt{t} + 45. P(9)=11.259+45=11.25(3)+45=33.75+45=78.75P(9) = 11.25\sqrt{9} + 45 = 11.25(3) + 45 = 33.75 + 45 = 78.75 The price at t=9t=9 months is $78.75.

Question 4

Let MC(q)MC(q) be a company's marginal cost function for producing qq units of a product. If the average value of the marginal cost function over the production interval [q1,q2][q_1, q_2] is denoted by MC‾\overline{MC}, what is the economic interpretation of the quantity MC‾×(q2−q1)\overline{MC} \times (q_2 - q_1)?

  1. The change in marginal cost when increasing production from q1q_1 to q2q_2 units.
  2. The total cost of producing q2q_2 units, assuming zero fixed costs.
  3. The average total cost per unit for production levels between q1q_1 and q2q_2.
  4. The total increase in cost when production is increased from q1q_1 to q2q_2 units. (correct answer)
Explanation: The definition of the average value MC‾\overline{MC} of the function MC(q)MC(q) over the interval [q1,q2][q_1, q_2] is: MC‾=1q2−q1∫q1q2MC(q) dq\overline{MC} = \frac{1}{q_2 - q_1} \int_{q_1}^{q_2} MC(q) \, dq Multiplying both sides by (q2−q1)(q_2 - q_1) gives: MC‾×(q2−q1)=∫q1q2MC(q) dq\overline{MC} \times (q_2 - q_1) = \int_{q_1}^{q_2} MC(q) \, dq The integral of the marginal cost function, ∫MC(q) dq\int MC(q) \, dq, gives the total cost function, C(q)C(q). Therefore, the definite integral ∫q1q2MC(q) dq\int_{q_1}^{q_2} MC(q) \, dq represents the total change in cost as quantity changes from q1q_1 to q2q_2. This is equal to C(q2)−C(q1)C(q_2) - C(q_1), which is the total variable cost of the additional units produced, or simply the total increase in cost for that production increase.

Question 5

The daily productivity of a factory is modeled by P(t)=−3t2+30t+150P(t) = -3t^2 + 30t + 150 units per day, where tt is the number of days after the start of a production run, for 0≤t≤50 \le t \le 5. According to the Mean Value Theorem for Integrals, there is a time t∗t^* in the interval (0,5)(0, 5) at which the daily productivity is exactly equal to the average productivity over that 5-day period. What is the value of t∗t^*?

  1. 2.11 days (correct answer)
  2. 2.50 days
  3. 7.89 days
  4. 200.00 days
Explanation: First, we must calculate the average productivity over the interval [0,5][0, 5]. Then, we set the productivity function P(t)P(t) equal to this average value and solve for tt.
  1. Calculate the average productivity: Pˉ=15−0∫05(−3t2+30t+150) dt\bar{P} = \frac{1}{5-0} \int_0^5 (-3t^2 + 30t + 150) \, dt Pˉ=15[−t3+15t2+150t]05\bar{P} = \frac{1}{5} \left[ -t^3 + 15t^2 + 150t \right]_0^5 Pˉ=15(−(53)+15(52)+150(5))−0\bar{P} = \frac{1}{5} \left( -(5^3) + 15(5^2) + 150(5) \right) - 0 Pˉ=15(−125+15(25)+750)=15(−125+375+750)=10005=200\bar{P} = \frac{1}{5} (-125 + 15(25) + 750) = \frac{1}{5} (-125 + 375 + 750) = \frac{1000}{5} = 200
  2. Find t∗t^* such that P(t∗)=200P(t^*) = 200: −3(t∗)2+30t∗+150=200-3(t^*)^2 + 30t^* + 150 = 200 −3(t∗)2+30t∗−50=0-3(t^*)^2 + 30t^* - 50 = 0 3(t∗)2−30t∗+50=03(t^*)^2 - 30t^* + 50 = 0
  3. Solve the quadratic equation using the quadratic formula, t∗=−b±b2−4ac2at^* = \frac{-b \pm \sqrt{b^2-4ac}}{2a}: t∗=30±(−30)2−4(3)(50)2(3)=30±900−6006=30±3006t^* = \frac{30 \pm \sqrt{(-30)^2 - 4(3)(50)}}{2(3)} = \frac{30 \pm \sqrt{900 - 600}}{6} = \frac{30 \pm \sqrt{300}}{6} 300=100⋅3=103≈17.32\sqrt{300} = \sqrt{100 \cdot 3} = 10\sqrt{3} \approx 17.32 t∗=30±17.326t^* = \frac{30 \pm 17.32}{6} This gives two possible solutions: t1=30+17.326=47.326≈7.89t_1 = \frac{30 + 17.32}{6} = \frac{47.32}{6} \approx 7.89 and t2=30−17.326=12.686≈2.11t_2 = \frac{30 - 17.32}{6} = \frac{12.68}{6} \approx 2.11.
Since we are looking for a value within the interval [0,5][0, 5], the correct answer is t∗≈2.11t^* \approx 2.11 days.

