- The total cost in dollars for a company to produce
x units of a specialized component is C(x) = \frac{1}{3}x^3 - 15x^2 + 250x + 1000. The company's analysis shows there are two distinct production levels at which the marginal cost is exactly $50 per unit. What is the sum of these two production levels? Answer: 30
First, find the marginal cost function C'(x) by taking the derivative of the cost function C(x): C'(x) = x^2 - 30x + 250. Set the marginal cost equal to $50: x^2 - 30x + 250 = 50. Rearrange this into a standard quadratic equation: x^2 - 30x + 200 = 0. Factor the quadratic: (x - 10)(x - 20) = 0. The two production levels are x = 10 and x = 20. The question asks for the sum of these levels, which is 10 + 20 = 30.
(A) This is the production level where marginal cost is minimized (C''(x) = 2x-30 = 0 \implies x=15).
(B) This is the minimum value of the marginal cost, C'(15)=25.
(D) This is the product of the two production levels (10 \times 20 = 200).
- A company's marginal cost to produce
x units is C'(x) = 0.06x^2 - 2x + 150 dollars per unit. The product is sold at a fixed price of $200 per unit. What is the approximate change in profit if the company increases its sales from 100 to 101 units? Answer: A decrease of $350
The approximate change in profit from selling the 101st unit is the marginal profit P'(100). Marginal profit is marginal revenue minus marginal cost, P'(x) = R'(x) - C'(x). Since the price is fixed at $200, the revenue function is R(x) = 200x, and the marginal revenue is R'(x) = 200. We need to find the marginal cost at x=100: C'(100) = 0.06(100)^2 - 2(100) + 150 = 0.06(10000) - 200 + 150 = 600 - 200 + 150 = 550. Now, calculate marginal profit: P'(100) = R'(100) - C'(100) = 200 - 550 = -350. A negative value indicates a decrease in profit. Thus, the profit decreases by approximately $350.
- An online retailer's daily revenue function is R(x)=50x−0.1x2 dollars, where x is the number of items sold. The daily cost function is C(x)=200+10x+0.05x2 dollars. At what production level does marginal profit begin to decrease?
Answer: Immediately, since marginal profit decreases for all values of x>0
Profit function: P(x)=R(x)−C(x)=50x−0.1x2−200−10x−0.05x2=40x−0.15x2−200. Marginal profit: P′(x)=40−0.3x. The second derivative: P′′(x)=−0.3<0 for all x. Since P′′(x) is always negative, marginal profit is always decreasing. Choice A confuses the break-even point with the question asked. Choice B refers to revenue maximization, not profit. Choice D incorrectly states the second derivative is never zero (it's constantly −0.3). - A manufacturing company's total revenue from producing x thousand units is R(x)=80x−2x2 thousand dollars. If the marginal revenue at the current production level is $24,000 per thousand units, and the company increases production by 500 units, what is the approximate change in total revenue?
Answer: An increase of $12,000, calculated as half of the marginal revenue
Marginal revenue represents the change in revenue per thousand units. Since the company is increasing production by 500 units (which is 0.5 thousand units), the approximate change in revenue is 0.5×$24,000=$12,000. Choice B ignores that the increase is only 500 units, not 1000. Choice C incorrectly calculates the proportion. Choice D assumes we're past the maximum without justification (we'd need to know the current production level and verify R′(x)<0). - The profit function for a company is
P(x), where x is the number of units produced. An analyst has determined that P'(500) = 0 and that the marginal profit function, P'(x), is a strictly decreasing function for all x > 0. Which of the following statements must be true? Answer: The company's profit is at a local maximum at a production level of 500 units.
This question tests your understanding of critical points and the relationship between a function and its derivative. When you see problems involving profit functions and their derivatives, think about how the first derivative test reveals where maximum and minimum values occur.
Since P′(500)=0, the point x=500 is a critical point where the slope of the profit function equals zero. To determine whether this critical point represents a maximum or minimum, you need to examine the behavior of P′(x) around this point. The key insight is that P′(x) is strictly decreasing for all x>0. This means that just before x=500, we have P′(x)>0 (positive slope), and just after x=500, we have P′(x)<0 (negative slope). By the first derivative test, when the derivative changes from positive to negative at a critical point, the function has a local maximum there.
