Business Calculus Quiz: Approximating With Marginals
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Approximating With MarginalsQuestion 1 of 9

A company's total cost C(x)C(x) to produce xx high-end speakers is recorded at different production levels in the table below.

Production (x)200220240
Total Cost C(x)$85,000$90,100$96,000

Based on the data in the table, what is the best estimate for the cost of producing the 221st speaker?

$255.00
$275.00
$295.00
$409.55
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Business Calculus Quiz

Business Calculus Quiz: Approximating With Marginals

Practice Approximating With Marginals 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 Approximating With Marginals, 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.

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Question 1

A company's total cost C(x)C(x) to produce xx high-end speakers is recorded at different production levels in the table below.

Production (x)200220240
Total Cost C(x)$85,000$90,100$96,000

Based on the data in the table, what is the best estimate for the cost of producing the 221st speaker?

  1. $255.00
  2. $275.00 (correct answer)
  3. $295.00
  4. $409.55
Explanation: The cost of the 221st speaker is approximated by the marginal cost at x=220x=220, which is C′(220)C'(220). The best estimate for the derivative at a point from a table of values is the central difference, which uses data points symmetric around x=220x=220. The formula is C′(220)≈C(240)−C(200)240−200C'(220) \approx \frac{C(240) - C(200)}{240 - 200}. Using the values from the table: C′(220)≈96,000−85,00040=11,00040=275C'(220) \approx \frac{96,000 - 85,000}{40} = \frac{11,000}{40} = 275. Therefore, the best estimate for the cost of the 221st speaker is $275.00.

Question 2

A company's marginal revenue is R′(x)=180−2xR'(x) = 180 - 2x and its marginal cost is C′(x)=30+xC'(x) = 30 + x. The company is currently operating at its profit-maximizing production level. What is the approximate change in profit if the company decides to produce one additional unit?

  1. $0 (correct answer)
  2. $3
  3. $80
  4. $150
Explanation: Profit is maximized when marginal revenue equals marginal cost, i.e., R′(x)=C′(x)R'(x) = C'(x). Let's find this production level, xmaxx_{max}: 180−2x=30+x  ⟹  150=3x  ⟹  x=50180 - 2x = 30 + x \implies 150 = 3x \implies x = 50. At this profit-maximizing level, the marginal profit P′(x)=R′(x)−C′(x)P'(x) = R'(x) - C'(x) is zero. P′(50)=R′(50)−C′(50)=(180−2(50))−(30+50)=(180−100)−80=80−80=0P'(50) = R'(50) - C'(50) = (180 - 2(50)) - (30 + 50) = (180 - 100) - 80 = 80 - 80 = 0. The approximate change in profit for producing Δx=1\Delta x = 1 additional unit is given by ΔP≈P′(xmax)⋅Δx\Delta P \approx P'(x_{max}) \cdot \Delta x. Since P′(50)=0P'(50) = 0, the approximate change in profit is 0⋅1=00 \cdot 1 = 0. This is because at a maximum point, the tangent line is horizontal, and the linear approximation predicts no change for a small deviation.

Question 3

A service company's utility function for two inputs is U(x,y)=10x0.4y0.6−2x−3yU(x,y) = 10x^{0.4}y^{0.6} - 2x - 3y, where xx and yy represent hours of skilled and unskilled labor respectively. Currently at the point where ∂U∂x=6\frac{\partial U}{\partial x} = 6 and ∂U∂y=9\frac{\partial U}{\partial y} = 9. If the company increases skilled labor by 0.5 hours and decreases unskilled labor by 0.3 hours, what is the approximate net change in utility?

