Concavity & Inflection Points - Business Calculus

1. The total cost for a warehouse to store $x$ tons of a commodity is given by $C(x) = 0.0002x^3 - 0.24x^2 + 120x + 5000$ for $0 \le x \le 600$. Due to capacity constraints and handling difficulties, the marginal storage cost eventually begins to increase. At what storage level $x$ does this occur?
   Answer: $x = 400$ tons The marginal storage cost is the first derivative, $C'(x)$. We want to find where this marginal cost begins to increase. This occurs at the minimum point of the $C'(x)$ function, which corresponds to an inflection point of the $C(x)$ function where the concavity changes from down to up. We find this point by setting the second derivative, $C''(x)$, to zero. 1. Marginal Cost: $C'(x) = 0.0006x^2 - 0.48x + 120$. 2. Second Derivative: $C''(x) = 0.0012x - 0.48$. 3. Set $C''(x) = 0$: $0.0012x - 0.48 = 0 \implies 0.0012x = 0.48 \implies x = \frac{0.48}{0.0012} = 400$. At $x=400$, the concavity of $C(x)$ changes, and it is the point where the marginal cost $C'(x)$ is at its minimum and begins to increase.
2. A manufacturing company's profit function is $P(x) = -0.5x^3 + 15x^2 - 120x + 500$, where $x$ represents the number of units produced (in thousands) and $P(x)$ is profit in thousands of dollars. The company's financial analyst needs to determine the production level where the rate of profit increase begins to slow down most significantly. At what production level does this occur?
   Answer: $x = 10$ thousand units, where profit transitions from increasing at an increasing rate to increasing at a decreasing rate To find where the rate of profit increase begins to slow down most significantly, we need the inflection point. First, find P'(x) = -1.5x² + 30x - 120 and P''(x) = -3x + 30. Setting P''(x) = 0: -3x + 30 = 0, so x = 10. At x = 10, P'(10) = -1.5(100) + 30(10) - 120 = 30 > 0, so profit is increasing. For x < 10, P''(x) > 0 (concave up), and for x > 10, P''(x) < 0 (concave down). This means profit changes from increasing at an increasing rate to increasing at a decreasing rate at x = 10.
3. A fintech company's user acquisition cost varies with marketing intensity according to $U(m) = m^4 - 8m^3 + 18m^2 - 8m + 50$, where $m$ represents marketing intensity (scale 0-5) and $U(m)$ is cost per user in dollars. The growth team needs to identify all points where the rate of change in acquisition cost transitions between accelerating and decelerating behaviors. How many such transition points exist in the domain $0 \leq m \leq 5$?
   Answer: Two transition points, corresponding to changes in the concavity of the cost function Transition points between accelerating and decelerating behaviors occur at inflection points where U''(m) = 0 and changes sign. U'(m) = 4m³ - 24m² + 36m - 8 and U''(m) = 12m² - 48m + 36 = 12(m² - 4m + 3) = 12(m - 1)(m - 3). Setting U''(m) = 0: m = 1 and m = 3. Both values are in [0,5]. We can verify sign changes: for m < 1, U''(m) > 0; for 1 < m < 3, U''(m) < 0; for m > 3, U''(m) > 0. This confirms two inflection points at m = 1 and m = 3, where the concavity changes and thus the acceleration/deceleration behavior transitions.
4. Let $P(t)$ represent the number of units a new employee produces on day $t$ of their employment. An efficiency expert observes that for a particular employee on day $t_0=10$, $P'(10) > 0$ and $P''(10) < 0$. Which of the following is the best interpretation of this observation?
   Answer: On day 10, the employee's production is increasing, but the rate of increase is slowing. $P(t)$ is the production function. $P'(t)$ is the rate of change of production (the slope). $P''(t)$ describes the concavity, or the rate of change of the slope. - $P'(10) > 0$ means that at $t=10$, production is increasing. - $P''(10) < 0$ means that at $t=10$, the function is concave down. This implies that the slope, $P'(t)$, is decreasing. Combining these two facts, we conclude that on day 10, the employee's production is still increasing, but the rate of that increase is slowing down. This is a classic example of diminishing returns on experience.
5. The total number of subscribers to a streaming service, $N(t)$, in millions, is modeled by the function $N(t) = -0.5t^3 + 9t^2 + 20t$ where $t$ is the number of years since the service launched, for $0 \le t \le 15$. At what time is the subscriber base growing the fastest?
