IB Mathematics: Applications and Interpretation Quiz: Differentiation Applications

20 questions · exam conditions

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Differentiation ApplicationsQuestion 1 of 20

The total cost, in euros, for a factory to produce $x$ units of a product is given by $C(x) = 0.01x^3 - 4.5x^2 + 800x + 12000$ , for $x > 0$ .

Find the approximate number of units that should be produced to minimize the average cost per unit.

150

200

225

300

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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Differentiation Applications

Practice Differentiation Applications in IB Mathematics: Applications and Interpretation 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 Differentiation Applications, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

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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.

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Question 1

The total cost, in euros, for a factory to produce $x$ units of a product is given by $C(x) = 0.01x^3 - 4.5x^2 + 800x + 12000$ , for $x > 0$ .

Find the approximate number of units that should be produced to minimize the average cost per unit.

150
200
225 (correct answer)
300

Explanation: The average cost per unit is

A(x) = \frac{C(x)}{x} = 0.01x^2 - 4.5x + 800 + \frac{12000}{x}

. To minimize the average cost, we find its derivative and set it to zero:

A'(x) = 0.02x - 4.5 - \frac{12000}{x^2}

. Setting

A'(x) = 0

gives

0.02x - 4.5 = \frac{12000}{x^2}

, which leads to the cubic equation

0.02x^3 - 4.5x^2 - 12000 = 0

. Using a graphing calculator's solver function, we find the positive real root is

x \approx 225.4

. The nearest whole number of units is 225.

Question 2

A rectangular field with an area of 5000 m² is to be enclosed by a fence. A straight river runs along one side of the field, so fencing is only required for the other three sides.

What is the minimum length of fencing required, to the nearest meter?

100 m
141 m
250 m
200 m (correct answer)

Explanation: This is a classic optimization problem where you need to minimize perimeter given a fixed area constraint. When you see problems about minimizing fencing or materials, think calculus optimization or recognize standard geometric relationships. Let's set up the problem systematically. If the field has width

w

(perpendicular to the river) and length

l

(parallel to the river), then the area constraint gives us

l \times w = 5000

, so

l = \frac{5000}{w}

. Since the river provides one long side, you only need fencing for two widths and one length:

F = 2w + l = 2w + \frac{5000}{w}

. To minimize this, take the derivative and set it equal to zero:

\frac{dF}{dw} = 2 - \frac{5000}{w^2} = 0

. Solving:

2 = \frac{5000}{w^2}

, so

w^2 = 2500

and

w = 50

meters. This gives

l = \frac{5000}{50} = 100

meters, and minimum fencing of

F = 2(50) + 100 = 200

meters. Answer A (100 m) represents just the length parallel to the river, forgetting the two perpendicular sides. Answer B (141 m) might come from incorrectly assuming a square field (

\sqrt{5000} \approx 71

, then

2 \times 71 = 142

). Answer C (250 m) could result from using the wrong optimization setup or assuming all four sides need fencing. Strategy tip: In three-sided fencing problems, the optimal rectangle always has the unfenced side twice as long as each fenced side. Memorizing this 2:1 ratio can save time on similar problems.

Question 3

The price $p$ , in dollars, for a concert ticket is related to the quantity of tickets sold, $q$ , by the demand function $p(q) = 400 - 0.2q$ .

What is the maximum possible revenue from ticket sales?

$200
$1,000
$72,000
$200,000 (correct answer)

Explanation: Revenue

R

is price times quantity:

R(q) = p \cdot q = (400 - 0.2q)q = 400q - 0.2q^2

. To maximize revenue, find the derivative:

R'(q) = 400 - 0.4q

. Set

R'(q)=0

to find the critical point:

400 - 0.4q = 0 \implies q = 1000

. The maximum revenue is

R(1000) = 400(1000) - 0.2(1000)^2 = 400,000 - 200,000 = 200,000

Question 4

The cost of fuel per hour for a cargo ship is proportional to the cube of its speed. The cost is $1350 per hour when the speed is 15 knots. Other fixed costs amount to $6400 per hour regardless of speed.

What speed, in knots, will minimize the total cost per nautical mile?

