Average Value in Context - Business Calculus

A warehouse's inventory of a seasonal product over the course of a year is modeled by the function $I(t) = 2000 + 500\sin(\frac{\pi t}{12})$ , where $t$ 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 ( $t=0$ to $t=6$ )?
Answer: 2318 units The average inventory level is the average value of the function $I(t)$ on the interval $[0, 6]$ . 1. Set up the integral for the average value: $\text{Average Inventory} = \frac{1}{6-0} \int_0^6 \left(2000 + 500\sin\left(\frac{\pi t}{12}\right)\right) \, dt$ 2. Find the antiderivative of $I(t)$ : $\int \left(2000 + 500\sin\left(\frac{\pi t}{12}\right)\right) \, dt = 2000t - 500 \cdot \frac{12}{\pi}\cos\left(\frac{\pi t}{12}\right)$ 3. Evaluate the definite integral from $0$ to $6$ : $\left[ 2000t - \frac{6000}{\pi}\cos\left(\frac{\pi t}{12}\right) \right]_0^6$ $= \left(2000(6) - \frac{6000}{\pi}\cos\left(\frac{6\pi}{12}\right)\right) - \left(2000(0) - \frac{6000}{\pi}\cos(0)\right)$ $= \left(12000 - \frac{6000}{\pi}\cos\left(\frac{\pi}{2}\right)\right) - \left(0 - \frac{6000}{\pi}(1)\right)$ $= \left(12000 - \frac{6000}{\pi}(0)\right) - \left(-\frac{6000}{\pi}\right) = 12000 + \frac{6000}{\pi}$ 4. Divide by the length of the interval (6 months): $\text{Average Inventory} = \frac{1}{6} \left(12000 + \frac{6000}{\pi}\right) = 2000 + \frac{1000}{\pi}$ 5. Approximate the value: $2000 + \frac{1000}{3.14159...} \approx 2000 + 318.31 \approx 2318.31$ The closest answer is 2318 units.
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?
Answer: The average rate equals $41 per hour, and this instantaneous rate occurs at exactly $t = 3.000$ hours First, find the average rate: $\frac{1}{6}\int_0^6(2t^3 - 18t^2 + 48t + 5)dt = \frac{1}{6}[\frac{t^4}{2} - 6t^3 + 24t^2 + 5t]_0^6 = \frac{1}{6}[648 - 1296 + 864 + 30] = \frac{246}{6} = 41$ . By the Mean Value Theorem for integrals, there exists some $c \in [0,6]$ where $R(c) = 41$ . Solving $2c^3 - 18c^2 + 48c + 5 = 41$ : $2c^3 - 18c^2 + 48c - 36 = 0$ , or $c^3 - 9c^2 + 24c - 18 = 0$ . Testing $c = 3$ : $27 - 81 + 72 - 18 = 0$ ✓. So the average rate is $41 per hour, occurring at exactly $t = 3$ 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.
The marginal profit for a product is given by $MP(q) = -0.3q^2 + 18q + 200$ dollars per unit, where $q$ is the number of units produced. What is the average marginal profit as production is increased from $q=20$ to $q=50$ units?
Answer: $440 To find the average marginal profit, we must calculate the average value of the function $MP(q)$ on the interval $[20, 50]$ . 1. The formula for the average value is $\frac{1}{b-a} \int_a^b f(x) \, dx$ . Here, $f(q) = MP(q)$ , $a=20$ , and $b=50$ . The length of the interval is $b-a = 30$ . 2. Set up the integral: $\text{Average MP} = \frac{1}{30} \int_{20}^{50} (-0.3q^2 + 18q + 200) \, dq$ 3. Find the antiderivative: $\int (-0.3q^2 + 18q + 200) \, dq = -0.1q^3 + 9q^2 + 200q$ 4. Evaluate the definite integral: $\left[ -0.1q^3 + 9q^2 + 200q \right]_{20}^{50}$ $= (-0.1(50^3) + 9(50^2) + 200(50)) - (-0.1(20^3) + 9(20^2) + 200(20))$ $= (-12500 + 22500 + 10000) - (-800 + 3600 + 4000)$ $= (20000) - (6800) = 13200$ 5. Divide by the interval length: $\text{Average MP} = \frac{13200}{30} = 440$ The average marginal profit is $440 per unit.