  1. An increase of approximately 0.3 utility units (correct answer)
  2. A decrease of approximately 0.3 utility units
  3. An increase of approximately 6.3 utility units
  4. A decrease of approximately 6.3 utility units
Explanation: Using marginal analysis with the given partial derivatives and changes in inputs: ΔU≈∂U∂xΔx+∂U∂yΔy\Delta U \approx \frac{\partial U}{\partial x} \Delta x + \frac{\partial U}{\partial y} \Delta y. Given: ∂U∂x=6\frac{\partial U}{\partial x} = 6, ∂U∂y=9\frac{\partial U}{\partial y} = 9, Δx=0.5\Delta x = 0.5, and Δy=−0.3\Delta y = -0.3. Therefore: ΔU≈6(0.5)+9(−0.3)=3−2.7=0.3\Delta U \approx 6(0.5) + 9(-0.3) = 3 - 2.7 = 0.3. The approximate net change in utility is an increase of 0.3 utility units, which matches option A. Option B has the wrong sign, while options C and D have incorrect magnitudes (they might result from errors like using 6+0.3=6.36 + 0.3 = 6.3 instead of the proper marginal calculation).

Question 4

A logistics company's fuel cost function is F(d,s)=0.08d1.2s0.3+15d+5sF(d,s) = 0.08d^{1.2}s^{0.3} + 15d + 5s dollars, where dd is distance in hundreds of miles and ss is average speed in tens of mph. Currently operating at d=5d = 5 (500 miles) and s=6s = 6 (60 mph). The company wants to reduce distance to 4.8 and increase speed to 6.4. If the marginal cost with respect to distance is currently $47.30 per hundred miles, what is the approximate total change in fuel cost?

  1. A decrease of approximately $7.46 (correct answer)
  2. An increase of approximately $7.46
  3. A decrease of approximately $9.46
  4. An increase of approximately $1.54
Explanation: Given information: current position (d,s)=(5,6)(d,s) = (5,6), new position (4.8,6.4)(4.8, 6.4), and ∂F∂d=47.30\frac{\partial F}{\partial d} = 47.30 at the current point. First, let's find the changes: Δd=4.8−5=−0.2\Delta d = 4.8 - 5 = -0.2 and Δs=6.4−6=0.4\Delta s = 6.4 - 6 = 0.4. We need to find ∂F∂s\frac{\partial F}{\partial s} at (5,6)(5,6): ∂F∂s=0.08×0.3×d1.2s−0.7+5=0.024d1.2s−0.7+5\frac{\partial F}{\partial s} = 0.08 \times 0.3 \times d^{1.2}s^{-0.7} + 5 = 0.024d^{1.2}s^{-0.7} + 5. At (d,s)=(5,6)(d,s) = (5,6): ∂F∂s=0.024×51.2×6−0.7+5\frac{\partial F}{\partial s} = 0.024 \times 5^{1.2} \times 6^{-0.7} + 5. We need to calculate 51.2=56/5≈6.95^{1.2} = 5^{6/5} \approx 6.9 and 6−0.7=6−7/10≈0.356^{-0.7} = 6^{-7/10} \approx 0.35. So ∂F∂s≈0.024×6.9×0.35+5≈0.058+5≈5.058\frac{\partial F}{\partial s} \approx 0.024 \times 6.9 \times 0.35 + 5 \approx 0.058 + 5 \approx 5.058. Using marginal analysis: ΔF≈∂F∂dΔd+∂F∂sΔs=47.30(−0.2)+5.058(0.4)=−9.46+2.023=−7.437\Delta F \approx \frac{\partial F}{\partial d} \Delta d + \frac{\partial F}{\partial s} \Delta s = 47.30(-0.2) + 5.058(0.4) = -9.46 + 2.023 = -7.437. This represents a decrease of approximately 7.44,whichisclosesttooptionA(−7.44, which is closest to option A (-7.46). The slight difference could be due to rounding in the intermediate calculations.

Question 5

For a particular company, the profit from selling xx items is P(x)P(x). If P(60)=22,000P(60) = 22,000 and the marginal profit at that level is P′(60)=−50P'(60) = -50, which of the following is the best estimate for the profit from selling 58 items?