Question 6

The rate of customers entering a store is modeled by the function f(t)=80+30cos⁡(πt4)f(t) = 80 + 30\cos(\frac{\pi t}{4}), where tt is the number of hours after the store opens. If the store is open for 8 hours, what is the average number of customers entering the store per hour during the day?

  1. 50 customers per hour
  2. 110 customers per hour
  3. 80 customers per hour (correct answer)
  4. 640 customers per hour
Explanation: When you encounter a rate function and need to find an average value over an interval, you're looking at a classic average value of a function problem. The key insight is that the average value of a continuous function over an interval equals the total area under the curve divided by the length of the interval. To find the average rate, you need to calculate 18−0∫08[80+30cos⁡(πt4)]dt\frac{1}{8-0}\int_0^8 [80 + 30\cos(\frac{\pi t}{4})] dt. Breaking this integral into parts: 18[∫0880dt+∫0830cos⁡(πt4)dt]\frac{1}{8}[\int_0^8 80 dt + \int_0^8 30\cos(\frac{\pi t}{4}) dt] The first integral gives you 80×8=64080 \times 8 = 640. For the second integral, the antiderivative of 30cos⁡(πt4)30\cos(\frac{\pi t}{4}) is 30×4πsin⁡(πt4)=120πsin⁡(πt4)\frac{30 \times 4}{\pi}\sin(\frac{\pi t}{4}) = \frac{120}{\pi}\sin(\frac{\pi t}{4}). Evaluating from 0 to 8: 120π[sin⁡(2π)−sin⁡(0)]=120π[0−0]=0\frac{120}{\pi}[\sin(2\pi) - \sin(0)] = \frac{120}{\pi}[0 - 0] = 0 Therefore, the average value is 18[640+0]=80\frac{1}{8}[640 + 0] = 80 customers per hour. Answer A (50) likely comes from incorrectly averaging the minimum and maximum values without considering the full function. Answer B (110) might result from adding the constant term and amplitude incorrectly. Answer D (640) represents the total customers over 8 hours, not the hourly average. Remember: when finding average values of periodic functions over complete periods, the oscillating terms often integrate to zero, leaving you with just the constant term.

Question 7

A delivery company's fuel efficiency varies throughout the day according to E(t)=25+8cos⁡(π(t−8)8)E(t) = 25 + 8\cos(\frac{\pi(t-8)}{8}) miles per gallon, where tt is hours after midnight. The efficiency peaks at 8 AM and 8 PM. If the company operates from 6 AM to 10 PM, during which 4-hour window within this period is their average fuel efficiency highest?

  1. From 6 AM to 10 AM, with an average efficiency of approximately 30.2 mpg during peak morning hours (correct answer)
  2. From 6 PM to 10 PM, with an average efficiency of approximately 30.2 mpg during peak evening hours
  3. From 8 AM to 12 PM, with an average efficiency of approximately 28.6 mpg starting from the morning peak
  4. From 4 PM to 8 PM, with an average efficiency of approximately 28.6 mpg approaching the evening peak
Explanation: The function E(t)=25+8cos⁡(π(t−8)8)E(t) = 25 + 8\cos(\frac{\pi(t-8)}{8}) has period 16 hours and peaks at t = 8 (8 AM) and t = 20 (8 PM). To find the highest average over any 4-hour window, we need the window that captures the most area under the curve. Since the cosine peaks at t = 8, the 4-hour window [6,10] (6 AM to 10 AM) will have the highest average because it's centered around the morning peak. For t ∈ [6,10]: Average = 14∫610[25+8cos⁡(π(t−8)8)]dt≈30.2\frac{1}{4}\int_6^{10}[25 + 8\cos(\frac{\pi(t-8)}{8})]dt ≈ 30.2 mpg. Choice B gives the same value but this window spans the evening peak equally well. However, both options A and B are mathematically equivalent due to symmetry around their respective peaks.

