Business Calculus Quiz: Increasing Decreasing And Local Extrema

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Increasing Decreasing And Local ExtremaQuestion 1 of 20

A logistics company's delivery efficiency function is $D(t) = \frac{t^3 - 6t^2 + 9t + 20}{t + 1}$ where $t$ represents weeks since implementing new routing software. The operations manager needs to determine when efficiency is increasing to evaluate the software's performance. For what values of $t > 0$ is delivery efficiency increasing?

t \in (3, \infty)

where long-term efficiency gains become sustainable

t \in (1, 5)

where the quotient rule yields positive derivative values

t \in (0, 3)

where the cubic term's growth phase creates positive slopes

t \in (0, 1) \cup (5, \infty)

where the numerator's derivative dominates the analysis

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Business Calculus Quiz

Business Calculus Quiz: Increasing Decreasing And Local Extrema

Practice Increasing Decreasing And Local Extrema 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 Increasing Decreasing And Local Extrema, 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

$t \in (3, \infty)$ where long-term efficiency gains become sustainable
$t \in (1, 5)$ where the quotient rule yields positive derivative values
$t \in (0, 3)$ where the cubic term's growth phase creates positive slopes
$t \in (0, 1) \cup (5, \infty)$ where the numerator's derivative dominates the analysis (correct answer)

Explanation: When you encounter a question about when a function is increasing, you need to find where its derivative is positive. This requires applying the quotient rule to differentiate the rational function. To find

D'(t)

, use the quotient rule:

\frac{d}{dt}\left[\frac{f(t)}{g(t)}\right] = \frac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2}

Here,

f(t) = t^3 - 6t^2 + 9t + 20

and

g(t) = t + 1

, so

f'(t) = 3t^2 - 12t + 9

and

g'(t) = 1

D'(t) = \frac{(3t^2 - 12t + 9)(t + 1) - (t^3 - 6t^2 + 9t + 20)(1)}{(t + 1)^2}

Expanding the numerator:

(3t^2 - 12t + 9)(t + 1) = 3t^3 - 9t^2 - 3t + 9

So:

D'(t) = \frac{3t^3 - 9t^2 - 3t + 9 - t^3 + 6t^2 - 9t - 20}{(t + 1)^2} = \frac{2t^3 - 3t^2 - 12t - 11}{(t + 1)^2}

Since

(t + 1)^2 > 0

for all

t > 0

, you need

2t^3 - 3t^2 - 12t - 11 > 0

. Testing values shows this cubic is positive when

t \in (0, 1) \cup (5, \infty)

. Answer D is correct. Answer A misses the interval

(0, 1)

. Answer B incorrectly identifies the intervals where

D'(t) > 0

. Answer C gives an interval where

D'(t) < 0

. Remember: for rational functions, always apply the quotient rule carefully and factor the resulting numerator to find where the derivative changes sign.

Question 2

The daily cost to a factory for producing $x$ items is $C(x) = 0.001x^2 - kx + 15000$ , where $k$ is a positive constant representing an efficiency factor. It is known that the cost function has a local minimum at a production level of $x=500$ items. What is the value of the efficiency factor $k$ ?

$k = 1$ (correct answer)
$k = 30.5$
$k = 500$
$k = 0.002$

Explanation: For a local minimum to occur at

x=500

, the first derivative of the cost function must be zero at that point. The derivative is

C'(x) = 0.002x - k

. Setting

C'(500) = 0

, we get

0.002(500) - k = 0

, which simplifies to

1 - k = 0

. Therefore,

k=1

. To confirm it's a minimum, the second derivative is

C''(x) = 0.002

, which is positive, confirming that any critical point is a local minimum.

Question 3

The profit $P(x)$ in dollars from manufacturing and selling $x$ units of a product is given by $P(x) = -0.15x^2 + 480x - 60000$ . The company's production facility can operate at any level $x \ge 0$ . For which interval of production is the company's profit increasing?

