Business Calculus Quiz: Accumulated Cost Revenue

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Accumulated Cost RevenueQuestion 1 of 13

A factory's marginal cost to produce $x$ widgets is $C'(x) = 0.003x^2 - 0.8x + 120$ dollars per widget. If the factory is currently producing 200 widgets per day, what would be the additional cost to increase daily production to 250 widgets?

$4,000

$4,625

$4,688

$20,625

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Business Calculus Quiz

Business Calculus Quiz: Accumulated Cost Revenue

Practice Accumulated Cost Revenue 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 Accumulated Cost Revenue, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

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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.

All questions

Question 1

$4,000
$4,625 (correct answer)
$4,688
$20,625

Explanation: The additional cost is the definite integral of the marginal cost function from the current production level (200) to the new level (250).

\int_{200}^{250} (0.003x^2 - 0.8x + 120) dx

The antiderivative is

F(x) = 0.001x^3 - 0.4x^2 + 120x

. Evaluate at the limits of integration:

F(250) = 0.001(250)^3 - 0.4(250)^2 + 120(250) = 15625 - 25000 + 30000 = 20625

F(200) = 0.001(200)^3 - 0.4(200)^2 + 120(200) = 8000 - 16000 + 24000 = 16000

. The additional cost is

F(250) - F(200) = 20625 - 16000 = \

4,625.

Question 2

The marginal revenue for a company's product is $R'(x) = 500 - 0.4x - 0.03x^2$ dollars per unit for $x$ units sold per week. What is the increase in total weekly revenue when sales are increased from 50 to 60 units per week?

$3,865
$3,870 (correct answer)
$4,050
$23,250

Explanation: The increase in revenue from selling 50 to 60 units is the definite integral of the marginal revenue function from

x=50

x=60

\int_{50}^{60} (500 - 0.4x - 0.03x^2) dx

The antiderivative is

F(x) = 500x - 0.2x^2 - 0.01x^3

. Evaluate at the limits of integration:

F(60) = 500(60) - 0.2(60)^2 - 0.01(60)^3 = 30000 - 720 - 2160 = 27120

F(50) = 500(50) - 0.2(50)^2 - 0.01(50)^3 = 25000 - 500 - 1250 = 23250

. The increase in revenue is

F(60) - F(50) = 27120 - 23250 = \

3,870.

Question 3

The marginal revenue function for a software product is $R'(x) = \frac{80}{x+2}$ , where $x$ is the number of licenses sold. Assuming revenue is zero when no licenses are sold, what is the total revenue from selling the first 18 licenses?

$4
$72
$80 \ln(10)$ (correct answer)
$80 \ln(20)$

Explanation: Total revenue from the first 18 units is found by integrating the marginal revenue function from

x=0

x=18

. Since

R(0)=0

, this is

R(18) = \int_{0}^{18} R'(x) dx

\int_{0}^{18} \frac{80}{x+2} dx = [80 \ln(x+2)]_{0}^{18}

Evaluate at the limits:

= 80 \ln(18+2) - 80 \ln(0+2)

= 80 \ln(20) - 80 \ln(2)

Using the logarithm property

\ln(a) - \ln(b) = \ln(a/b)

= 80 \ln(\frac{20}{2}) = 80 \ln(10)

Question 4

A company's marginal cost is $C'(x) = 40$ dollars per unit for the first 500 units produced and $C'(x) = 60$ dollars per unit for any additional units. Fixed costs are $5,000. What is the total cost of producing 700 units?

$32,000
$33,000
$37,000 (correct answer)
$47,000

Explanation: The total cost is the sum of the fixed costs and the total variable costs. The variable costs must be calculated in two parts due to the piecewise marginal cost function. Total Cost = Fixed Cost + (Cost of first 500 units) + (Cost of units from 501 to 700). Total Cost =

5,000 + \int_0^{500} 40 dx + \int_{500}^{700} 60 dx

Calculate the integrals:

\int_0^{500} 40 dx = [40x]_0^{500} = 40(500) - 0 = 20,000

\int_{500}^{700} 60 dx = [60x]_{500}^{700} = 60(700) - 60(500) = 42,000 - 30,000 = 12,000

. Total Cost = $5,000 + 20,000 + 12,000 = $37,000.

Question 5

A company's marginal profit from selling $x$ items is $P'(x) = \frac{100}{\sqrt{x+1}}$ . The company's total profit from selling the first 15 items was $500. What is the total profit from selling the first 99 items?

