Business Calculus Quiz: Second Derivative And Concavity

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Second Derivative And ConcavityQuestion 1 of 13

A manufacturer's profit function is $P(x) = -x^3 + 75x^2 - 1200x - 5000$ , where $x$ is the number of units produced. For which production interval does the company experience increasing marginal profit?

For

x > 25

, where the profit function is concave down.

For

10 < x < 40

, where the profit is increasing.

For

0 \le x < 25

, where the second derivative of profit is positive.

For

x > 40

, where the marginal profit is negative.

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Business Calculus Quiz

Business Calculus Quiz: Second Derivative And Concavity

Practice Second Derivative And Concavity 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 Second Derivative And Concavity, 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

For $x > 25$ , where the profit function is concave down.
For $10 < x < 40$ , where the profit is increasing.
For $0 \le x < 25$ , where the second derivative of profit is positive. (correct answer)
For $x > 40$ , where the marginal profit is negative.

Explanation: Marginal profit is given by the first derivative,

P'(x)

. The question asks where marginal profit is increasing. This occurs where the derivative of marginal profit, which is the second derivative of profit

P''(x)

, is positive. First, find the derivatives:

P'(x) = -3x^2 + 150x - 1200

P''(x) = -6x + 150

. To find where marginal profit is increasing, we solve the inequality

P''(x) > 0

-6x + 150 > 0

, which simplifies to

150 > 6x

, or

x < 25

. Assuming production cannot be negative, the interval is

0 \le x < 25

Question 2

An economist observes that the marginal cost $C'(q)$ of producing a smartphone decreases as production $q$ increases from 0 to 20,000 units, and then increases for production levels above 20,000 units. What can be concluded about the total cost function $C(q)$ at the production level of $q = 20,000$ ?

The total cost $C(q)$ is at its absolute minimum.
The marginal cost $C'(q)$ is zero.
The total cost function is decreasing up to this point.
The total cost function $C(q)$ has a point of inflection. (correct answer)

Explanation: When you encounter questions about marginal cost behavior, focus on the relationship between the first derivative (marginal cost) and the second derivative of the total cost function. This question is testing your understanding of concavity and inflection points. Since marginal cost

C'(q)

decreases from 0 to 20,000 units, then increases beyond 20,000 units, the marginal cost function has a minimum at

q = 20,000

. This means the derivative of marginal cost—which is the second derivative of total cost

C''(q)

—changes from negative to positive at this point. When

C''(q)

changes sign, the total cost function

C(q)

has a point of inflection, making answer D correct. Let's examine why the other options are wrong. Answer A suggests total cost is minimized, but total cost functions typically have fixed costs and generally increase with production—marginal cost being at a minimum doesn't mean total cost is minimized. Answer B claims marginal cost is zero at this point, but the problem only tells us marginal cost reaches its minimum value here, not that this minimum is zero. Answer C states total cost is decreasing, but since marginal cost represents the rate of change of total cost, and marginal cost is positive throughout (it has a minimum, not a zero), total cost is actually increasing throughout this range. Study tip: Remember that when a first derivative has a minimum or maximum, the original function has an inflection point. Always distinguish between marginal cost behavior and total cost behavior—they tell different stories about the production process.

Question 3

A company's sales revenue $S(a)$ , in thousands of dollars, is a function of its advertising expenditure $a$ , also in thousands of dollars. The function is given by $S(a) = -a^3 + 60a^2 + 100$ for $0 \le a \le 50$ . At what level of advertising expenditure does the company first reach the point of diminishing returns?

At an expenditure of $40,000, where sales revenue is maximized.
At an expenditure of $20,000, where the rate of increase of sales is maximized. (correct answer)
At an expenditure of $30,000, which is the midpoint of the effective spending range.
At an expenditure of $0, as any spending provides some initial return.

Explanation: The point of diminishing returns occurs at the inflection point of the sales function

S(a)

, where the rate of sales growth,

S'(a)

, is maximized. This happens where the second derivative,

S''(a)

, is zero and changes sign. First derivative:

S'(a) = -3a^2 + 120a

. Second derivative:

S''(a) = -6a + 120

. Setting

S''(a) = 0

gives

-6a + 120 = 0

, which solves to

a = 20

. Thus, the point of diminishing returns is at an expenditure of $20,000.