The monthly sales rate for a new phone is modeled by $S(t) = 200 + 36t - 3t^2$ units per month, where $t$ is the number of months since its launch, for $0 \le t \le 10$ . How does the average monthly sales rate during the first five months ( $[0, 5]$ ) compare to the average monthly sales rate during the second five months ( $[5, 10]$ )?
Answer: The average rate in the second five months is 30 units higher. We need to compute the average value of $S(t)$ on two separate intervals, $[0, 5]$ and $[5, 10]$ , and then compare them. 1. Average sales rate for the first five months ( $[0, 5]$ ): $\bar{S}_1 = \frac{1}{5} \int_0^5 (200 + 36t - 3t^2) \, dt = \frac{1}{5} \left[ 200t + 18t^2 - t^3 \right]_0^5$ $= \frac{1}{5} (200(5) + 18(5^2) - 5^3) = \frac{1}{5} (1000 + 18(25) - 125) = \frac{1}{5} (1000 + 450 - 125) = \frac{1325}{5} = 265$ 2. Average sales rate for the second five months ( $[5, 10]$ ): $\bar{S}_2 = \frac{1}{5} \int_5^{10} (200 + 36t - 3t^2) \, dt = \frac{1}{5} \left[ 200t + 18t^2 - t^3 \right]_5^{10}$ $= \frac{1}{5} \left[ (200(10) + 18(10^2) - 10^3) - (200(5) + 18(5^2) - 5^3) \right]$ We already calculated the value at $t=5$ to be 1325. $= \frac{1}{5} \left[ (2000 + 1800 - 1000) - 1325 \right] = \frac{1}{5} (2800 - 1325) = \frac{1475}{5} = 295$ 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 $295 - 265 = 30$ . Therefore, the average rate in the second five months is 30 units higher.
The daily productivity of a factory is modeled by $P(t) = -3t^2 + 30t + 150$ units per day, where $t$ is the number of days after the start of a production run, for $0 \le t \le 5$ . According to the Mean Value Theorem for Integrals, there is a time $t^*$ in the interval $(0, 5)$ at which the daily productivity is exactly equal to the average productivity over that 5-day period. What is the value of $t^*$ ?
Answer: 2.11 days First, we must calculate the average productivity over the interval $[0, 5]$ . Then, we set the productivity function $P(t)$ equal to this average value and solve for $t$ . 1. Calculate the average productivity: $\bar{P} = \frac{1}{5-0} \int_0^5 (-3t^2 + 30t + 150) \, dt$ $\bar{P} = \frac{1}{5} \left[ -t^3 + 15t^2 + 150t \right]_0^5$ $\bar{P} = \frac{1}{5} \left( -(5^3) + 15(5^2) + 150(5) \right) - 0$ $\bar{P} = \frac{1}{5} (-125 + 15(25) + 750) = \frac{1}{5} (-125 + 375 + 750) = \frac{1000}{5} = 200$ 2. Find $t^*$ such that $P(t^*) = 200$ : $-3(t^*)^2 + 30t^* + 150 = 200$ $-3(t^*)^2 + 30t^* - 50 = 0$ $3(t^*)^2 - 30t^* + 50 = 0$ 3. Solve the quadratic equation using the quadratic formula, $t^* = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ : $t^* = \frac{30 \pm \sqrt{(-30)^2 - 4(3)(50)}}{2(3)} = \frac{30 \pm \sqrt{900 - 600}}{6} = \frac{30 \pm \sqrt{300}}{6}$ $\sqrt{300} = \sqrt{100 \cdot 3} = 10\sqrt{3} \approx 17.32$ $t^* = \frac{30 \pm 17.32}{6}$ This gives two possible solutions: $t_1 = \frac{30 + 17.32}{6} = \frac{47.32}{6} \approx 7.89$ and $t_2 = \frac{30 - 17.32}{6} = \frac{12.68}{6} \approx 2.11$ . Since we are looking for a value within the interval $[0, 5]$ , the correct answer is $t^* \approx 2.11$ days.
The price of a company's stock over a 20-day period is modeled by the function $S(t) = 50 - 2t + 0.1t^2$ for $t \in [0, 20]$ . How does the continuous average price over the 20-day period compare to the price at the midpoint of the interval, $S(10)$ ?