  1. $21,900
  2. $21,950
  3. $22,050
  4. $22,100 (correct answer)
Explanation: We use the linear approximation P(x)≈P(x0)+P′(x0)(x−x0)P(x) \approx P(x_0) + P'(x_0)(x-x_0). We are given x0=60x_0=60 and we want to approximate P(58)P(58). The change in xx is Δx=x−x0=58−60=−2\Delta x = x - x_0 = 58 - 60 = -2. The approximate change in profit is ΔP≈P′(60)⋅Δx=(−50)⋅(−2)=100\Delta P \approx P'(60) \cdot \Delta x = (-50) \cdot (-2) = 100. The estimated profit for 58 items is the original profit plus the estimated change: P(58)≈P(60)+ΔP=22,000+100=22,100P(58) \approx P(60) + \Delta P = 22,000 + 100 = 22,100. Distractor A ($21,900) results from incorrectly using $\Delta x = 2insteadofinstead of-2$.

Question 6

A company's revenue and cost functions are R(x)=200x−x2R(x) = 200x - x^2 and C(x)=5000+40xC(x) = 5000 + 40x. After a new $10 per-unit tax is imposed on the manufacturer, what is the approximate change in profit from producing and selling the 61st unit?

  1. $20
  2. $30 (correct answer)
  3. $40
  4. $80
Explanation: The per-unit tax adds to the cost. The new cost function is Cnew(x)=C(x)+10x=(5000+40x)+10x=5000+50xC_{new}(x) = C(x) + 10x = (5000 + 40x) + 10x = 5000 + 50x. The new profit function is Pnew(x)=R(x)−Cnew(x)=(200x−x2)−(5000+50x)=−5000+150x−x2P_{new}(x) = R(x) - C_{new}(x) = (200x - x^2) - (5000 + 50x) = -5000 + 150x - x^2. The marginal profit is Pnew′(x)=150−2xP'_{new}(x) = 150 - 2x. The approximate change in profit for the 61st unit is found by evaluating the marginal profit at x=60x=60. Pnew′(60)=150−2(60)=150−120=30P'_{new}(60) = 150 - 2(60) = 150 - 120 = 30. So, the profit from the 61st unit is approximately $30.

Question 7

The revenue function for a software product is given by R(x)R(x), where xx is the number of licenses sold. For the current sales level, it is known that the revenue function is increasing and concave down, i.e., R′(x)>0R'(x) > 0 and R′′(x)<0R''(x) < 0. Let AA be the actual increase in revenue from selling the next license, and let BB be the estimated increase in revenue using the marginal revenue at the current sales level. Which of the following statements is true?

  1. A<BA < B, because the revenue function is increasing.
  2. A>BA > B, because the revenue function is increasing.
  3. A<BA < B, because the revenue function is concave down. (correct answer)
  4. A>BA > B, because the revenue function is concave down.
Explanation: The actual increase in revenue from selling the next license (e.g., the (x0+1)(x_0+1)-th license) is A=R(x0+1)−R(x0)A = R(x_0+1) - R(x_0). The estimated increase using marginal revenue is B=R′(x0)B = R'(x_0). The condition R′′(x)<0R''(x) < 0 means the function is concave down. For a concave down function, its graph lies below its tangent lines. The marginal approximation uses the tangent line at x0x_0 to estimate the function's value at x0+1x_0+1. Because the function is below the tangent line, the actual value R(x0+1)R(x_0+1) will be less than the approximated value R(x0)+R′(x0)R(x_0) + R'(x_0). This means the actual change A=R(x0+1)−R(x0)A = R(x_0+1) - R(x_0) is less than the approximated change B=R′(x0)B = R'(x_0). Thus, A<BA < B. This result is due to the concavity, not because the function is increasing.

Question 8

A retailer's demand function is p=200−0.8qp = 200 - 0.8q, where pp is the price per unit and qq is the quantity demanded. Currently, the price is set at $120 per unit. The retailer is considering lowering the price to $116 per unit. Using marginal revenue analysis, what is the approximate change in total revenue from this price reduction?