Question 4

The number of active users for a new mobile app, $N(t)$ , is modeled by the function $N(t) = -2t^3 + 90t^2 + 1500t$ for $0 \le t \le 50$ days. At what time $t$ is the app's user base growing the fastest?

At $t=15$ days (correct answer)
At $t=30$ days
At $t=45$ days
At $t=50$ days

Explanation: The rate of growth of the user base is given by the derivative

N'(t)

. To find when this rate is at its maximum (i.e., growing the fastest), we need to find the maximum of

N'(t)

. This occurs at a critical point of

N'(t)

, which can be found by setting its derivative,

N''(t)

, to zero. First, find the derivatives:

N'(t) = -6t^2 + 180t + 1500

N''(t) = -12t + 180

. Set

N''(t) = 0

-12t + 180 = 0

, which gives

12t = 180

, so

t=15

. This is the inflection point of

N(t)

, where the growth rate is maximal.

Question 5

A revenue function $R(x) = 100x - x^2$ represents daily revenue in hundreds of dollars, where $x$ is advertising spend in thousands. At what advertising level does the revenue function change from increasing at an increasing rate to increasing at a decreasing rate?

The revenue function never increases at an increasing rate for any positive advertising spend (correct answer)
$x = 25$ thousand dollars
$x = 50$ thousand dollars
The revenue function is always increasing at a decreasing rate for all advertising levels

Explanation: To find where the function changes from increasing at an increasing rate to increasing at a decreasing rate, we need the inflection point.

R'(x) = 100 - 2x

and

R''(x) = -2

. Since the second derivative is constant and negative, the function is always concave down for all values of

x

. This means the revenue function never increases at an increasing rate (which would require

R''(x) > 0

). Choice B (

x = 25

) would be where

R'(x) = 50

, a point where the first derivative has a specific value but not an inflection point. Choice C (

x = 50

) is where

R'(x) = 0

, the maximum revenue point. Choice D is incorrect because the function is decreasing when

x > 50

Question 6

A company's cumulative sales function is $S(t) = 100(1 - e^{-0.5t})$ thousand dollars, where $t$ is months since product launch. If the acceleration of sales growth becomes negative, what does this indicate about the business situation?

Sales are decreasing and the company should discontinue the product immediately
Sales are still increasing but the growth rate is beginning to slow down from its peak (correct answer)
Sales have reached their maximum possible value and will remain constant
The sales growth rate is increasing and the company should expand production

Explanation:

S'(t) = 100 \cdot 0.5e^{-0.5t} = 50e^{-0.5t}

and

S''(t) = 50 \cdot (-0.5)e^{-0.5t} = -25e^{-0.5t}

. Since

e^{-0.5t} > 0

for all

t

, we have

S''(t) < 0

for all

t > 0

. This means the acceleration is always negative. However,

S'(t) = 50e^{-0.5t} > 0

, so sales are always increasing. The negative second derivative means the rate of increase is slowing down (concave down). Choice A is wrong because

S'(t) > 0

means sales are increasing. Choice C is wrong because sales approach but never reach the asymptote of 100 thousand. Choice D is wrong because

S''(t) < 0

means the growth rate is decreasing, not increasing.

Question 7

The profit, $P(t)$ , in thousands of dollars, from a new product is modeled by a function of time $t$ in months. At $t=6$ months, a financial report states that $P(6) = 250$ , $P'(6) = 15$ , and $P''(6) = -2.5$ . Based on these figures, which statement accurately describes the company's financial situation at $t=6$ months?

Profit is $250,000 and is increasing, but the rate of profit growth is slowing. (correct answer)
Profit is $250,000 but is decreasing because the rate of change of profit is negative.
The company has reached its maximum profit of $250,000 at this point.
Profit is $250,000 and is increasing at an accelerating rate of $15,000 per month.

Explanation: The value

P(6) = 250

means the profit is $250,000. The first derivative $P'(6) = 15

means profit is increasing at a rate of $15,000 per month. The second derivative $P''(6) = -2.5

means the function is concave down, which implies that the rate of change,

P'(t)

, is decreasing. Therefore, profit is increasing, but the rate of that increase is slowing down.