Answer: The average price is greater than the midpoint price by $3.33. This question requires calculating two values: the average value of $S(t)$ on $[0, 20]$ and the value of $S(10)$ . 1. Calculate the average price over $[0, 20]$ : $\bar{S} = \frac{1}{20} \int_0^{20} (50 - 2t + 0.1t^2) \, dt$ $\bar{S} = \frac{1}{20} \left[ 50t - t^2 + \frac{0.1}{3}t^3 \right]_0^{20}$ $\bar{S} = \frac{1}{20} \left( 50(20) - (20)^2 + \frac{0.1}{3}(20)^3 \right) = \frac{1}{20} \left( 1000 - 400 + \frac{0.1(8000)}{3} \right)$ $\bar{S} = \frac{1}{20} \left( 600 + \frac{800}{3} \right) = 30 + \frac{40}{3} = 30 + 13.33... = 43.33...$ So, the average price is $43.33. 2. Calculate the price at the midpoint, $t=10$ : $S(10) = 50 - 2(10) + 0.1(10)^2 = 50 - 20 + 0.1(100) = 30 + 10 = 40$ 3. Compare the two values: The average price ($43.33) is greater than the midpoint price ($40). The difference is $43.33 - 40 = 3.33$ . The function is concave up, so it is expected that the average value is greater than the midpoint value.
The rate of customers entering a store is modeled by the function $f(t) = 80 + 30\cos(\frac{\pi t}{4})$ , where $t$ 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?
Answer: 80 customers per hour 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 $\frac{1}{8-0}\int_0^8 [80 + 30\cos(\frac{\pi t}{4})] dt$ . Breaking this integral into parts: $\frac{1}{8}[\int_0^8 80 dt + \int_0^8 30\cos(\frac{\pi t}{4}) dt]$ The first integral gives you $80 \times 8 = 640$ . For the second integral, the antiderivative of $30\cos(\frac{\pi t}{4})$ is $\frac{30 \times 4}{\pi}\sin(\frac{\pi t}{4}) = \frac{120}{\pi}\sin(\frac{\pi t}{4})$ . Evaluating from 0 to 8: $\frac{120}{\pi}[\sin(2\pi) - \sin(0)] = \frac{120}{\pi}[0 - 0] = 0$ Therefore, the average value is $\frac{1}{8}[640 + 0] = 80$ 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.
Let $MC(q)$ be a company's marginal cost function for producing $q$ units of a product. If the average value of the marginal cost function over the production interval $[q_1, q_2]$ is denoted by $\overline{MC}$ , what is the economic interpretation of the quantity $\overline{MC} \times (q_2 - q_1)$ ?
Answer: The total increase in cost when production is increased from $q_1$ to $q_2$ units. The definition of the average value $\overline{MC}$ of the function $MC(q)$ over the interval $[q_1, q_2]$ is: $\overline{MC} = \frac{1}{q_2 - q_1} \int_{q_1}^{q_2} MC(q) \, dq$ Multiplying both sides by $(q_2 - q_1)$ gives: $\overline{MC} \times (q_2 - q_1) = \int_{q_1}^{q_2} MC(q) \, dq$ The integral of the marginal cost function, $\int MC(q) \, dq$ , gives the total cost function, $C(q)$ . Therefore, the definite integral $\int_{q_1}^{q_2} MC(q) \, dq$ represents the total change in cost as quantity changes from $q_1$ to $q_2$ . This is equal to $C(q_2) - C(q_1)$ , which is the total variable cost of the additional units produced, or simply the total increase in cost for that production increase.
The rate of production at a new manufacturing plant is modeled by $R(t) = 6t^2 - 4t + 50$ units per hour, where $t$ 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 $t=0$ to $t=8$ ?
Answer: 162 units per hour The average value of a function $f(x)$ on an interval $[a, b]$ is given by the formula $\frac{1}{b-a} \int_a^b f(x) \, dx$ . Here, we need to find the average value of $R(t)$ on the interval $[0, 8]$ . 1. Set up the integral for the average value: $\text{Average Rate} = \frac{1}{8-0} \int_0^8 (6t^2 - 4t + 50) \, dt$ 2. Find the antiderivative of $R(t)$ : $\int (6t^2 - 4t + 50) \, dt = \frac{6t^3}{3} - \frac{4t^2}{2} + 50t = 2t^3 - 2t^2 + 50t$ 3. Evaluate the definite integral: $\left[ 2t^3 - 2t^2 + 50t \right]_0^8 = (2(8^3) - 2(8^2) + 50(8)) - (0)$ $= (2(512) - 2(64) + 400) = 1024 - 128 + 400 = 1296$ 4. Divide by the length of the interval: $\text{Average Rate} = \frac{1}{8} (1296) = 162$ The average rate of production during the 8-hour shift is 162 units per hour.