$1,400
$1,700 (correct answer)
$1,800
$2,300

Explanation: This is a multi-step problem. First, find the general profit function

P(x)

by integrating

P'(x)

. Then use the given information

P(15)=500

to find the constant of integration. Finally, use the complete profit function to find

P(99)

Integrate $P'(x) = 100(x+1)^{-1/2}$ : $P(x) = \int 100(x+1)^{-1/2} dx = 100 \frac{(x+1)^{1/2}}{1/2} + K = 200\sqrt{x+1} + K$ .
Use $P(15) = 500$ to find $K$ : $500 = 200\sqrt{15+1} + K$ $500 = 200\sqrt{16} + K$ $500 = 200(4) + K$ $500 = 800 + K \implies K = -300$ . So, the total profit function is $P(x) = 200\sqrt{x+1} - 300$ .
Calculate $P(99)$ : $P(99) = 200\sqrt{99+1} - 300$ $P(99) = 200\sqrt{100} - 300$ $P(99) = 200(10) - 300 = 2000 - 300 = 1700$ .

The total profit is $1,700.

Question 6

An e-commerce company's marginal revenue from online sales is $R'(x) = \frac{500}{\sqrt{x + 25}}$ dollars per unit, where $x$ is the number of units sold above their baseline of 25 units. What is the total additional revenue generated by increasing sales from 75 units above baseline to 175 units above baseline?

$4,000
$5,000 (correct answer)
$6,000
$7,000

Explanation: Additional revenue =

\int_{75}^{175} \frac{500}{\sqrt{x + 25}} dx = 500 \int_{75}^{175} (x + 25)^{-1/2} dx = 500[2\sqrt{x + 25}]_{75}^{175} = 1000[\sqrt{200} - \sqrt{100}] = 1000[\sqrt{200} - 10] = 1000[14.14 - 10] = 1000(4.14) ≈ 5000

. Choice A uses incorrect antiderivative. Choice C forgets the factor of 2. Choice D adds instead of subtracting boundary values.

Question 7

The marginal cost of manufacturing a specialized electronic component is given by $C'(x) = 50 + \frac{1000}{(x+10)^2}$ dollars per unit. Find the total cost of producing the 11th through the 40th components.

$1,470
$1,478
$1,530 (correct answer)
$1,980

Explanation: The cost of producing the 11th through the 40th components corresponds to the accumulated cost as

x

increases from 10 to 40. This is found by the definite integral

\int_{10}^{40} C'(x) dx

. First, find the antiderivative of

C'(x) = 50 + 1000(x+10)^{-2}

F(x) = \int (50 + 1000(x+10)^{-2}) dx = 50x + \frac{1000(x+10)^{-1}}{-1} = 50x - \frac{1000}{x+10}

. Next, evaluate the definite integral:

[50x - \frac{1000}{x+10}]_{10}^{40} = (50(40) - \frac{1000}{40+10}) - (50(10) - \frac{1000}{10+10})

= (2000 - \frac{1000}{50}) - (500 - \frac{1000}{20})

= (2000 - 20) - (500 - 50)

= 1980 - 450 = 1530

. The total cost is $1,530.

Question 8

A company's marginal revenue is $R'(x) = 150 - 2x$ and its marginal cost is $C'(x) = x + 30$ dollars per unit at a production level of $x$ units. The company's fixed costs are $1,200. What is the change in total profit if production is increased from 30 to 40 units?

An increase of $150 (correct answer)
An increase of $1,200
A decrease of $1,050
A decrease of $30

Explanation: The change in profit is the integral of the marginal profit function,

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

. The fixed costs do not affect the change in profit over an interval. First, find the marginal profit function:

P'(x) = (150 - 2x) - (x + 30) = 120 - 3x

. Next, integrate

P'(x)

from 30 to 40:

\int_{30}^{40} (120 - 3x) dx = [120x - 1.5x^2]_{30}^{40}

Evaluate at the limits of integration: At

x=40: 120(40) - 1.5(40)^2 = 4800 - 1.5(1600) = 4800 - 2400 = 2400

. At

x=30: 120(30) - 1.5(30)^2 = 3600 - 1.5(900) = 3600 - 1350 = 2250

. The change in profit is $2400 - 2250 = $150.

Question 9

The marginal cost function for a particular product is given by $C'(x) = 0.06x^2 - 0.8x + 15$ , where $x$ is the number of units produced. Fixed costs are $500. What is the exact cost of producing the 21st unit?

$23.00
$23.82 (correct answer)
$24.66
$823.82

Explanation: The cost of producing the 21st unit is the total cost of producing 21 units minus the total cost of producing 20 units, which is given by the definite integral of the marginal cost function from 20 to 21. The fixed costs are irrelevant as they cancel out in the subtraction

C(21) - C(20)

. First, find the antiderivative of

C'(x)

F(x) = \int (0.06x^2 - 0.8x + 15) dx = 0.02x^3 - 0.4x^2 + 15x

. Next, evaluate the definite integral:

\int_{20}^{21} C'(x) dx = F(21) - F(20)

F(21) = 0.02(21)^3 - 0.4(21)^2 + 15(21) = 0.02(9261) - 0.4(441) + 315 = 185.22 - 176.4 + 315 = 323.82

F(20) = 0.02(20)^3 - 0.4(20)^2 + 15(20) = 0.02(8000) - 0.4(400) + 300 = 160 - 160 + 300 = 300

. The cost is $323.82 - 300 = $23.82.