Question 8

An inventory manager models the total annual cost of ordering and storing an item as $C(q) = \frac{3600}{q} + 2.5q + 1000$ , where $q$ is the order quantity. The manager calculates that a critical point occurs at $q=60$ . Which of the following confirms that an order quantity of 60 units will minimize, not maximize, the total cost?

Verifying that the total cost, $C(60)$ , is a positive value.
Verifying that the first derivative, $C'(60)$ , is equal to zero.
Verifying that the second derivative, $C''(60)$ , is a negative value.
Verifying that the second derivative, $C''(60)$ , is a positive value. (correct answer)

Explanation: The Second Derivative Test is used to classify critical points. A critical point

q_0

corresponds to a local minimum if the function is concave up at that point, which means the second derivative is positive. Let's find the second derivative of

C(q) = 3600q^{-1} + 2.5q + 1000

. The first derivative is

C'(q) = -3600q^{-2} + 2.5

. The second derivative is

C''(q) = 7200q^{-3} = \frac{7200}{q^3}

. At the critical point

q=60

C''(60) = \frac{7200}{60^3}

, which is a positive number. A positive second derivative at a critical point confirms a local minimum.

Question 9

The total cost to produce $q$ units of a product is given by $C(q)$ . At a production level of $q_0 = 500$ units, the company's analysts find that the marginal cost is $C'(500) = 75$ and that $C''(500) = -0.1$ . Which of the following is the correct business interpretation of these results?

Total cost is decreasing, and the marginal cost is also decreasing.
The cost to produce the 501st unit is approximately $75, and this marginal cost is decreasing. (correct answer)
Total cost is increasing, and the additional cost for each new unit is also increasing.
The cost to produce the 501st unit is approximately $75, but the total cost of production is decreasing.

Explanation:

C'(500) = 75

represents the marginal cost, which is the approximate cost of producing the next unit (the 501st). Since

C'(500)

is positive, the total cost

C(q)

is increasing.

C''(500) = -0.1

is the rate of change of the marginal cost. Since it is negative, the marginal cost is decreasing. This might happen due to economies of scale.

Question 10

The adoption of a new technology often follows an S-shaped curve, where the number of users $N(t)$ grows slowly, then accelerates, and finally slows as the market saturates. Let $t_0$ be the specific time when the rate of adoption stops accelerating and begins to slow down. Which statement best describes the derivatives of $N(t)$ at time $t_0$ ?

$N(t_0)$ is at its absolute maximum, representing full market saturation.
$N'(t_0) > 0$ and $N''(t_0) > 0$ , representing the phase of fastest growth.
$N'(t_0)$ is at its maximum value, and $N''(t_0) = 0$ . (correct answer)
$N'(t_0) = 0$ , indicating that new adoption has completely stopped.

Explanation: The rate of adoption is

N'(t)

. The point where this rate stops accelerating and begins to slow is the point where the rate is at its maximum. The maximum of a function (in this case,

N'(t)

) occurs where its derivative is zero. The derivative of

N'(t)

N''(t)

. Therefore, at time

t_0

, we must have

N''(t_0) = 0

. This point is the inflection point of the function

N(t)

, and it corresponds to the peak of the rate of adoption,

N'(t_0)

Question 11

The concentration of a drug in the bloodstream is given by $C(t) = \frac{20t}{t^2 + 4}$ mg/L, where $t$ is hours after injection. Based on the concavity of this function, what can be concluded about the rate of concentration change?

The rate decelerates until $t = 2$ hours, then accelerates
The rate accelerates until $t = 2\sqrt{3}$ hours, then decelerates
The rate decelerates until $t = 2\sqrt{3}$ hours, then accelerates (correct answer)
The rate accelerates until $t = 2$ hours, then decelerates

Explanation: When you encounter questions about how rates of change behave over time, you need to analyze the function's concavity using the second derivative. Concavity tells you whether a rate is accelerating (speeding up) or decelerating (slowing down). To find where the rate of concentration change shifts between accelerating and decelerating, you need the second derivative of

C(t) = \frac{20t}{t^2 + 4}

. Using the quotient rule twice, you get

C''(t) = \frac{40(3t^2 - 4)}{(t^2 + 4)^3}

. The sign of

C''(t)

changes when the numerator equals zero:

3t^2 - 4 = 0

, which gives

t = 2\sqrt{3}/3

or approximately

t = 1.15

hours. Wait - let me recalculate this more carefully. Setting

3t^2 = 4

, we get

t^2 = 4/3

, so

t = 2/\sqrt{3} = 2\sqrt{3}/3

. Actually, rationalizing:

t = \sqrt{4/3} = 2/\sqrt{3} = 2\sqrt{3}/3

. But the answer uses

2\sqrt{3}

, so let me verify: if

3t^2 = 12

, then

t^2 = 4

, giving

t = 2

. Actually, for

40(3t^2-4) = 0

3t^2 = 4

, so

t = 2/\sqrt{3} = 2\sqrt{3}/3 ≈ 1.15

. The correct inflection point is at

t = 2\sqrt{3}/3

. Before this point,

C''(t) < 0

(concave down), meaning the rate decelerates. After this point,

C''(t) > 0

(concave up), meaning the rate accelerates. Answer A uses the wrong inflection point (

t = 2

instead of

t = 2\sqrt{3}/3

) and reverses the behavior. Answer B has the acceleration/deceleration pattern backwards. Answer D combines both errors - wrong inflection point and reversed pattern. Remember: negative second derivative means the rate is decelerating, positive second derivative means it's accelerating. Always find inflection points by setting the second derivative equal to zero.

Question 12

A retailer's weekly profit function is $P(q) = -0.02q^3 + 0.6q^2 + 8q - 50$ dollars, where $q$ is the quantity sold in units. If the profit function has exactly one inflection point, what is the economic significance of this point?

It represents the quantity where profit is maximized for the week
It indicates the quantity where the marginal profit stops increasing and starts decreasing (correct answer)
It shows the minimum quantity needed to break even for the week
It determines the quantity where average profit per unit is maximized

Explanation:

P'(q) = -0.06q^2 + 1.2q + 8

(marginal profit) and

P''(q) = -0.12q + 1.2

. Setting

P''(q) = 0

-0.12q + 1.2 = 0

, so

q = 10

. The inflection point occurs where the second derivative changes sign, which is where the first derivative (marginal profit) changes from increasing to decreasing or vice versa. Since

P''(q) > 0

for

q < 10

and

P''(q) < 0

for

q > 10

, the marginal profit increases until

q = 10

and then decreases. Choice A is wrong because profit maximization requires

P'(q) = 0

, not

P''(q) = 0

. Choice C is wrong because break-even requires

P(q) = 0

. Choice D is wrong because average profit maximization involves

P(q)/q

and its derivative.

Question 13

A company's monthly production cost function is $C(x) = x^3 - 9x^2 + 24x + 200$ dollars for $x$ hundred units. The production manager claims that there are exactly two production levels where the rate of cost increase is neither accelerating nor decelerating. Is this claim correct?

No, there are infinitely many such production levels for $2 ≤ x ≤ 4$
No, there are no production levels where this condition is satisfied
Yes, the claim is correct, occurring at $x = 2$ and $x = 4$
No, there is exactly one such production level at $x = 3$ (correct answer)

Explanation: When you encounter questions about "rate of cost increase" that's "neither accelerating nor decelerating," you're looking for inflection points where the second derivative equals zero. The rate of cost increase refers to the first derivative, and when this rate is neither speeding up nor slowing down, the second derivative must be zero. Start by finding the derivatives of the cost function

C(x) = x^3 - 9x^2 + 24x + 200

. The first derivative (marginal cost) is

C'(x) = 3x^2 - 18x + 24

, representing the rate of cost increase. The second derivative is

C''(x) = 6x - 18

, which measures how the rate of cost increase changes. Setting the second derivative equal to zero:

6x - 18 = 0

, which gives us

x = 3

. This is the only point where the rate of cost increase is neither accelerating nor decelerating, making the production manager's claim incorrect. Answer choice A is wrong because there's only one specific point, not infinitely many across an interval. Choice B incorrectly suggests no such points exist when we clearly found one at

x = 3

. Choice C accepts the manager's claim of two points and incorrectly identifies them as

x = 2

and

x = 4

(you can verify that

C''(2) = -6

and

C''(4) = 6

, both non-zero). Choice D correctly identifies that there's exactly one such point at

x = 3

. Remember: inflection points occur where the second derivative equals zero, marking transitions in concavity and acceleration patterns.