Choice A is incorrect because P′(x) represents marginal profit, not profit itself, and we're told it's strictly decreasing everywhere, so it has no maximum. Choice B is wrong because the derivative changes from positive to negative, indicating a maximum, not a minimum. Choice C is incorrect because breaking even means P(x)=0, but we only know that P′(500)=0, which tells us about the slope, not the actual profit value.
Remember: when the first derivative equals zero and changes from positive to negative, you've found a local maximum. Focus on how the derivative behaves around critical points, not just at them. - A company is analyzing its production of electric scooters. At the current production level of 500 scooters per month, the financial analyst determines that the marginal revenue is $125 per scooter and the marginal cost is $95 per scooter. Assuming the profit function is concave down near this production level, what action should the company take to increase its profit?
Answer: Increase the production level.
Profit is increased by producing more units as long as the marginal revenue (MR) from each additional unit is greater than the marginal cost (MC). In this case, at 500 units, MR = $125 and MC = $95. Since MR > MC, producing and selling another scooter will add $125 - $95 = $30 to the total profit. Therefore, the company should increase its production level.
(B) Decreasing production would be advisable if MC > MR.
(C) Keeping production the same is optimal when MR = MC.
(D) While decreasing fixed costs always helps profit, the data on marginals provides direct guidance on production levels.
- A firm producing
x units of a commodity has a total revenue function R(x) = 120x - 0.1x^2 and a total cost function C(x) = 4000 + 20x + 0.2x^2. The firm is currently producing 150 units. To improve its profit, which of the following actions should the firm take? Answer: Increase production, because marginal revenue exceeds marginal cost at this level.
When you encounter profit optimization problems, the key insight is that firms should produce where marginal revenue equals marginal cost. If marginal revenue exceeds marginal cost, the firm can increase profit by producing more units.
Let's find the marginal functions by taking derivatives. Marginal revenue is MR(x)=R′(x)=120−0.2x and marginal cost is MC(x)=C′(x)=20+0.4x.
At the current production level of 150 units:
- MR(150)=120−0.2(150)=120−30=90
- MC(150)=20+0.4(150)=20+60=80
Since marginal revenue ($90) exceeds marginal cost ($80), the firm should increase production. Each additional unit adds $90 in revenue but only costs $80 to produce, generating $10 in additional profit.
Answer A is wrong because profit is maximized when MR=MC, which occurs at 120−0.2x=20+0.4x, solving to x=0.6100≈167 units, not 150.
Answer B incorrectly states that marginal cost exceeds marginal revenue—we calculated the opposite.
Answer C uses faulty reasoning. While total revenue may exceed total cost (indicating positive profit), this doesn't determine the direction for profit improvement. You need to compare marginal values, not totals.
Remember: for profit optimization, always compare marginal revenue to marginal cost. Increase production when MR > MC, decrease when MR < MC, and you've found the optimum when MR = MC. - The marginal profit for a new software product is given by
P'(x) = -0.4x + 100 dollars per unit, where x is the number of units sold. The company's fixed costs for developing and marketing the software are $3,000. Assuming there are no other costs, what is the total profit if 200 units are sold? Answer: $9,000
To find the total profit function P(x), we must integrate the marginal profit function: P(x) = \int (-0.4x + 100) dx = -0.2x^2 + 100x + K. The constant of integration K is determined by the initial conditions. At x=0 units sold, revenue is $0 and cost is the fixed cost of $3,000. Thus, profit at x=0 is P(0) = 0 - 3000 = -3000. Using our integrated function, P(0) = K, so K = -3000. The full profit function is P(x) = -0.2x^2 + 100x - 3000. Now, evaluate at x=200: P(200) = -0.2(200)^2 + 100(200) - 3000 = -0.2(40000) + 20000 - 3000 = -8000 + 20000 - 3000 = $9,000.
(A) This is the marginal profit at x=200, P'(200).
(C) This result comes from incorrectly omitting the fixed costs (K=0).
(D) This result comes from incorrectly adding the fixed costs instead of subtracting (K=3000).