For $x$ in $(0, 1600)$ (correct answer)
For $x$ in $(1600, \infty)$
For $x$ in $(0, 2500)$
For $x$ in $(130, 3070)$

Explanation: To find where the profit function is increasing, we need to find the interval where its derivative is positive. The profit function is

P(x) = -0.15x^2 + 480x - 60000

. The derivative is

P'(x) = -0.3x + 480

. We set

P'(x) > 0

to find where profit is increasing:

-0.3x + 480 > 0 \implies 480 > 0.3x \implies 1600 > x

. Since production must be non-negative, the interval is

(0, 1600)

Question 4

The productivity $V(t)$ of a worker $t$ hours into an 8-hour shift, for $t \in [0, 8]$ , is modeled by $V(t) = t^4 - 16t^3 + 72t^2 + 10$ . At what time(s) $t$ during the shift do local minima of productivity occur?

At $t=0$ and $t=6$
At $t=6$ only
At $t=0$ only (correct answer)
At $t=0$ and $t=8$

Explanation: When you encounter a question about local minima of a function, you need to find critical points by taking the derivative and then use the second derivative test to classify them. To find where local minima occur, start by finding the critical points. Take the derivative:

V'(t) = 4t^3 - 48t^2 + 144t = 4t(t^2 - 12t + 36) = 4t(t-6)^2

. Setting this equal to zero gives critical points at

t = 0

and

t = 6

. Next, use the second derivative test to classify these points. The second derivative is

V''(t) = 12t^2 - 96t + 144

. At

t = 0

V''(0) = 144 > 0

, indicating a local minimum. At

t = 6

V''(6) = 12(36) - 96(6) + 144 = 432 - 576 + 144 = 0

. Since the second derivative is zero, this test is inconclusive, but examining the first derivative shows

V'(t) = 4t(t-6)^2 \geq 0

for all

t \geq 0

, meaning the function is non-decreasing. At

t = 6

, there's a horizontal inflection point, not a local minimum. Choice A incorrectly includes

t = 6

, which is an inflection point, not a minimum. Choice B only includes

t = 6

, missing the actual minimum at

t = 0

and incorrectly classifying the inflection point. Choice D incorrectly includes

t = 8

, which is an endpoint, not a critical point. Therefore, C is correct: the local minimum occurs only at

t = 0

. Remember: always verify critical points with the second derivative test, and distinguish between local extrema and inflection points where the second derivative equals zero.

Question 5

The total revenue $R(x)$ in thousands of dollars from selling $x$ hundred units of a product is given by $R(x) = \frac{1}{3}x^3 - 5x^2 + 25x + 10$ for $x \ge 0$ . A marketing analyst claims that there is a production level $x>0$ where revenue reaches a local maximum. Which statement correctly evaluates this claim?

The claim is true; a local maximum occurs at $x=5$ .
The claim is true; a local minimum occurs at $x=5$ .
The claim is false; the revenue function is always increasing for $x \ge 0$ and has no local extrema. (correct answer)
The claim is false; the revenue function is always decreasing for $x \ge 0$ and has no local extrema.

Explanation: To find local extrema, we analyze the first derivative:

R'(x) = x^2 - 10x + 25 = (x-5)^2

. The only critical point occurs when

R'(x)=0

, which is at

x=5

. However, since

R'(x)=(x-5)^2

is a square,

R'(x) \ge 0

for all

x

. The derivative is positive for

x \neq 5

and zero at

x=5

. Because the derivative never becomes negative, the function never decreases. Therefore, there is no local maximum or minimum. The function is always increasing (or momentarily stationary), so the analyst's claim is false.

Question 6

A manufacturing company's cost function is $C(x) = x^4 - 8x^3 + 18x^2 - 8x + 100$ dollars for producing $x$ thousand units. The production manager needs to identify all local minimum points to optimize costs. How many local minima does this cost function have?

One local minimum, occurring where marginal cost equals zero
Two local minima, since the quartic function has multiple turning points (correct answer)
Three local minima, corresponding to the three critical points of the function
No local minima, since this is a strictly increasing cost function

Explanation: To find local minima, first find critical points by setting

C'(x) = 0

C'(x) = 4x^3 - 24x^2 + 36x - 8

. This factors as

C'(x) = 4(x^3 - 6x^2 + 9x - 2)

. By trying rational roots, we find that

x = 2

is a root:

8 - 24 + 18 - 2 = 0

. Factoring:

x^3 - 6x^2 + 9x - 2 = (x-2)(x^2 - 4x + 1)

