Second Derivative Test - Business Calculus

1. A company's marginal cost function is $MC(x) = 6x^2 - 36x + 54$. To find where the marginal cost is minimized, the critical points are found by setting $MC'(x) = 0$, which gives $x = 3$. What additional information is needed to confirm this is indeed a minimum using the second derivative test?
   Answer: Verify that $MC''(3) > 0$ to confirm the marginal cost curve is concave up For the second derivative test, we need $MC''(3) > 0$ to confirm that $x = 3$ gives a local minimum. Since $MC'(x) = 12x - 36$, we have $MC''(x) = 12$, so $MC''(3) = 12 > 0$, confirming a minimum. Choice B tests the wrong condition - the sign of the function value doesn't determine if it's a minimum. Choice C describes verifying the critical point but not applying the second derivative test. Choice D would indicate a maximum, not a minimum.
2. The value of a commodity, $t$ months after its release, is modeled by $V(t) = t^3 - 12t^2 + 36t + 50$ for $t > 0$. A local maximum value was reached in the first year. Using the second derivative test to identify the correct time, what was this local maximum value?
   Answer: $82 First, find the critical points of the value function $V(t)$ by setting its derivative to zero. $V'(t) = 3t^2 - 24t + 36$. Set $V'(t)=0$: $3(t^2 - 8t + 12) = 0 \implies 3(t-2)(t-6) = 0$. The critical points are $t=2$ and $t=6$. Next, use the second derivative test to classify them. $V''(t) = 6t - 24$. For $t=2$: $V''(2) = 6(2) - 24 = 12 - 24 = -12$. Since $V''(2) < 0$, this is a local maximum. For $t=6$: $V''(6) = 6(6) - 24 = 36 - 24 = 12$. Since $V''(6) > 0$, this is a local minimum. The question asks for the local maximum value. This occurs at $t=2$. We must now calculate the value of the function at this time: $V(2) = (2)^3 - 12(2)^2 + 36(2) + 50 = 8 - 12(4) + 72 + 50 = 8 - 48 + 72 + 50 = 82$.
3. Let $f(x)$ be a profit function with derivatives $f'(x)$ and $f''(x)$. Analysis shows that for the interval $[0, 50]$, critical points exist at $x=10$ and $x=40$. It is also known that $f''(10) = -2.5$ and $f''(40) = 1.8$. Which statement correctly interprets these findings for identifying a local maximum profit in the given interval?
   Answer: A local maximum profit occurs at $x=10$ because $f''(10)$ is negative. When analyzing profit functions, you're looking for points where profit is maximized or minimized. This requires understanding how the second derivative test works at critical points. At critical points (where $f'(x) = 0$), the second derivative tells you the nature of that point. If $f''(x) < 0$ at a critical point, the function is concave down there, creating a local maximum—like the peak of a hill. If $f''(x) > 0$, the function is concave up, creating a local minimum—like the bottom of a valley. Here, at $x = 10$, we have $f''(10) = -2.5 < 0$, which means the profit function is concave down at this critical point, confirming a local maximum. This makes answer A correct. Answer B is wrong because $f''(40) = 1.8 > 0$ indicates concave up behavior at $x = 40$, which means this critical point is actually a local minimum, not a maximum. Answer C misunderstands what the second derivative test tells us. Having non-zero second derivatives doesn't make both points maxima—the sign matters crucially. One is a maximum (negative second derivative) and one is a minimum (positive second derivative). Answer D incorrectly suggests an inflection point. Inflection points occur where $f''(x) = 0$, not where the second derivative is simply positive or negative. Remember: at critical points, negative second derivative means local maximum, positive second derivative means local minimum. This is essential for optimizing business functions like profit, cost, and revenue.
4. An investment portfolio's value is modeled by $V(t) = t^5 - 5t^4 + 5t^3 + 10$, where $t$ represents years. The critical points satisfy $V'(t) = 5t^2(t^2 - 4t + 3) = 0$, giving $t = 0$, $t = 1$, and $t = 3$. For the economically relevant critical points $t = 1$ and $t = 3$, what does the second derivative test determine?