   Answer: $t = 6$ years The rate of growth of the subscriber base is the first derivative, $N'(t)$. To find when this rate is fastest (maximized), we need to find the maximum of the function $N'(t)$. This occurs where its derivative, $N''(t)$, is zero. 1. Find the rate of growth: $N'(t) = -1.5t^2 + 18t + 20$. 2. Find the rate of change of the growth rate: $N''(t) = -3t + 18$. 3. Set $N''(t) = 0$ to find the critical point of $N'(t)$: $-3t + 18 = 0 \implies 3t = 18 \implies t = 6$. This corresponds to the inflection point of the original function $N(t)$. Since $N'''(t) = -3 < 0$, this point is a maximum for the growth rate $N'(t)$.
6. Let $C(x)$ be a differentiable total cost function for producing $x$ units of a product, with $x>0$. If the average cost function, $AC(x) = C(x)/x$, reaches its minimum value at a production level $x_0$, which of the following statements about the total cost function $C(x)$ at $x_0$ must be true?
   Answer: $C(x)$ is concave up at $x_0$. The average cost $AC(x)$ is minimized when $AC'(x_0) = 0$ and $AC''(x_0) > 0$. The derivative of average cost is $AC'(x) = \frac{x C'(x) - C(x)}{x^2}$. Setting $AC'(x_0)=0$ gives the well-known result that marginal cost equals average cost at this point: $C'(x_0) = C(x_0)/x_0 = AC(x_0)$. To ensure it is a minimum, we examine the second derivative, $AC''(x)$. A detailed calculation shows that at the point where $AC'(x_0)=0$, the formula simplifies to $AC''(x_0) = \frac{C''(x_0)}{x_0}$. For $AC(x_0)$ to be a minimum, we need $AC''(x_0) > 0$. Since production $x_0$ must be positive, this implies that $C''(x_0)$ must also be positive. A positive second derivative means the function $C(x)$ is concave up at $x_0$.
7. A firm's total revenue, $R(x)$, from selling $x$ thousand units is given by $R(x) = -0.02x^3 + 3.6x^2 + 50x$ for $0 \le x \le 150$. Over which production interval is the firm experiencing accelerating growth in revenue?
   Answer: Between 0 and 60 thousand units. When you encounter questions about "accelerating growth," you're looking for where the rate of change is itself increasing - this means finding where the second derivative is positive. To find where revenue growth is accelerating, you need to examine $R''(x)$. Start by finding the first derivative: $R'(x) = -0.06x^2 + 7.2x + 50$. This represents the marginal revenue (rate of revenue change). Then find the second derivative: $R''(x) = -0.12x + 7.2$. This tells you whether marginal revenue is increasing or decreasing. For accelerating growth, you need $R''(x) > 0$: $-0.12x + 7.2 > 0$ $7.2 > 0.12x$ $x < 60$ Since the domain starts at $x = 0$, revenue experiences accelerating growth from 0 to 60 thousand units. Looking at the wrong answers: Choice A (0 to 125) is incorrect because accelerating growth stops at 60 thousand units, not 125. Choice B (60 to 125) represents the interval where $R''(x) < 0$, meaning revenue growth is actually decelerating. Choice C (125 to 150) continues in the decelerating region - at $x = 125$, $R''(125) = -7.8 < 0$. Remember that "accelerating growth" doesn't just mean revenue is increasing - it means the rate of increase is getting faster. Always check the second derivative's sign to distinguish between accelerating growth ($R'' > 0$) and decelerating growth ($R'' < 0$).
8. A financial analyst determines that for a specific investment of $x$ dollars, the profit function $P(x)$ has the properties $P''(5000) = 0$ and $P'''(5000) = -2$. Based on the third derivative test, what is the most definitive conclusion about the profit function at an investment level of $x = $5,000?
   Answer: The profit function has an inflection point at $x = $5,000. The third derivative test for inflection points states that if $f''(c) = 0$ and $f'''(c) \neq 0$, then the function $f(x)$ has an inflection point at $x=c$. In this case, we are given $P''(5000) = 0$ and $P'''(5000) = -2 \neq 0$. Therefore, we can definitively conclude that the profit function $P(x)$ has an inflection point at $x = $5,000. This point often represents where the marginal profit is maximized (the point of diminishing returns).
9. The marginal profit for a product is found to be increasing for production levels between 0 and 1,000 units. For production levels above 1,000 units, the marginal profit is positive but decreasing. Which statement provides the most accurate description of the total profit function, $P(q)$, at the production level $q=1,000$?