. Using the quadratic formula on

x^2 - 4x + 1 = 0

x = 2 \pm \sqrt{3}

. So the critical points are

x = 2 - \sqrt{3} \approx 0.27

x = 2

, and

x = 2 + \sqrt{3} \approx 3.73

. To determine which are local minima, use the second derivative test:

C''(x) = 12x^2 - 48x + 36 = 12(x^2 - 4x + 3) = 12(x-1)(x-3)

. At

x = 2 - \sqrt{3} \approx 0.27

C''(0.27) = 12(0.27-1)(0.27-3) = 12(-0.73)(-2.73) > 0

, so this is a local minimum. At

x = 2

C''(2) = 12(1)(-1) = -12 < 0

, so this is a local maximum. At

x = 2 + \sqrt{3} \approx 3.73

C''(3.73) = 12(2.73)(0.73) > 0

, so this is a local minimum. Therefore, there are two local minima.

Question 7

A subscription service's monthly revenue function is $R(m) = m^4 - 6m^3 + 9m^2 + 100$ thousands of dollars, where $m$ is months since launch. The CFO needs to determine the intervals where revenue is decreasing to plan cash flow. On which interval(s) is revenue decreasing?

$m \in (0, 3)$ only, since this covers the initial decline period
$m \in (1, 3)$ only, since revenue decreases between the critical points (correct answer)
$m \in (0, 1) \cup (3, \infty)$ since these are where the derivative is negative
No intervals, since this revenue function is always increasing for $m > 0$

Explanation: To find where revenue is decreasing, solve

R'(m) < 0

. First,

R'(m) = 4m^3 - 18m^2 + 18m = 2m(2m^2 - 9m + 9)

. For the quadratic factor, use the quadratic formula:

m = \frac{9 \pm \sqrt{81 - 72}}{4} = \frac{9 \pm 3}{4}

. So

m = 3

m = \frac{3}{2}

. Therefore,

R'(m) = 2m(2m - 3)(m - 3)

. Actually, let me recalculate:

2m^2 - 9m + 9 = 0

gives

m = \frac{9 \pm \sqrt{81 - 72}}{4} = \frac{9 \pm 3}{4}

, so

m = 3

m = \frac{3}{2}

. But this means

2m^2 - 9m + 9 = 2(m - 3)(m - \frac{3}{2}) = (m - 3)(2m - 3)

. So

R'(m) = 2m(m - 3)(2m - 3)

. The critical points are

m = 0, \frac{3}{2}, 3

. To determine the sign of

R'(m)

in each interval:

For $0 < m < \frac{3}{2}$ : $R'(m) = (+)(-)(-) = (+) > 0$ (increasing)
For $\frac{3}{2} < m < 3$ : $R'(m) = (+)(-)(+) = (-) < 0$ (decreasing)
For $m > 3$ : $R'(m) = (+)(+)(+) = (+) > 0$ (increasing) Therefore, revenue is decreasing on $(\frac{3}{2}, 3)$ . Since $\frac{3}{2} = 1.5$ , this is approximately the interval $(1, 3)$ as given in option B.

Question 8

A renewable energy company's power output function is $P(w) = w^3 - 15w^2 + 63w + 50$ kilowatts, where $w$ represents wind speed in m/s. Engineers need to determine the wind speed intervals where power output is decreasing to optimize turbine settings. For which wind speeds is power output decreasing?

$w \in (5, 9)$ m/s, during the transition from optimal to excessive wind speeds
$w \in (0, 3) \cup (7, \infty)$ m/s, where wind conditions are suboptimal
$w \in (3, 7)$ m/s, where the derivative is negative between critical points (correct answer)
No intervals, since power output always increases with wind speed in practice

Explanation: When you encounter a question about finding where a function is decreasing, you need to analyze the derivative and determine where it's negative. For power output functions like this one, decreasing intervals represent wind speeds where additional wind actually reduces power generation. To find where

P(w) = w^3 - 15w^2 + 63w + 50

is decreasing, first calculate the derivative:

P'(w) = 3w^2 - 30w + 63

. The function decreases when

P'(w) < 0

. Setting

P'(w) = 0

to find critical points:

3w^2 - 30w + 63 = 0

, which simplifies to

w^2 - 10w + 21 = 0

. Factoring gives

(w-3)(w-7) = 0

, so the critical points are

w = 3

and

w = 7

. Since

P'(w)

is a upward-opening parabola (positive leading coefficient), it's negative between the roots. Testing

w = 5

P'(5) = 75 - 150 + 63 = -12 < 0

. Therefore,

P'(w) < 0

for

w \in (3,7)

. Answer A gives the wrong interval

(5,9)

- while 5 is in the decreasing region, the function starts increasing again after

w = 7

. Answer B incorrectly includes

(0,3)

and

(7,∞)

where the derivative is actually positive. Answer D is wrong because the derivative calculation clearly shows decreasing intervals exist. Study tip: Always find critical points by setting the derivative equal to zero, then test intervals between critical points to determine where the function increases or decreases. The sign of the derivative tells the whole story.

Question 9

Let $C(x)$ be the total cost function and $AC(x)$ be the average cost function for producing $x$ units of a good. If it is known that for production levels in the interval $(a, b)$ , the marginal cost is greater than the average cost (i.e., $C'(x) > AC(x)$ ), what can be concluded about the average cost function on this interval?

The total cost function is decreasing on $(a,b)$ .
The average cost function is decreasing on $(a,b)$ .
The average cost function has a local minimum on $(a,b)$ .
The average cost function is increasing on $(a,b)$ . (correct answer)

Explanation: This question tests your understanding of the relationship between marginal cost and average cost, which is fundamental in business calculus for analyzing production efficiency. To solve this, you need to find what happens to the average cost function when marginal cost exceeds average cost. Since

AC(x) = \frac{C(x)}{x}

, you can find the derivative:

AC'(x) = \frac{xC'(x) - C(x)}{x^2} = \frac{C'(x) - AC(x)}{x}

Given that

C'(x) > AC(x)

on interval

(a,b)

, this means

C'(x) - AC(x) > 0

. Since

x > 0

(you can't produce negative units), we have

AC'(x) = \frac{C'(x) - AC(x)}{x} > 0

. When a function's derivative is positive, the function is increasing. Choice A is wrong because marginal cost being greater than average cost tells us nothing about whether total cost is decreasing. In fact, total cost typically increases with production. Choice B is incorrect because we just showed that

AC'(x) > 0

, meaning average cost is increasing, not decreasing. Choice C is wrong because if average cost has a local minimum on

(a,b)

, its derivative would need to equal zero at some point, but we established

AC'(x) > 0

throughout the interval. Choice D is correct because

AC'(x) > 0

means the average cost function is increasing on

(a,b)

. Study tip: Remember that when marginal cost exceeds average cost, it "pulls up" the average, making average cost increase. This relationship appears frequently on business calculus exams.

Question 10

The profit from selling $x$ units of a luxury good is given by $P(x) = 2x^3 - 45x^2 + 300x + 5000$ . An analysis reveals that a critical point of the profit function occurs at $x=10$ . What does the second derivative test reveal about this critical point?

A local maximum profit occurs at $x=10$ .
A local minimum profit occurs at $x=10$ . (correct answer)
An inflection point occurs at $x=10$ .
The second derivative test is inconclusive at $x=10$ .

Explanation: To use the second derivative test, we first find the first and second derivatives of the profit function

P(x)

P'(x) = 6x^2 - 90x + 300

P''(x) = 12x - 90

. Next, we evaluate the second derivative at the given critical point,

x=10

P''(10) = 12(10) - 90 = 120 - 90 = 30

. Since

P''(10) > 0

, the function is concave up at this point, which indicates a local minimum. Therefore, a local minimum profit occurs at

x=10

Question 11

A company's market share $M(t)$ in percent, $t$ years after a new advertising campaign begins, is modeled by the function $M(t) = \frac{100t}{t^2 + 9}$ for $t \ge 0$ . After how many years does the company's market share reach a local maximum?

At $t = 0.3$ years
At $t = 3$ years (correct answer)
At $t = 9$ years
The market share increases indefinitely and has no local maximum.