Question 8

A financial analyst is studying the volatility of a stock price over a trading day. The rate of price change is given by R(t)=2t3−18t2+48t+5R(t) = 2t^3 - 18t^2 + 48t + 5 dollars per hour, where tt represents hours after market opening (0≤t≤60 \leq t \leq 6).

The analyst wants to determine if the stock's average rate of change over the full trading day equals the instantaneous rate at any specific time during the day. If such a time exists, what would be the average rate of change over the entire 6-hour period?

  1. The average rate equals $23 per hour, and this instantaneous rate occurs at approximately $t=3.464t = 3.464 $ hours
  2. The average rate equals $29 per hour, and this instantaneous rate occurs at approximately $t=2.536t = 2.536 $ hours
  3. The average rate equals $41 per hour, and this instantaneous rate occurs at exactly $t=3.000t = 3.000 $ hours (correct answer)
  4. No such time exists because the average rate of $35 per hour never equals any instantaneous rate during trading
Explanation: First, find the average rate: 16∫06(2t3−18t2+48t+5)dt=16[t42−6t3+24t2+5t]06=16[648−1296+864+30]=2466=41\frac{1}{6}\int_0^6(2t^3 - 18t^2 + 48t + 5)dt = \frac{1}{6}[\frac{t^4}{2} - 6t^3 + 24t^2 + 5t]_0^6 = \frac{1}{6}[648 - 1296 + 864 + 30] = \frac{246}{6} = 41. By the Mean Value Theorem for integrals, there exists some c∈[0,6]c \in [0,6] where R(c)=41R(c) = 41. Solving 2c3−18c2+48c+5=412c^3 - 18c^2 + 48c + 5 = 41: 2c3−18c2+48c−36=02c^3 - 18c^2 + 48c - 36 = 0, or c3−9c2+24c−18=0c^3 - 9c^2 + 24c - 18 = 0. Testing c=3c = 3: 27−81+72−18=027 - 81 + 72 - 18 = 0 ✓. So the average rate is $41 per hour, occurring at exactly $t=3t = 3 $ hours. Choice A gives the wrong average and time. Choice B also has incorrect values. Choice D incorrectly claims no such time exists, contradicting the Mean Value Theorem.

Question 9

A continuous revenue stream for a company is modeled by the function R(t)=120e0.05tR(t) = 120e^{0.05t}, where RR is in thousands of dollars per year and tt is in years from the present. What is the approximate average annual revenue over the second five-year period (from the start of year 5 to the end of year 10)?

  1. $136,300
  2. $175,100 (correct answer)
  3. $176,000
  4. $875,300
Explanation: The second five-year period corresponds to the time interval [5,10][5, 10]. We need to find the average value of R(t)R(t) on this interval.
  1. Set up the average value formula: Average Revenue=110−5∫510120e0.05t dt\text{Average Revenue} = \frac{1}{10-5} \int_5^{10} 120e^{0.05t} \, dt
  2. Find the antiderivative of R(t)R(t): ∫120e0.05t dt=1200.05e0.05t=2400e0.05t\int 120e^{0.05t} \, dt = \frac{120}{0.05}e^{0.05t} = 2400e^{0.05t}
  3. Evaluate the definite integral: [2400e0.05t]510=2400e0.05(10)−2400e0.05(5)=2400e0.5−2400e0.25\left[ 2400e^{0.05t} \right]_5^{10} = 2400e^{0.05(10)} - 2400e^{0.05(5)} = 2400e^{0.5} - 2400e^{0.25} =2400(e0.5−e0.25)= 2400(e^{0.5} - e^{0.25}) Using a calculator, e0.5≈1.64872e^{0.5} \approx 1.64872 and e0.25≈1.28403e^{0.25} \approx 1.28403. 2400(1.64872−1.28403)=2400(0.36469)≈875.2562400(1.64872 - 1.28403) = 2400(0.36469) \approx 875.256
  4. Divide by the length of the interval (5 years): Average Revenue=875.2565≈175.051\text{Average Revenue} = \frac{875.256}{5} \approx 175.051 Since R(t)R(t) is in thousands of dollars, the average annual revenue is approximately 175.1175.1 thousand dollars, or $175,100.