A delivery company's fuel efficiency varies throughout the day according to $E(t) = 25 + 8\cos(\frac{\pi(t-8)}{8})$ miles per gallon, where $t$ 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?
Answer: From 6 AM to 10 AM, with an average efficiency of approximately 30.2 mpg during peak morning hours The function $E(t) = 25 + 8\cos(\frac{\pi(t-8)}{8})$ 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 = $\frac{1}{4}\int_6^{10}[25 + 8\cos(\frac{\pi(t-8)}{8})]dt ≈ 30.2$ 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.
The price of a commodity in dollars is modeled by the function $P(t) = k\sqrt{t} + P_0$ , where $t$ 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?
Answer: $78.75 This is a multi-step problem. First, we find $P_0$ , then solve for $k$ using the average value, and finally calculate $P(9)$ . 1. Find $P_0$ : The initial price ( $t=0$ ) was $45. So, $P(0) = k\sqrt{0} + P_0 = 45$ , which implies $P_0 = 45$ . The price function is $P(t) = k\sqrt{t} + 45$ . 2. Use the average value to find $k$ : The average price over $[0, 4]$ was $60. $$ \frac{1}{4-0} \int_0^4 (k $t^{1/2}$ + 45) \, dt = 60 $$ $\frac{1}{4} \left[ k \frac{t^{3/2}}{3/2} + 45t \right]_0^4 = 60$ $\frac{1}{4} \left[ \frac{2k}{3} t^{3/2} + 45t \right]_0^4 = 60$ $\frac{1}{4} \left( (\frac{2k}{3} (4^{3/2}) + 45(4)) - (0) \right) = 60$ $4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$ . $\frac{1}{4} \left( \frac{2k}{3}(8) + 180 \right) = 60$ Multiply by 4: $\frac{16k}{3} + 180 = 240$ $\frac{16k}{3} = 60$ $k = \frac{60 \cdot 3}{16} = \frac{180}{16} = \frac{45}{4} = 11.25$ 3. Find the price at $t=9$ : Now we have the full price function $P(t) = 11.25\sqrt{t} + 45$ . $P(9) = 11.25\sqrt{9} + 45 = 11.25(3) + 45 = 33.75 + 45 = 78.75$ The price at $t=9$ months is $78.75.
An online retailer's hourly profit rate is modeled by $P(t) = 120 + 40\sin(\frac{\pi t}{12})$ dollars per hour, where $t$ 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?
Answer: Both averages are exactly equal at $120 per hour, so the difference is $0 For 6 AM to 6 PM (t = 6 to t = 18): Average = $\frac{1}{12}\int_6^{18}[120 + 40\sin(\frac{\pi t}{12})]dt = \frac{1}{12}[120t - \frac{480}{\pi}\cos(\frac{\pi t}{12})]_6^{18}$ . The sine function $\sin(\frac{\pi t}{12})$ 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 $120$ 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.
A continuous revenue stream for a company is modeled by the function $R(t) = 120e^{0.05t}$ , where $R$ is in thousands of dollars per year and $t$ 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)?
Answer: $175,100 The second five-year period corresponds to the time interval $[5, 10]$ . We need to find the average value of $R(t)$ on this interval. 1. Set up the average value formula: $\text{Average Revenue} = \frac{1}{10-5} \int_5^{10} 120e^{0.05t} \, dt$ 2. Find the antiderivative of $R(t)$ : $\int 120e^{0.05t} \, dt = \frac{120}{0.05}e^{0.05t} = 2400e^{0.05t}$ 3. Evaluate the definite integral: $\left[ 2400e^{0.05t} \right]_5^{10} = 2400e^{0.05(10)} - 2400e^{0.05(5)} = 2400e^{0.5} - 2400e^{0.25}$ $= 2400(e^{0.5} - e^{0.25})$ Using a calculator, $e^{0.5} \approx 1.64872$ and $e^{0.25} \approx 1.28403$ . $2400(1.64872 - 1.28403) = 2400(0.36469) \approx 875.256$ 4. Divide by the length of the interval (5 years): $\text{Average Revenue} = \frac{875.256}{5} \approx 175.051$ Since $R(t)$ is in thousands of dollars, the average annual revenue is approximately $175.1$ thousand dollars, or $175,100.

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