15.0
18.9
20.0 (correct answer)
23.8

Explanation: Let

C_f

be fuel cost and

v

be speed.

C_f = kv^3

. Given

1350 = k(15^3)

, so

k = 1350/3375 = 0.4

. Total cost per hour is

C_h(v) = 0.4v^3 + 6400

. The cost per nautical mile,

C_{nm}

, is the cost per hour divided by the speed (miles per hour):

C_{nm}(v) = \frac{0.4v^3 + 6400}{v} = 0.4v^2 + \frac{6400}{v}

. To minimize this, find the derivative:

C'_{nm}(v) = 0.8v - \frac{6400}{v^2}

. Set to zero:

0.8v = \frac{6400}{v^2} \implies v^3 = \frac{6400}{0.8} = 8000

. Thus,

v = \sqrt[3]{8000} = 20

knots.

Question 5

An open-top box is to be made from a square piece of cardboard measuring 30 cm on each side, by cutting out equal squares from each of the four corners and folding up the sides.

What is the maximum possible volume of the box, in cm³?

5
800
2000 (correct answer)
3125

Explanation: Let

x

be the side length of the squares cut from the corners. The dimensions of the box will be length

(30-2x)

, width

(30-2x)

, and height

x

. The volume is

V(x) = x(30-2x)^2 = 4x^3 - 120x^2 + 900x

. To maximize the volume, find the derivative:

V'(x) = 12x^2 - 240x + 900

. Set

V'(x) = 0

and solve for

x

, which gives

x=5

x=15

. The domain is

0 < x < 15

, so

x=5

is the only valid solution. The maximum volume is

V(5) = 5(30-2(5))^2 = 5(20)^2 = 2000

cm³.

Question 6

The altitude of a drone, in meters, is given by the function $A(t) = -0.1t^3 + 1.5t^2 + 5$ , where $t$ is the time in seconds after launch, for $0 \le t \le 12$ .

For which time interval is the drone's altitude increasing?

$0 < t < 5$
$0 < t < 10$ (correct answer)
$5 < t < 10$
$10 < t < 12$

Explanation: The altitude is increasing when its rate of change, the derivative

A'(t)

, is positive.

A'(t) = -0.3t^2 + 3t

. To find where

A'(t) > 0

, we first find the roots of

A'(t) = 0

-0.3t(t - 10) = 0

, so the roots are

t=0

and

t=10

. Since the graph of

A'(t)

is a downward-opening parabola, it is positive between its roots. Therefore, the altitude is increasing for the interval

0 < t < 10

Question 7

The value of a particular cryptocurrency, $V$ , in US dollars, is modelled by a differentiable function $V(t)$ , where $t$ is the number of weeks since its launch.

Which of the following is the best interpretation of the statement $V'(10) = -250$ ?

After 10 weeks, the value of the cryptocurrency was $250.
The value of the cryptocurrency decreased by $250 during the 10th week.
The total loss in value after 10 weeks was $250.
At the 10-week mark, the value of the cryptocurrency was decreasing at a rate of $250 per week. (correct answer)

Explanation: When you encounter derivative notation like

V'(10) = -250

, you're dealing with instantaneous rates of change. The derivative

V'(t)

tells you how fast the value is changing at any specific moment in time, not cumulative changes over a period. Let's break down what

V'(10) = -250

means: at exactly

t = 10

weeks, the derivative (rate of change) of the cryptocurrency's value is

-250

dollars per week. The negative sign indicates the value is decreasing, and the magnitude tells you the rate of that decrease. This matches option D perfectly. Now let's see why the other options miss the mark. Option A confuses the derivative with the original function—

V'(10)

is not the same as

V(10)

. The derivative gives you the rate of change, not the actual value. Option B misinterprets the derivative as a total change over the entire 10th week, but derivatives measure instantaneous rates, not cumulative changes over time intervals. Option C suggests this represents the total accumulated loss after 10 weeks, which would require integrating the derivative function, not just evaluating it at one point. Remember this key distinction:

V(t)

gives you the value at time

t

, while

V'(t)

gives you how fast that value is changing at time

t

. On IB exams, questions often test whether you can distinguish between a function's value and its rate of change—always check whether you're looking at

f(x)

f'(x)

Question 8

The value of a company's shares, in dollars, is modelled by $V(t) = 0.1t^3 - 1.8t^2 + 7.2t + 20$ , where $t$ is the number of months after the company went public, for $0 \le t \le 10$ .