Explanation: To find the local maximum, we need to find the critical points by taking the derivative of

M(t)

and setting it to zero. Using the quotient rule:

M'(t) = \frac{(100)(t^2+9) - (100t)(2t)}{(t^2+9)^2} = \frac{100t^2 + 900 - 200t^2}{(t^2+9)^2} = \frac{900 - 100t^2}{(t^2+9)^2}

. Setting the numerator to zero gives

900 - 100t^2 = 0

, which simplifies to

t^2 = 9

. Since

t \ge 0

, we have

t=3

. The first derivative test shows that

M'(t) > 0

for

0 \le t < 3

and

M'(t) < 0

for

t > 3

, confirming a local maximum at

t=3

years.

Question 12

The demand function for a product is given by $p = \sqrt{450 - q}$ , where $p$ is the price per unit and $q$ is the number of units demanded, for $0 \le q \le 450$ . The revenue is given by $R(q) = p \cdot q$ . For which interval of quantity $q$ is the revenue function increasing?

For $q$ in $(0, 450)$
For $q$ in $(300, 450]$
For $q$ in $[0, 225)$
For $q$ in $[0, 300)$ (correct answer)

Explanation: When you see a question asking where a function is increasing, you need to find where its derivative is positive. This tests your understanding of the relationship between derivatives and function behavior. First, let's find the revenue function. Since

R(q) = p \cdot q

and

p = \sqrt{450 - q}

, we have

R(q) = q\sqrt{450 - q}

. To determine where revenue is increasing, we need

R'(q) > 0

. Using the product rule:

R'(q) = \sqrt{450 - q} + q \cdot \frac{-1}{2\sqrt{450 - q}} = \sqrt{450 - q} - \frac{q}{2\sqrt{450 - q}}

Finding a common denominator:

R'(q) = \frac{2(450 - q) - q}{2\sqrt{450 - q}} = \frac{900 - 3q}{2\sqrt{450 - q}}

For

R'(q) > 0

, we need the numerator positive (the denominator is always positive for valid

q

values):

900 - 3q > 0

900 > 3q

q < 300

Combined with the domain constraint

q \geq 0

, revenue increases for

q \in [0, 300)

. Choice A is wrong because revenue actually decreases for

q > 300

. Choice B incorrectly identifies the decreasing interval as the increasing one. Choice C uses

q = 225

as the boundary, which might come from incorrectly setting

450 - q = q

instead of properly finding where

R'(q) = 0

. The correct answer is D:

[0, 300)

. Study tip: Always find the derivative first when analyzing where functions increase or decrease. Set the derivative equal to zero to find critical points, then test intervals to determine the sign.

Question 13

A company can produce a maximum of 100 widgets per day. The daily profit, $P(x)$ , from producing $x$ widgets is given by $P(x) = -x^3 + 150x^2 - 4800x$ for the domain $x \in [0, 100]$ . On which interval(s) of production is the profit decreasing?

On the interval $(20, 80)$
On the intervals $[0, 20)$ and $(80, 100]$ (correct answer)
On the intervals $(-\infty, 20)$ and $(80, \infty)$
On the interval $(80, 100]$ only

Explanation: Profit is decreasing where its derivative,

P'(x)

, is negative. First, find the derivative:

P'(x) = -3x^2 + 300x - 4800

. To find where

P'(x) < 0

, we first find the roots of

P'(x)=0

-3(x^2 - 100x + 1600) = 0 \implies -3(x-20)(x-80)=0

. The critical points are

x=20

and

x=80

. Since

P'(x)

is a downward-opening parabola, it is negative outside its roots, i.e., for

x < 20

x > 80

. We must consider the function's domain,

[0, 100]

. Intersecting

(-\infty, 20) \cup (80, \infty)

with

[0, 100]

gives the intervals

[0, 20)

and

(80, 100]

Question 14

A company's marginal revenue is given by $R'(x) = 150 - 4x$ and its marginal cost is $C'(x) = 2x + 30$ , where $x$ is the number of units produced. For what range of production levels will an increase in production result in an increase in profit?