Question 10

The monthly sales rate for a new phone is modeled by S(t)=200+36t−3t2S(t) = 200 + 36t - 3t^2 units per month, where tt is the number of months since its launch, for 0≤t≤100 \le t \le 10. How does the average monthly sales rate during the first five months ([0,5][0, 5]) compare to the average monthly sales rate during the second five months ([5,10][5, 10])?

  1. The average rate in the second five months is 150 units lower.
  2. The average rate in the second five months is 30 units lower.
  3. The average rates for both five-month periods are equal.
  4. The average rate in the second five months is 30 units higher. (correct answer)
Explanation: We need to compute the average value of S(t)S(t) on two separate intervals, [0,5][0, 5] and [5,10][5, 10], and then compare them.
  1. Average sales rate for the first five months ([0,5][0, 5]): Sˉ1=15∫05(200+36t−3t2) dt=15[200t+18t2−t3]05\bar{S}_1 = \frac{1}{5} \int_0^5 (200 + 36t - 3t^2) \, dt = \frac{1}{5} \left[ 200t + 18t^2 - t^3 \right]_0^5 =15(200(5)+18(52)−53)=15(1000+18(25)−125)=15(1000+450−125)=13255=265= \frac{1}{5} (200(5) + 18(5^2) - 5^3) = \frac{1}{5} (1000 + 18(25) - 125) = \frac{1}{5} (1000 + 450 - 125) = \frac{1325}{5} = 265
  2. Average sales rate for the second five months ([5,10][5, 10]): Sˉ2=15∫510(200+36t−3t2) dt=15[200t+18t2−t3]510\bar{S}_2 = \frac{1}{5} \int_5^{10} (200 + 36t - 3t^2) \, dt = \frac{1}{5} \left[ 200t + 18t^2 - t^3 \right]_5^{10} =15[(200(10)+18(102)−103)−(200(5)+18(52)−53)]= \frac{1}{5} \left[ (200(10) + 18(10^2) - 10^3) - (200(5) + 18(5^2) - 5^3) \right] We already calculated the value at t=5t=5 to be 1325. =15[(2000+1800−1000)−1325]=15(2800−1325)=14755=295= \frac{1}{5} \left[ (2000 + 1800 - 1000) - 1325 \right] = \frac{1}{5} (2800 - 1325) = \frac{1475}{5} = 295
  3. Compare the two averages: The average rate in the second five months is 295, while the average rate in the first five months is 265. The difference is 295−265=30295 - 265 = 30. Therefore, the average rate in the second five months is 30 units higher.

Question 11

The rate of production at a new manufacturing plant is modeled by R(t)=6t2−4t+50R(t) = 6t^2 - 4t + 50 units per hour, where tt is the number of hours since the start of a shift. What is the average rate of production during a single 8-hour shift, from t=0t=0 to t=8t=8?

  1. 130 units per hour
  2. 162 units per hour (correct answer)
  3. 226 units per hour
  4. 1296 units per hour
Explanation: The average value of a function f(x)f(x) on an interval [a,b][a, b] is given by the formula 1b−a∫abf(x) dx\frac{1}{b-a} \int_a^b f(x) \, dx. Here, we need to find the average value of R(t)R(t) on the interval [0,8][0, 8].
  1. Set up the integral for the average value: Average Rate=18−0∫08(6t2−4t+50) dt\text{Average Rate} = \frac{1}{8-0} \int_0^8 (6t^2 - 4t + 50) \, dt
  2. Find the antiderivative of R(t)R(t): ∫(6t2−4t+50) dt=6t33−4t22+50t=2t3−2t2+50t\int (6t^2 - 4t + 50) \, dt = \frac{6t^3}{3} - \frac{4t^2}{2} + 50t = 2t^3 - 2t^2 + 50t
  3. Evaluate the definite integral: [2t3−2t2+50t]08=(2(83)−2(82)+50(8))−(0)\left[ 2t^3 - 2t^2 + 50t \right]_0^8 = (2(8^3) - 2(8^2) + 50(8)) - (0) =(2(512)−2(64)+400)=1024−128+400=1296= (2(512) - 2(64) + 400) = 1024 - 128 + 400 = 1296
  4. Divide by the length of the interval: Average Rate=18(1296)=162\text{Average Rate} = \frac{1}{8} (1296) = 162
The average rate of production during the 8-hour shift is 162 units per hour.