   Answer: $t = 1$ is a local maximum and $t = 3$ is a local minimum for portfolio optimization When you encounter critical points in business calculus, the second derivative test helps you classify them as local maxima or minima—crucial for understanding when a portfolio reaches peak or trough values. To apply the second derivative test, you need to find $V''(t)$ and evaluate it at each critical point. Starting with $V'(t) = 5t^4 - 20t^3 + 15t^2$, the second derivative is $V''(t) = 20t^3 - 60t^2 + 30t = 10t(2t^2 - 6t + 3)$. At $t = 1$: $V''(1) = 10(1)(2 - 6 + 3) = 10(-1) = -10 < 0$, indicating a local maximum. At $t = 3$: $V''(3) = 10(3)(18 - 18 + 3) = 10(3)(3) = 90 > 0$, indicating a local minimum. This confirms answer choice D. Answer A is wrong because $V''(1) = -10 \neq 0$, so the test isn't inconclusive at $t = 1$. Answer B incorrectly swaps the classifications—it has the maximum and minimum backwards. Answer C claims both points are minima, but $V''(1) < 0$ clearly indicates a maximum, not a minimum. The key insight is that negative second derivatives signal local maxima (the portfolio value peaks), while positive second derivatives signal local minima (portfolio value bottoms out). Always compute the second derivative carefully and remember: negative means maximum, positive means minimum.
5. A firm's profit function is $P(x) = -x^3 + ax^2 - bx$, where $x$ is the production level and $x>0$. A critical point for profit is known to occur at $x=12$. For the second derivative test to confirm that this critical point represents a local maximum profit, what condition must be true for the parameter $a$?
   Answer: $a < 18$ First, we find the first and second derivatives of the profit function $P(x)$. $P'(x) = -3x^2 + 2ax - b$. $P''(x) = -6x + 2a$. We are given that $x=12$ is a critical point, which means $P'(12)=0$. While this gives a relationship between $a$ and $b$, it is not needed to answer the question about the condition on $a$. For the second derivative test to confirm a local maximum at $x=12$, the second derivative must be negative at that point: $P''(12) < 0$. Substitute $x=12$ into the expression for $P''(x)$: $P''(12) = -6(12) + 2a < 0$. $-72 + 2a < 0$. $2a < 72$. $a < 18$. Thus, the parameter $a$ must be less than 18.
6. The average cost per unit, $\bar{C}(q)$, for a manufacturing process is given by $\bar{C}(q) = 0.02q^2 - 1.6q + 50$ for $q > 0$. A manager wishes to find the production quantity $q$ that minimizes the average cost. After finding the critical point of $\bar{C}(q)$, the manager uses the second derivative test to confirm it is a minimum. What is the correct quantity and justification?
   Answer: $q=40$ units, because $\bar{C}''(40)$ is positive. To find the quantity that minimizes average cost, first find the critical points of $\bar{C}(q)$ by taking the first derivative and setting it to zero. $\bar{C}'(q) = 0.04q - 1.6$. Setting $\bar{C}'(q) = 0$ gives $0.04q = 1.6$, which solves to $q = 1.6 / 0.04 = 40$. Next, apply the second derivative test. The second derivative is $\bar{C}''(q) = 0.04$. Since $\bar{C}''(q)$ is a positive constant (0.04 > 0) for all $q$, the function is always concave up. This confirms that any critical point must be a local (and in this case, global) minimum. At the critical point $q=40$, $\bar{C}''(40) = 0.04$, which is positive. Therefore, $q=40$ units minimizes the average cost.
7. A company's profit from producing $x$ units of a specialized component is given by $P(x) = -\frac{1}{3}x^3 + 13x^2 - 120x + 800$ for $x > 0$. The first derivative, $P'(x)$, indicates two critical values where profit might be optimized. To ensure profit is maximized, which production level should the company choose based on the second derivative test?
   Answer: 20 units, because it is a critical value and the second derivative is negative at this point. First, find the critical points by setting the first derivative of the profit function $P(x)$ to zero. $P'(x) = -x^2 + 26x - 120$. Setting $P'(x) = 0$: $x^2 - 26x + 120 = 0 \implies (x-6)(x-20) = 0$. The critical points are $x=6$ and $x=20$. Next, use the second derivative test to classify these points. The second derivative is $P''(x) = -2x + 26$. Evaluate $P''(x)$ at each critical point: For $x=6$: $P''(6) = -2(6) + 26 = 14$. Since $P''(6) > 0$, the function is concave up, and $x=6$ is a local minimum. For $x=20$: $P''(20) = -2(20) + 26 = -14$. Since $P''(20) < 0$, the function is concave down, and $x=20$ is a local maximum. Therefore, to maximize profit, the company should choose the production level of 20 units.
8. The marginal revenue function for a product is $R'(x) = -0.3x^2 + 12x - 90$, and its derivative is $R''(x) = -0.6x + 12$. A critical point for the total revenue function $R(x)$ occurs at $x=10$. Without finding $R(x)$, determine the nature of this critical point using the information provided.