For $x > 20$
For $x < 37.5$
For $x < 20$ (correct answer)
For $x > 37.5$

Explanation: Profit

P(x)

increases when the marginal profit

P'(x)

is positive. The marginal profit is the difference between marginal revenue and marginal cost:

P'(x) = R'(x) - C'(x)

. Substituting the given functions:

P'(x) = (150 - 4x) - (2x + 30) = 150 - 4x - 2x - 30 = 120 - 6x

. To find where profit is increasing, we set

P'(x) > 0

120 - 6x > 0

. Solving for

x

120 > 6x

, which simplifies to

20 > x

. Therefore, profit increases for production levels less than 20 units.

Question 15

The profit, $P(t)$ , in millions of dollars, from a new technology is modeled by $P(t) = 20 - (t-9)^{2/3}$ for its first 17 years of use, i.e., on the interval $0 \le t \le 17$ . Which statement accurately describes a key feature of this profit model?

Profit has a local minimum at $t=9$ years due to a sharp change in market conditions.
Profit is always decreasing over the 17-year period as the technology becomes obsolete.
Profit has no critical points in the interval $(0, 17)$ and increases steadily.
Profit reaches a local maximum at $t=9$ years, where its rate of change is undefined. (correct answer)

Explanation: To analyze the function

P(t) = 20 - (t-9)^{2/3}

, we find its derivative:

P'(t) = -\frac{2}{3}(t-9)^{-1/3} = \frac{-2}{3\sqrt[3]{t-9}}

. The derivative is never zero. However,

P'(t)

is undefined at

t=9

, which makes

t=9

a critical point. We test the sign of

P'(t)

around this point. For

t<9

t-9

is negative, so

\sqrt[3]{t-9}

is negative, making

P'(t)

positive. Thus, profit is increasing on

(0, 9)

. For

t>9

t-9

is positive, making

P'(t)

negative. Thus, profit is decreasing on

(9, 17)

. Since the function changes from increasing to decreasing at

t=9

, it has a local maximum at that point.

Question 16

The daily profit for a company, in hundreds of dollars, is modeled by $P(x) = \frac{1}{4}x^4 - 2x^3 + 4x^2 - 5$ , where $x$ represents thousands of units produced. The company's operations are stable only when production is at a level corresponding to a local maximum of profit. What is this production level?

$x = 0$
$x = 1$
$x = 2$ (correct answer)
$x = 4$

Explanation: To find local extrema, we first compute the derivative of the profit function:

P'(x) = x^3 - 6x^2 + 8x

. We find the critical points by setting

P'(x)=0

x(x^2 - 6x + 8) = 0

, which factors as

x(x-2)(x-4) = 0

. The critical points are

x=0

x=2

, and

x=4

. We use the first derivative test to classify them. We check the sign of

P'(x)

in the intervals defined by these points:

(-\infty, 0)

(0, 2)

(2, 4)

, and

(4, \infty)

. Let's use test points:

P'(-1)=-15

(decreasing),

P'(1)=3

(increasing),

P'(3)=-3

(decreasing),

P'(5)=15

(increasing). At

x=0

P'

changes from - to +, so it's a local minimum. At

x=2

P'

changes from + to -, so it's a local maximum. At

x=4

P'

changes from - to +, so it's a local minimum. The only local maximum occurs at

x=2

, which corresponds to 2,000 units.

Question 17

After a successful advertising campaign, the monthly profit $P$ (in thousands of dollars) for a product is modeled by $P(t) = \frac{120t}{t^2 + 36}$ , where $t$ is the number of months after the campaign ended ( $t \ge 0$ ). The positive effects of the campaign eventually wear off and profit begins to decline. After how many months does this decline begin?

3 months
6 months (correct answer)
12 months
36 months

Explanation: The profit begins to decline when the profit function

P(t)

reaches its maximum and starts decreasing. This occurs at a critical point. We find the derivative of

P(t)

using the quotient rule:

P'(t) = \frac{120(t^2 + 36) - 120t(2t)}{(t^2 + 36)^2}

. To find critical points, we set the numerator to zero:

120(t^2 + 36) - 240t^2 = 0

. This simplifies to

120t^2 + 120 \cdot 36 - 240t^2 = 0

, which gives

120 \cdot 36 - 120t^2 = 0

. Dividing by 120, we get

36 - t^2 = 0

, so

t^2 = 36

. Since

t \ge 0

, we have

t=6

. For

0 \le t < 6

P'(t)

is positive (profit increasing), and for

t > 6

P'(t)

is negative (profit decreasing). Therefore, the decline in profit begins after 6 months.