Question 12

The price of a company's stock over a 20-day period is modeled by the function S(t)=50−2t+0.1t2S(t) = 50 - 2t + 0.1t^2 for t∈[0,20]t \in [0, 20]. How does the continuous average price over the 20-day period compare to the price at the midpoint of the interval, S(10)S(10)?

  1. The average price is greater than the midpoint price by $3.33. (correct answer)
  2. The midpoint price is greater than the average price by $3.33.
  3. The average price is equal to the midpoint price.
  4. The average price is equal to the average of the initial and final prices.
Explanation: This question requires calculating two values: the average value of S(t)S(t) on [0,20][0, 20] and the value of S(10)S(10).
  1. Calculate the average price over [0,20][0, 20]: Sˉ=120∫020(50−2t+0.1t2) dt\bar{S} = \frac{1}{20} \int_0^{20} (50 - 2t + 0.1t^2) \, dt Sˉ=120[50t−t2+0.13t3]020\bar{S} = \frac{1}{20} \left[ 50t - t^2 + \frac{0.1}{3}t^3 \right]_0^{20} Sˉ=120(50(20)−(20)2+0.13(20)3)=120(1000−400+0.1(8000)3)\bar{S} = \frac{1}{20} \left( 50(20) - (20)^2 + \frac{0.1}{3}(20)^3 \right) = \frac{1}{20} \left( 1000 - 400 + \frac{0.1(8000)}{3} \right) Sˉ=120(600+8003)=30+403=30+13.33...=43.33...\bar{S} = \frac{1}{20} \left( 600 + \frac{800}{3} \right) = 30 + \frac{40}{3} = 30 + 13.33... = 43.33... So, the average price is $43.33.
  2. Calculate the price at the midpoint, $t=10:: S(10)=50−2(10)+0.1(10)2=50−20+0.1(100)=30+10=40S(10) = 50 - 2(10) + 0.1(10)^2 = 50 - 20 + 0.1(100) = 30 + 10 = 40 $
  3. Compare the two values: The average price (43.33)isgreaterthanthemidpointprice(43.33) is greater than the midpoint price (40). The difference is 43.33−40=3.3343.33 - 40 = 3.33. The function is concave up, so it is expected that the average value is greater than the midpoint value.

Question 13

An online retailer's hourly profit rate is modeled by P(t)=120+40sin⁡(πt12)P(t) = 120 + 40\sin(\frac{\pi t}{12}) dollars per hour, where tt is hours after midnight. The company wants to compare their average profit rate from 6 AM to 6 PM with their average profit rate from 6 PM to 6 AM the next day. What is the difference between these two 12-hour average rates?

  1. The daytime average exceeds the nighttime average by approximately $25.46 per hour
  2. The nighttime average exceeds the daytime average by approximately $25.46 per hour
  3. Both averages are exactly equal at $120 per hour, so the difference is $0 (correct answer)
  4. The daytime average exceeds the nighttime average by approximately $40.00 per hour
Explanation: For 6 AM to 6 PM (t = 6 to t = 18): Average = 112∫618[120+40sin⁡(πt12)]dt=112[120t−480πcos⁡(πt12)]618\frac{1}{12}\int_6^{18}[120 + 40\sin(\frac{\pi t}{12})]dt = \frac{1}{12}[120t - \frac{480}{\pi}\cos(\frac{\pi t}{12})]_6^{18}. The sine function sin⁡(πt12)\sin(\frac{\pi t}{12}) has period 24, so over any 12-hour interval, the integral of the sine term is zero (it completes exactly half a cycle). Therefore, both 12-hour averages equal 120120 dollars per hour. Choice A is wrong because it suggests an asymmetric difference. Choice B is wrong for the same reason with opposite sign. Choice D is wrong because it suggests the maximum amplitude becomes the difference, ignoring that sine averages to zero over half-periods.