When is the value of the shares decreasing most rapidly?

t = 2 months
t = 6 months (correct answer)
t = 8 months
t = 10 months

Explanation: The rate of change of the value is

V'(t) = 0.3t^2 - 3.6t + 7.2

. The value is decreasing most rapidly when this rate is at its minimum. To find the minimum of

V'(t)

, we find its derivative,

V''(t)

, and set it to zero.

V''(t) = 0.6t - 3.6

. Setting

V''(t) = 0

gives

0.6t = 3.6

, so

t = 6

. This is the vertex of the parabola

V'(t)

, which is its minimum point. Thus, the value decreases most rapidly at

t=6

months.

Question 9

A company is designing a closed cylindrical can that must hold 500 cm³ of soup. The company wants to use the minimum amount of metal to make the can.

What should be the height of the can, in cm, to minimize the surface area, correct to one decimal place?

8.6 cm (correct answer)
5.4 cm
4.3 cm
10.8 cm

Explanation: This is a classic optimization problem where you need to minimize surface area while maintaining a fixed volume constraint. When you see "minimize material" for containers, you're looking for the point where the derivative equals zero. Start by setting up your equations. For a cylinder with radius

r

and height

h

: Volume =

\pi r^2 h = 500

, so

h = \frac{500}{\pi r^2}

. Surface area =

2\pi r^2 + 2\pi rh

(top/bottom plus sides). Substitute the height expression into the surface area formula:

A = 2\pi r^2 + 2\pi r \cdot \frac{500}{\pi r^2} = 2\pi r^2 + \frac{1000}{r}

. To minimize, take the derivative and set it equal to zero:

\frac{dA}{dr} = 4\pi r - \frac{1000}{r^2} = 0

. Solving:

4\pi r = \frac{1000}{r^2}

, which gives

r^3 = \frac{250}{\pi}

, so

r = \sqrt[3]{\frac{250}{\pi}} ≈ 4.3

cm. Now find the height:

h = \frac{500}{\pi(4.3)^2} ≈ 8.6

cm. Looking at the wrong answers: B (5.4 cm) might come from incorrectly using the radius value as the height. C (4.3 cm) is actually the radius, not the height—a common mix-up. D (10.8 cm) could result from calculation errors in the optimization process. For optimization problems, always double-check which variable the question asks for. The minimum surface area occurs when height equals the diameter, but make sure you're reporting the right measurement.

Question 10

The value of a piece of industrial machinery, in thousands of dollars, $t$ years after purchase is given by the function $V(t) = 150e^{-0.15t} + 10$ .

At what rate is the machinery depreciating 5 years after purchase, in dollars per year?

$10,580 per year (correct answer)
$15,870 per year
$22,500 per year
$33,750 per year

Explanation: The rate of change of value is given by the derivative

V'(t) = 150(-0.15)e^{-0.15t} = -22.5e^{-0.15t}

. At

t=5

, the rate is

V'(5) = -22.5e^{-0.75} \approx -10.58

thousand dollars per year. Since this is negative, the machinery is losing value. The depreciation rate (positive value) is approximately $10,580 per year.

Question 11

An open-top box is to be made from a square piece of cardboard measuring 30 cm on each side, by cutting out equal squares from each of the four corners and folding up the sides.

What is the maximum possible volume of the box, in cm³?

5
800
2000 (correct answer)
3125

Explanation: Let

x

be the side length of the squares cut from the corners. The dimensions of the box will be length

(30-2x)

, width

(30-2x)

, and height

x

. The volume is

V(x) = x(30-2x)^2 = 4x^3 - 120x^2 + 900x

. To maximize the volume, find the derivative:

V'(x) = 12x^2 - 240x + 900

. Set

V'(x) = 0

and solve for

x

, which gives

x=5

x=15

. The domain is

0 < x < 15

, so

x=5

is the only valid solution. The maximum volume is

V(5) = 5(30-2(5))^2 = 5(20)^2 = 2000

cm³.