- The weekly demand function for a brand of noise-canceling headphones is given by
p = 500 - 2x, where p is the price in dollars and x is the number of units. The weekly cost function is C(x) = 20000 + 80x + x^2. What price should the company charge for the headphones to maximize its weekly profit? Answer: $360
This is a multi-step problem. First, find the revenue function R(x) = x \cdot p = x(500 - 2x) = 500x - 2x^2. Second, find the marginal revenue R'(x) = 500 - 4x and marginal cost C'(x) = 80 + 2x. Third, set R'(x) = C'(x) to find the profit-maximizing quantity: 500 - 4x = 80 + 2x, which gives 420 = 6x, so x = 70. Finally, substitute this quantity back into the demand function to find the optimal price: p = 500 - 2(70) = 500 - 140 = $360.
(A) This is the profit-maximizing quantity x, not the price p.
(B) This is the price that maximizes revenue, not profit.
(C) This price is derived from mistakenly setting the demand function equal to marginal cost (p(x) = C'(x)).
- A manufacturer's profit function is P(x)=−2x3+90x2−1200x+5000 dollars, where x is hundreds of units produced. If marginal profit equals zero at x=10 and x=20, what can be concluded about profit optimization?
Answer: Additional analysis of the second derivative is needed to determine which point maximizes profit
When marginal profit P′(x)=0, we have critical points, but we cannot determine whether they represent maxima or minima without further analysis. The second derivative test: P′′(x)=−12x+180. At x=10: P′′(10)=60>0 (local minimum). At x=20: P′′(20)=−60<0 (local maximum). Choice A assumes the first critical point is always a maximum. Choice B incorrectly assumes the maximum occurs at the midpoint. Choice C makes an assumption about cubic behavior without proper analysis. - The demand for a particular e-reader is given by the price-demand equation
p = 2000 - 5x, and the cost function to produce x e-readers is C(x) = 10000 + 200x + x^2. At what production level is the profit maximized? Answer: 150 units
Profit is maximized when marginal revenue equals marginal cost (MR = MC). First, find the revenue function: R(x) = x \cdot p = x(2000 - 5x) = 2000x - 5x^2. Now, find the marginal functions: MR(x) = R'(x) = 2000 - 10x and MC(x) = C'(x) = 200 + 2x. Set them equal: 2000 - 10x = 200 + 2x. Solving for x: 1800 = 12x, which gives x = 150.
(A) This is the production level that minimizes average cost.
(C) This is the production level that maximizes revenue (R'(x)=0).
(D) This is approximately one of the break-even points where total revenue equals total cost (R(x)=C(x)).
- The demand function for a certain luxury good is
p(x) = \sqrt{600 - x}, where x is the number of units sold. What is the marginal revenue when 200 units are sold? Answer: $15
First, define the revenue function, R(x) = x \cdot p(x) = x(600 - x)^{1/2}. To find the marginal revenue, R'(x), we must use the product rule: R'(x) = (1) \cdot (600 - x)^{1/2} + x \cdot \frac{1}{2}(600 - x)^{-1/2}(-1). This simplifies to R'(x) = \sqrt{600 - x} - \frac{x}{2\sqrt{600 - x}}. Now, evaluate at x=200: R'(200) = \sqrt{600 - 200} - \frac{200}{2\sqrt{600 - 200}} = \sqrt{400} - \frac{200}{2\sqrt{400}} = 20 - \frac{200}{2(20)} = 20 - \frac{200}{40} = 20 - 5 = 15.
(A) This is the value of p'(200), the rate of change of price.
(C) This is the price p(200) when 200 units are sold, not the marginal revenue.
(D) This results from a sign error in the product rule.
- A manufacturer's cost function for producing
x smartwatches is C(x) = 5000 + 30x + 0.05x^2. The company is currently producing 200 watches per week. Using marginal analysis, what is the estimated cost of producing the 201st watch? Answer: $50.00
The marginal cost function C'(x) estimates the cost of producing the next (x+1) unit. The derivative of the cost function is C'(x) = 30 + 0.1x. To estimate the cost of the 201st watch, we evaluate the marginal cost at x=200: C'(200) = 30 + 0.1(200) = 30 + 20 = $50.00.
(B) $50.05 is the exact cost of the 201st watch, C(201) - C(200), not the marginal cost estimate.
(C) $65.00 is the average cost per watch at a production level of 200, C(200)/200.