   Answer: A local minimum in revenue, because $R''(10)$ is positive. The second derivative test uses the sign of the second derivative at a critical point to determine if the point is a local maximum or minimum. We are given that $x=10$ is a critical point of the total revenue function $R(x)$, which means $R'(10) = 0$. To apply the test, we evaluate the second derivative, $R''(x)$, at this critical point: $R''(10) = -0.6(10) + 12 = -6 + 12 = 6$. Since $R''(10) = 6$, which is positive, the revenue function $R(x)$ is concave up at $x=10$. According to the second derivative test, a critical point where the function is concave up corresponds to a local minimum. Therefore, a local minimum in revenue occurs at $x=10$.
9. Let $C(q)$ be the total cost function for producing a quantity $q$ of a certain product. For a production level of $q=500$ units, it is determined that $C'(500) = 0$ and $C''(500) = 1.2$. What is the most accurate business interpretation of these mathematical results?
   Answer: Producing 500 units results in a local minimum cost for the company. $C'(500) = 0$ indicates that the rate of change of cost with respect to quantity is zero at $q=500$. This means $q=500$ is a critical point of the cost function. $C''(500) = 1.2$, being a positive value, indicates that the cost function $C(q)$ is concave up at $q=500$. According to the second derivative test, if $f'(c)=0$ and $f''(c)>0$, then $f$ has a local minimum at $c$. Therefore, a production level of 500 units corresponds to a local minimum in the total cost.
10. Consider the cost function $C(x) = x^3 - 9x^2 + 24x + 100$. The average cost function is $AC(x) = \frac{C(x)}{x} = x^2 - 9x + 24 + \frac{100}{x}$. To minimize average cost, critical points are found where $AC'(x) = 0$. If $x = 5$ is a critical point, what does the second derivative test indicate?
    Answer: Local minimum since $AC''(5) = 2 + \frac{200}{125} = 3.6 > 0$ confirming optimal production When you encounter average cost minimization problems, you're applying the second derivative test to determine whether a critical point represents a minimum (cost-efficient production level) or maximum. This requires careful calculation of the second derivative and correct interpretation of the result. To find $AC''(x)$, you differentiate the average cost function $AC(x) = x^2 - 9x + 24 + \frac{100}{x}$. The first derivative is $AC'(x) = 2x - 9 - \frac{100}{x^2}$, and the second derivative is $AC''(x) = 2 + \frac{200}{x^3}$. At $x = 5$: $AC''(5) = 2 + \frac{200}{125} = 2 + 1.6 = 3.6 > 0$. Since the second derivative is positive, this confirms a local minimum, which represents the optimal production level. Choice A contains a calculation error, computing $\frac{200}{125}$ as $\frac{40}{25}$, though it reaches the correct conclusion about having a minimum. Choice B makes the same calculation error as A but incorrectly interprets a positive second derivative as indicating a maximum, which contradicts the second derivative test. Choice C incorrectly claims the second derivative equals zero, suggesting the test is inconclusive when it actually gives a clear result. Remember that for average cost functions, a positive second derivative at a critical point always indicates a local minimum, representing the most cost-efficient production level. Double-check your arithmetic when evaluating derivatives at specific points, as calculation errors are common in these problems.
11. A demand function is given by $D(p) = 100e^{-0.1p}$, where $p$ is price. The revenue function $R(p) = p \cdot D(p) = 100pe^{-0.1p}$ has a critical point where $R'(p) = 0$. If this critical point occurs at $p = 10$, what does the second derivative test reveal about this point?
    Answer: It is a local maximum since $R''(10) = -10e^{-1} < 0$ for revenue optimization $R'(p) = 100e^{-0.1p} + 100p(-0.1)e^{-0.1p} = 100e^{-0.1p}(1 - 0.1p)$. Setting $R'(p) = 0$ gives $1 - 0.1p = 0$, so $p = 10$. $R''(p) = 100(-0.1)e^{-0.1p}(1 - 0.1p) + 100e^{-0.1p}(-0.1) = -10e^{-0.1p}(1 - 0.1p) - 10e^{-0.1p} = -10e^{-0.1p}(2 - 0.1p)$. At $p = 10$: $R''(10) = -10e^{-1}(2 - 1) = -10e^{-1} < 0$, confirming a local maximum. Choice B has the wrong sign. Choice C incorrectly claims the second derivative is zero. Choice D has an incorrect coefficient.