Question 12

The population of a species of insect in a controlled environment is modelled by the function $P(t) = -t^3 + 45t^2 + 48t + 1000$ , where $t$ is the number of days after the start of the experiment, for $0 \le t \le 35$ .

At what time $t$ is the population's rate of growth at a maximum?

15 days (correct answer)
30.5 days
723 insects per day
15988 insects

Explanation: The rate of growth is the derivative of the population function,

R(t) = P'(t) = -3t^2 + 90t + 48

. To find when this rate is at a maximum, we need to find the derivative of

R(t)

and set it to zero.

R'(t) = P''(t) = -6t + 90

. Setting

R'(t) = 0

gives

-6t + 90 = 0

, which solves to

t=15

. This corresponds to the point of inflection of

P(t)

, where the gradient is steepest.

Question 13

A cylindrical can must have a volume of 1000 cm³. The material for the circular top and bottom costs $0.05 per cm², while the material for the curved side costs $0.03 per cm².

What is the radius of the can, in cm, that minimizes the total cost of the material, correct to two decimal places?

4.30
4.57 (correct answer)
5.42
15.24

Explanation: Let radius be

r

and height be

h

. Volume

V = \pi r^2 h = 1000

, so

h = \frac{1000}{\pi r^2}

. The cost function is

C(r) = 0.05(2\pi r^2) + 0.03(2\pi rh)

. Substituting for

h

, we get

C(r) = 0.1\pi r^2 + 0.06\pi r (\frac{1000}{\pi r^2}) = 0.1\pi r^2 + \frac{60}{r}

. To minimize cost, find the derivative:

C'(r) = 0.2\pi r - \frac{60}{r^2}

. Set

C'(r)=0

, so

0.2\pi r = \frac{60}{r^2}

, which gives

r^3 = \frac{60}{0.2\pi} = \frac{300}{\pi}

. Solving for

r

gives

r = \sqrt[3]{\frac{300}{\pi}} \approx 4.57

cm.

Question 14

The displacement, $s$ meters, of a particle from a fixed point O at time $t$ seconds ( $t \ge 0$ ) is given by $s(t) = t^3 - 9t^2 + 15t + 2$ .

What is the displacement of the particle when it first comes to a momentary rest?

1 m
2 m
9 m (correct answer)
-23 m

Explanation: The particle is at rest when its velocity is zero. Velocity is the derivative of displacement:

v(t) = s'(t) = 3t^2 - 18t + 15

. Set

v(t) = 0

3(t^2 - 6t + 5) = 0 \implies 3(t-1)(t-5) = 0

. The particle is at rest at

t=1

and

t=5

. It first comes to rest at

t=1

second. The displacement at this time is

s(1) = (1)^3 - 9(1)^2 + 15(1) + 2 = 1 - 9 + 15 + 2 = 9

meters.

Question 15

The price $p$ , in dollars, for a concert ticket is related to the quantity of tickets sold, $q$ , by the demand function $p(q) = 400 - 0.2q$ .

What is the maximum possible revenue from ticket sales?

$200
$1,000
$72,000
$200,000 (correct answer)

Explanation: Revenue

R

is price times quantity:

R(q) = p \cdot q = (400 - 0.2q)q = 400q - 0.2q^2

. To maximize revenue, find the derivative:

R'(q) = 400 - 0.4q

. Set

R'(q)=0

to find the critical point:

400 - 0.4q = 0 \implies q = 1000

. The maximum revenue is

R(1000) = 400(1000) - 0.2(1000)^2 = 400,000 - 200,000 = 200,000

Question 16

The total cost, in euros, for a factory to produce $x$ units of a product is given by $C(x) = 0.01x^3 - 4.5x^2 + 800x + 12000$ , for $x > 0$ .

Find the approximate number of units that should be produced to minimize the average cost per unit.