   Answer: The rate of change of total profit is maximized at $q=1,000$. Let $MP(q)$ be the marginal profit. We are given that $MP(q) = P'(q)$. The problem states that $MP(q)$ is increasing for $q < 1000$ and decreasing for $q > 1000$. This means that the function $MP(q)$ itself has a maximum at $q=1,000$. Since $MP(q)$ is the rate of change of total profit, this means the rate of change of total profit is maximized at $q=1,000$. This corresponds to an inflection point on the total profit curve $P(q)$, where the concavity changes from up ($P''(q)>0$ for $q<1000$) to down ($P''(q)<0$ for $q>1000$).
10. The total cost to produce $q$ units of a specialized electronic component is given by the function $C(q) = 0.05q^3 - 4.5q^2 + 150q + 1000$. At what production level is the marginal cost, $C'(q)$, at a minimum?
    Answer: $q = 30$ units The marginal cost is minimized when its derivative is zero and its second derivative is positive. The derivative of marginal cost, $C'(q)$, is the second derivative of the total cost function, $C''(q)$. 1. Find the marginal cost function: $C'(q) = 0.15q^2 - 9q + 150$. 2. To find the minimum of $C'(q)$, find its derivative, which is $C''(q)$: $C''(q) = 0.3q - 9$. 3. Set $C''(q) = 0$ to find the critical point of the marginal cost function: $0.3q - 9 = 0 \implies 0.3q = 9 \implies q = 30$. 4. Use the second derivative test on $C'(q)$ by examining $C'''(q)$. $C'''(q) = 0.3$. Since $C'''(q) > 0$, the critical point at $q=30$ is indeed a minimum for the marginal cost.
11. An e-commerce company's customer acquisition cost as a function of marketing spend is $A(s) = \frac{500s}{s^2 + 100} + 50$, where $s$ is marketing spend in thousands of dollars and $A(s)$ is the cost per customer in dollars. The marketing team wants to identify the spend level where the rate of change of acquisition cost transitions from concave behavior to convex behavior. What characterizes this transition point?
    Answer: The point where $A''(s) = 0$ and $A''(s)$ changes sign, indicating an inflection point The transition from concave to convex behavior occurs at an inflection point, which is characterized by A''(s) = 0 and a sign change in A''(s). First, A'(s) = 500(100-s²)/(s²+100)². Then A''(s) = 500(-2s)(s²+100)² - 500(100-s²)·2(s²+100)·2s / (s²+100)⁴. Simplifying: A''(s) = 1000s(s²-300)/(s²+100)³. Setting A''(s) = 0: s = 0 or s² = 300. Since s > 0 for marketing spend, s = √300 ≈ 17.32. We can verify the sign change occurs here, making this an inflection point.
12. A company models its profit, $P(x)$, in thousands of dollars, from spending $x$ thousand dollars on advertising as $P(x) = -x^3 + 21x^2 + 90x + 500$ for $x \ge 0$. The company wishes to identify the point of diminishing returns, which is the advertising expenditure at which the rate of growth of profit begins to decrease. What is this expenditure level?
    Answer: $x = 7$ The point of diminishing returns occurs at the inflection point of the profit function, where the concavity changes from upward to downward. This is found by setting the second derivative of the profit function to zero. 1. Find the first derivative (marginal profit): $P'(x) = -3x^2 + 42x + 90$. 2. Find the second derivative: $P''(x) = -6x + 42$. 3. Set the second derivative to zero and solve for $x$: $-6x + 42 = 0 \implies 6x = 42 \implies x = 7$. For $x < 7$, $P''(x) > 0$ (concave up), and for $x > 7$, $P''(x) < 0$ (concave down). Therefore, the point of diminishing returns is at an expenditure of $x=7$ thousand dollars.
13. Two competing advertising campaigns, A and B, generate profit over time $t$ (in weeks) according to the functions $P_A(t) = -2t^3 + 30t^2$ and $P_B(t) = -t^3 + 24t^2$. For what initial time interval is the profit from campaign A accelerating faster than the profit from campaign B?
    Answer: For $t$ between 0 and 2 weeks. The 'acceleration' of profit is represented by the second derivative of the profit function. We need to find the interval where the acceleration of campaign A is greater than that of campaign B, which means we must solve the inequality $P_A''(t) > P_B''(t)$. 1. Find the first and second derivatives for Campaign A: $P_A'(t) = -6t^2 + 60t$ $P_A''(t) = -12t + 60$ 2. Find the first and second derivatives for Campaign B: $P_B'(t) = -3t^2 + 48t$ $P_B''(t) = -6t + 48$ 3. Solve the inequality $P_A''(t) > P_B''(t)$: $-12t + 60 > -6t + 48$ $12 > 6t$ $2 > t$ Since time must be positive, the interval is $0 < t < 2$. So, for the first 2 weeks, campaign A's profit is accelerating faster.

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