Question 10

The revenue from a new online service is generated at a rate of $R'(t) = 200e^{0.05t}$ dollars per day, where $t$ is the number of days since the service launched. Find the total accumulated revenue during the first 30 days of operation.

$200(e^{1.5} - 1)$
$6000e^{1.5}$
$4000e^{1.5}$
$4000(e^{1.5} - 1)$ (correct answer)

Explanation: When you encounter a rate function like

R'(t) = 200e^{0.05t}

, you're looking at accumulated change over time. Since this represents revenue generated per day, finding total accumulated revenue requires integration - you need to "add up" all the daily revenue over the 30-day period. To find total revenue, integrate the rate function from day 0 to day 30:

\int_0^{30} 200e^{0.05t} \, dt

. The antiderivative of

200e^{0.05t}

\frac{200}{0.05}e^{0.05t} = 4000e^{0.05t}

. Applying the Fundamental Theorem of Calculus:

4000e^{0.05(30)} - 4000e^{0.05(0)} = 4000e^{1.5} - 4000(1) = 4000(e^{1.5} - 1)

. Choice A gives

200(e^{1.5} - 1)

, which incorrectly uses the original coefficient 200 instead of dividing by the exponent's coefficient (0.05). This is a common integration error with exponential functions. Choice B gives

6000e^{1.5}

, which fails to subtract the initial value and uses an incorrect coefficient, suggesting confusion about both integration and the definite integral evaluation. Choice C gives

4000e^{1.5}

, which has the correct coefficient from integration but forgets to subtract

4000e^0 = 4000

when evaluating the definite integral. Remember: when integrating

ae^{bt}

, divide by the coefficient in the exponent to get

\frac{a}{b}e^{bt}

, and always subtract both bounds when evaluating definite integrals, even when the lower bound seems "simple."

Question 11

A company's marginal revenue function is $R'(q) = 80 - 0.4q$ dollars per unit. If current weekly sales are 50 units and the company projects sales will increase to 120 units next week, what is the additional revenue generated by this sales increase, assuming the marginal revenue function remains constant?

$3,620 (correct answer)
$4,600
$5,580
$6,560

Explanation: Additional revenue =

\int_{50}^{120} (80 - 0.4q) dq = [80q - 0.2q^2]_{50}^{120} = (9600 - 2880) - (4000 - 500) = 6720 - 3500 = 3620

. Choice B uses incorrect coefficient (0.4 instead of 0.2 in antiderivative). Choice C evaluates at wrong limits. Choice D calculates total revenue instead of additional revenue.

Question 12

The marginal cost for producing $x$ units of a smart speaker is $C'(x) = 40 - 0.1x$ . If the total cost to produce the first 100 speakers is $5,500, what are the company's fixed costs?

$1,000
$2,000 (correct answer)
$3,500
$5,500

Explanation: The total cost function

C(x)

is the integral of the marginal cost function

C'(x)

. The constant of integration,

K

, represents the fixed costs.

C(x) = \int (40 - 0.1x) dx = 40x - 0.05x^2 + K

. We are given that

C(100) = 5500

. We can use this to solve for

K

C(100) = 40(100) - 0.05(100)^2 + K = 5500

4000 - 0.05(10000) + K = 5500

4000 - 500 + K = 5500

3500 + K = 5500

K = 5500 - 3500 = 2000

. The fixed costs are $2,000.

Question 13

The function $C'(t)$ gives the marginal cost, in dollars per unit, of producing the $t$ -th unit of a product. What is the economic interpretation of the quantity $1000 + \int_0^{50} C'(t) dt$ ?

The total cost to produce 50 units, assuming fixed costs are $1,000. (correct answer)
The total revenue from selling 50 units, assuming an initial revenue of $1,000.
The total cost to produce 50 units, assuming the cost of the first unit is $1,000.
The marginal cost of the 50th unit, plus fixed costs of $1,000.

Explanation: The definite integral

\int_0^{50} C'(t) dt

represents the total variable cost of producing the first 50 units, which is the change in cost from producing 0 units to 50 units (

C(50) - C(0)

). The total cost function is

C(x) = \int C'(x) dx + K

, where

K

is the fixed cost,

C(0)

. Therefore, the total cost of producing 50 units is

C(50) = \int_0^{50} C'(t) dt + C(0)

. The expression

1000 + \int_0^{50} C'(t) dt

represents the total cost to produce 50 units, with the fixed cost

C(0)

being $1,000.