(D) $50.10 is the marginal cost evaluated at x=201, C'(201), which is an estimate for the 202nd watch.
- A consulting firm's monthly cost function is C(x)=x3−15x2+100x+5000 dollars for x consulting projects. If the firm currently handles 8 projects per month, by approximately how much would total cost change if they accepted one additional project?
Answer: Increase by $52, calculated using C′(8)=192−240+100=52
Marginal cost is C′(x)=3x2−30x+100. At x=8: C′(8)=3(64)−30(8)+100=192−240+100=52. This means the cost increases by approximately $52 for one additional project. Choice A has the wrong calculation for the derivative. Choice B has the right calculation but wrong magnitude. Choice D uses exact calculation rather than marginal approximation and would give a different (larger) value. - A retail chain's profit function is P(x)=−0.5x3+12x2−50x−200 thousand dollars, where x is the number of stores (in tens). Current analysis shows that marginal profit is positive at x=5 stores but negative at x=15 stores. What strategy should the company pursue?
Answer: Expand from the current position toward the point where marginal profit becomes zero
When you encounter profit optimization questions, focus on marginal profit—the derivative of the profit function—which tells you whether each additional unit increases or decreases total profit.
The marginal profit is P′(x)=−1.5x2+24x−50. Since marginal profit is positive at x=5 (meaning profit increases with each additional store) but negative at x=15 (meaning profit decreases with each additional store), there must be a maximum profit point between these values where marginal profit equals zero.
Answer B correctly identifies that the company should expand toward this optimal point where marginal profit becomes zero. This represents the profit-maximizing number of stores—the sweet spot where adding one more store would start decreasing total profit.
Answer A misunderstands that while marginal profit might become positive again at very high store counts, this would occur at a much lower profit level than the maximum. Answer C incorrectly assumes 15 stores is optimal, but negative marginal profit there means the company has already passed the optimal point and is losing profit with each additional store. Answer D reflects a dangerous misconception: negative marginal profit doesn't mean the company is losing money overall, just that additional stores are reducing total profit from what it could be.
Remember: positive marginal profit means "keep expanding," negative marginal profit means "you've gone too far," and zero marginal profit indicates the optimal point. Always move toward where the marginal value equals zero for optimization problems. - For a particular monopoly, the marginal revenue function,
MR(x), is positive and strictly decreasing for all x > 0. The marginal cost function, MC(x), is positive and strictly increasing for all x > 0. At the current production level of x = 1000 units, the company finds that MR(1000) > MC(1000). Let x_{max} be the production level that maximizes profit. Which of the following conclusions is correct? Answer: x_{max} > 1000
Profit is maximized at the production level x_{max} where marginal revenue equals marginal cost, MR(x_{max}) = MC(x_{max}). We are given that at x = 1000, MR(1000) > MC(1000). This means that producing the 1001st unit will add more to revenue than to cost, so profit will increase. To reach the maximum profit point, the company must increase production. Since MR(x) is decreasing and MC(x) is increasing, the point where they are equal must occur at a production level greater than 1000. Therefore, x_{max} > 1000.
(A) This would be true if MR(1000) < MC(1000).
(B) This would be true if MR(1000) = MC(1000).
(D) The relationship can be determined by reasoning about the behavior of the marginal functions.
- A company's profit from selling
x units of a product is given by the function P(x) = -0.01x^2 + 80x - 5000. The marginal profit at a production level of 1000 units is calculated. How does this marginal profit value relate to the actual profit gained from selling the 1001st unit? Answer: The marginal profit is a close approximation of the actual profit from the 1001st unit.
The marginal profit P'(x) provides an approximation for the profit from selling the next unit, P(x+1) - P(x). We calculate the marginal profit at x=1000: P'(x) = -0.02x + 80. So, P'(1000) = -0.02(1000) + 80 = -20 + 80 = $60. The actual profit from the 1001st unit is P(1001) - P(1000). P(1000) = -10000 + 80000 - 5000 = 65000. P(1001) = -0.01(1001)^2 + 80(1001) - 5000 = -10020.01 + 80080 - 5000 = 65059.99. The actual profit is $65059.99 - $65000 = $59.99. The marginal profit of $60 is a close approximation.