12. A revenue function $R(x) = -x^4 + 4x^3 - 4x^2 + 16$ has critical points where $R'(x) = 0$. If one critical point occurs at $x = 2$, and $R''(2) = 0$, what can be concluded about the nature of this critical point?
    Answer: The second derivative test is inconclusive and higher-order tests are needed When $f''(c) = 0$ at a critical point $c$, the second derivative test fails to determine whether the point is a local maximum, minimum, or neither. Higher-order derivatives or other methods (like the first derivative test) must be used. Choice A incorrectly assumes the leading coefficient determines local behavior. Choice B makes an unfounded assumption about revenue functions. Choice D confuses the conditions for inflection points - a point can have $f''(c) = 0$ without being an inflection point.
13. A company models its production efficiency with $E(t) = t^4 - 8t^3 + 18t^2 - 27$, where $t$ is time in hours. Critical points occur at $t = 0$, $t = 3$, and $t = 6$. For practical purposes, only $t > 0$ is considered. Which statement correctly applies the second derivative test to the relevant critical points?
    Answer: $t = 3$ gives $E''(3) = 0$ making the test inconclusive, while $t = 6$ is a local minimum $E'(t) = 4t^3 - 24t^2 + 36t$ and $E''(t) = 12t^2 - 48t + 36$. At $t = 3$: $E''(3) = 12(9) - 48(3) + 36 = 108 - 144 + 36 = 0$, so the second derivative test is inconclusive. At $t = 6$: $E''(6) = 12(36) - 48(6) + 36 = 432 - 288 + 36 = 180 > 0$, indicating a local minimum. Choice A incorrectly determines the nature without calculation. Choice C incorrectly concludes about $t = 3$. Choice D incorrectly claims both have positive second derivatives.
14. The price per unit, $p$, for a product is given by the demand equation $p = 500e^{-0.02q}$, where $q$ is the number of units sold. The total revenue is $R(q) = q \cdot p$. What quantity $q$ maximizes total revenue, as confirmed by the second derivative test?
    Answer: $q=50$ units When you encounter a revenue maximization problem with a demand equation, you're looking for the quantity where marginal revenue equals zero. This requires finding the derivative of the revenue function and setting it equal to zero. Given $p = 500e^{-0.02q}$, the revenue function is $R(q) = q \cdot p = 500qe^{-0.02q}$. To find the maximum, take the derivative using the product rule: $R'(q) = 500e^{-0.02q} + 500q(-0.02)e^{-0.02q} = 500e^{-0.02q}(1 - 0.02q)$ Setting $R'(q) = 0$: Since $500e^{-0.02q} > 0$ for all $q$, we need $1 - 0.02q = 0$, which gives us $q = 50$. To confirm this is a maximum, check the second derivative: $R''(q) = 500e^{-0.02q}(-0.04 + 0.0004q)$. At $q = 50$: $R''(50) = 500e^{-1}(-0.04 + 0.02) = 500e^{-1}(-0.02) < 0$, confirming a maximum. Answer A ($q = 25$) gives $1 - 0.02(25) = 0.5 \neq 0$, so the derivative isn't zero here. Answer B ($q = 100$) gives $1 - 0.02(100) = -1 \neq 0$, and since this is negative, revenue is actually decreasing at this point. Answer D ($q = 500$) gives $1 - 0.02(500) = -9$, indicating revenue is decreasing rapidly. Remember: in optimization problems, always verify your critical point with the second derivative test to confirm whether you've found a maximum or minimum.
15. The profit for a product $t$ years after launch is modeled by $P(t) = (t-5)^4 + 200$. An analyst uses the second derivative test to find the time $t$ at which profit is minimized. What is the correct conclusion from applying this specific test at the function's critical point?
    Answer: The second derivative test is inconclusive and cannot determine the nature of the critical point. First, we find the critical point(s) by setting the first derivative of the profit function, $P(t)$, equal to zero. $P'(t) = 4(t-5)^3(1) = 4(t-5)^3$. Setting $P'(t) = 0$ gives $4(t-5)^3 = 0$, so the only critical point is $t=5$. Next, we apply the second derivative test. The second derivative is $P''(t) = 12(t-5)^2$. We evaluate the second derivative at the critical point $t=5$: $P''(5) = 12(5-5)^2 = 12(0)^2 = 0$. When the second derivative at a critical point is zero, the second derivative test is inconclusive. It does not provide any information about whether the point is a local maximum, a local minimum, or neither. (Note: The first derivative test would show that $t=5$ is a local minimum, but the question specifically asks for the conclusion from the second derivative test).

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