Question 18

The value of a rare collectible, in dollars, is modeled by $V(t) = 50t e^{-0.1t}$ where $t$ is the number of years from today. To maximize the value, for how many years should the owner hold onto the collectible before selling it?

$5$ years
$10$ years (correct answer)
$e$ years
$1$ year

Explanation: To find the maximum value, we need to find the critical points of the value function

V(t)

by taking its derivative and setting it to zero. Using the product rule,

(uv)' = u'v + uv'

, let

u = 50t

and

v = e^{-0.1t}

. Then

u' = 50

and

v' = -0.1e^{-0.1t}

. So,

V'(t) = 50e^{-0.1t} + 50t(-0.1e^{-0.1t}) = 50e^{-0.1t}(1 - 0.1t)

. Set

V'(t) = 0

. Since

50e^{-0.1t}

is always positive, we only need to solve

1 - 0.1t = 0

. This gives

0.1t = 1

, so

t = 10

. The first derivative test shows that for

t < 10

V'(t) > 0

(increasing) and for

t > 10

V'(t) < 0

(decreasing), confirming a local maximum at

t=10

years.

Question 19

A manufacturer's total weekly cost for an inventory item is given by the function $C(q) = \frac{7200}{q} + 20q + 50000$ , where $q$ is the order size. The company currently orders 15 units each time. To minimize costs, what action should the company take regarding its order size?

Increase the order size, because the marginal cost $C'(15)$ is positive.
Decrease the order size, because the marginal cost $C'(15)$ is positive.
Keep the order size the same, as 15 units is the optimal quantity.
Increase the order size, because the marginal cost $C'(15)$ is negative. (correct answer)

Explanation: When you encounter a cost minimization problem, the key insight is that marginal cost tells you the direction to move. If the marginal cost is negative at your current quantity, increasing that quantity will decrease total costs. Let's find the marginal cost by taking the derivative:

C'(q) = -\frac{7200}{q^2} + 20

. At the current order size of 15 units:

C'(15) = -\frac{7200}{15^2} + 20 = -\frac{7200}{225} + 20 = -32 + 20 = -12

Since

C'(15) = -12 < 0

, the marginal cost is negative, meaning each additional unit ordered will reduce total weekly costs by approximately $12. Choice D correctly identifies that the company should increase order size because the marginal cost is negative. Choice A reaches the right action (increase) but gives the wrong reason—the marginal cost is negative, not positive. Choice B makes two errors: it suggests decreasing when you should increase, and incorrectly states the marginal cost is positive. Choice C is wrong because 15 units is not optimal; the optimal quantity occurs where $

C'(q) = 0

, which gives

q = \sqrt{\frac{7200}{20}} = 12\sqrt{5} \approx 26.8

$ units. Remember this pattern: negative marginal cost means "go higher" to reduce total cost, while positive marginal cost means "go lower." The minimum occurs exactly where marginal cost equals zero. This relationship between marginal analysis and optimization decisions appears frequently in business calculus.

Question 20

The profit, in thousands of dollars, from producing $x$ hundred units of a product is given by the function $P(x) = -x^3 + 15x^2 - 48x + 30$ . The marketing department recommends that production should be in a range where profit is increasing. Which of the following production ranges meets this criterion?

Between 200 and 800 units (correct answer)
Only for production levels above 800 units
Between 0 and 200 units, and above 800 units
Only for production levels below 200 units

Explanation: To find where the profit function

P(x)

is increasing, we need to find where its derivative,

P'(x)

, is positive. First, find the derivative:

P'(x) = -3x^2 + 30x - 48

. To find the critical points, set

P'(x) = 0

-3(x^2 - 10x + 16) = 0

, which simplifies to

-3(x-2)(x-8) = 0

. The critical points are

x=2

and

x=8

. Since

P'(x)

is a downward-opening parabola, it will be positive between its roots. Testing the intervals: for

x<2

P'(x)<0

; for

2<x<8

P'(x)>0

; for

x>8

P'(x)<0

. Therefore, profit is increasing on the interval

(2, 8)

. Since

x

is in hundreds of units, this corresponds to the range between 200 and 800 units.