150
200
225 (correct answer)
300

Explanation: The average cost per unit is

A(x) = \frac{C(x)}{x} = 0.01x^2 - 4.5x + 800 + \frac{12000}{x}

. To minimize the average cost, we find its derivative and set it to zero:

A'(x) = 0.02x - 4.5 - \frac{12000}{x^2}

. Setting

A'(x) = 0

gives

0.02x - 4.5 = \frac{12000}{x^2}

, which leads to the cubic equation

0.02x^3 - 4.5x^2 - 12000 = 0

. Using a graphing calculator's solver function, we find the positive real root is

x \approx 225.4

. The nearest whole number of units is 225.

Question 17

The profit, $P$ , in thousands of dollars, from producing $x$ thousand units of a product is given by $P(x) = -x^3 + 27x^2 - 120x - 50$ . The factory has a maximum production capacity of 20 thousand units.

What production level, in thousands of units, maximizes the profit within the factory's capacity?

2.7
15.3 (correct answer)
18.0
20.0

Explanation: To find the maximum profit, we differentiate

P(x)

P'(x) = -3x^2 + 54x - 120

. We set

P'(x)=0

and solve for

x

using a GDC or the quadratic formula. The solutions are

x \approx 2.74

and

x \approx 15.26

. We check the second derivative,

P''(x) = -6x + 54

P''(2.74) > 0

(local minimum).

P''(15.26) < 0

(local maximum). We must also check the endpoints of the domain [0, 20].

P(0) = -50

P(15.26) \approx 485

P(20) = 430

. The local maximum at

x \approx 15.26

is the global maximum on the interval. The closest answer is 15.3.

Question 18

The value of a piece of industrial machinery, in thousands of dollars, $t$ years after purchase is given by the function $V(t) = 150e^{-0.15t} + 10$ .

At what rate is the machinery depreciating 5 years after purchase, in dollars per year?

$10,580 per year (correct answer)
$15,870 per year
$22,500 per year
$33,750 per year

Explanation: The rate of change of value is given by the derivative

V'(t) = 150(-0.15)e^{-0.15t} = -22.5e^{-0.15t}

. At

t=5

, the rate is

V'(5) = -22.5e^{-0.75} \approx -10.58

thousand dollars per year. Since this is negative, the machinery is losing value. The depreciation rate (positive value) is approximately $10,580 per year.

Question 19

The value of a company's shares, in dollars, is modelled by $V(t) = 0.1t^3 - 1.8t^2 + 7.2t + 20$ , where $t$ is the number of months after the company went public, for $0 \le t \le 10$ .

When is the value of the shares decreasing most rapidly?

t = 2 months
t = 6 months (correct answer)
t = 8 months
t = 10 months

Explanation: The rate of change of the value is

V'(t) = 0.3t^2 - 3.6t + 7.2

. The value is decreasing most rapidly when this rate is at its minimum. To find the minimum of

V'(t)

, we find its derivative,

V''(t)

, and set it to zero.

V''(t) = 0.6t - 3.6

. Setting

V''(t) = 0

gives

0.6t = 3.6

, so

t = 6

. This is the vertex of the parabola

V'(t)

, which is its minimum point. Thus, the value decreases most rapidly at

t=6

months.

Question 20

The daily temperature in a desert, in degrees Celsius, can be modelled by the function $T(t) = 0.02t^4 - 0.9t^3 + 12t^2 - 40t + 50$ , where $t$ is the number of hours after midnight ( $0 \le t \le 24$ ).

At what time of day, approximately, is the temperature the lowest?

2:00
4:00 (correct answer)
14:00
20:00

Explanation: To find the minimum temperature, we need to find the derivative of

T(t)

and find its roots.

T'(t) = 0.08t^3 - 2.7t^2 + 24t - 40

. Using a graphing calculator to solve

T'(t)=0

, we find critical points at approximately

t \approx 2.1

t \approx 4.0

, and

t \approx 27.6

. Since we're restricted to

0 \le t \le 24

, we evaluate

T(t)

at the critical points and endpoints:

T(0) = 50

T(2.1) \approx 18.3

T(4.0) \approx 18.0

T(24) \approx 166

. The minimum temperature occurs at

t \approx 4.0

, which corresponds to 4:00.