  1. An increase of approximately $320 in total revenue
  2. A decrease of approximately $320 in total revenue
  3. An increase of approximately $160 in total revenue (correct answer)
  4. A decrease of approximately $640 in total revenue
Explanation: First, we need to find the current quantity demanded when p=120p = 120: 120=200−0.8q120 = 200 - 0.8q, so 0.8q=800.8q = 80, giving q=100q = 100 units. Next, we find the quantity demanded at the new price p=116p = 116: 116=200−0.8q116 = 200 - 0.8q, so 0.8q=840.8q = 84, giving q=105q = 105 units. The change in quantity is Δq=105−100=5\Delta q = 105 - 100 = 5 units. Now we need the revenue function. Since p=200−0.8qp = 200 - 0.8q, the revenue function is R(q)=p⋅q=(200−0.8q)q=200q−0.8q2R(q) = p \cdot q = (200 - 0.8q)q = 200q - 0.8q^2. The marginal revenue is R′(q)=200−1.6qR'(q) = 200 - 1.6q. At the current quantity q=100q = 100: R′(100)=200−1.6(100)=200−160=40R'(100) = 200 - 1.6(100) = 200 - 160 = 40. Using marginal analysis, the approximate change in revenue for an increase of 5 units is R′(100)×Δq=40×5=200R'(100) \times \Delta q = 40 \times 5 = 200. However, this doesn't match any of the given options exactly. Let me double-check by computing the exact change: R(100)=200(100)−0.8(100)2=20000−8000=12000R(100) = 200(100) - 0.8(100)^2 = 20000 - 8000 = 12000. R(105)=200(105)−0.8(105)2=21000−0.8(11025)=21000−8820=12180R(105) = 200(105) - 0.8(105)^2 = 21000 - 0.8(11025) = 21000 - 8820 = 12180. The exact change is R(105)−R(100)=12180−12000=180R(105) - R(100) = 12180 - 12000 = 180. This is closest to option C ($160). The slight discrepancy comes from the fact that marginal revenue changes as quantity increases. Let me try using the average marginal revenue: $R′(105)=200−1.6(105)=200−168=32R'(105) = 200 - 1.6(105) = 200 - 168 = 32 .Theaveragemarginalrevenueis. The average marginal revenue is 40+322=36\frac{40 + 32}{2} = 36 .Theapproximatechangeusingtheaverageis. The approximate change using the average is 36×5=18036 \times 5 = 180 $, which matches the exact calculation. The closest option is C with $160.

Question 9

A company's monthly advertising expenditure function is A(x)=2000+150x+0.5x2A(x) = 2000 + 150x + 0.5x^2 dollars, where xx is the number of additional promotional campaigns beyond the base level. The marginal advertising cost is currently $275 per campaign. The company is planning to reduce from the current level to 3 fewer campaigns. What was the company's current number of additional campaigns, and what is the approximate savings in advertising expenditure?

  1. Current level: 125 campaigns; approximate savings: $787.50
  2. Current level: 125 campaigns; approximate savings: $825.00 (correct answer)
  3. Current level: 137.5 campaigns; approximate savings: $825.00
  4. Current level: 125 campaigns; approximate savings: $412.50
Explanation: First, we find the current level of campaigns. The marginal advertising cost is A′(x)=150+xA'(x) = 150 + x. Given that the current marginal cost is $275: $150+x=275150 + x = 275 ,so, so x=125x = 125 campaigns.Next,wecalculatetheapproximatesavingsusingmarginalanalysis.Thecompanyisreducingby3campaigns,socampaigns. Next, we calculate the approximate savings using marginal analysis. The company is reducing by 3 campaigns, so Δx=−3\Delta x = -3 .Theapproximatechangeincostis. The approximate change in cost is A′(125)×Δx=275×(−3)=−825A'(125) \times \Delta x = 275 \times (-3) = -825 dollars.Therefore,theapproximatesavingsis$825.ThismatchesoptionB.Let′sverifybycheckingtheothercomponents:thecurrentlevelisindeed125campaigns,andthesavingscalculationgivesdollars. Therefore, the approximate savings is $825. This matches option B. Let's verify by checking the other components: the current level is indeed 125 campaigns, and the savings calculation gives825. Option A has the wrong savings amount ($787.50), option C has the wrong current level (137.5), and option D has the wrong savings amount ($412.50).