Difficulty: Introductory-Intermediate | Prerequisites: Basic calculus (derivatives), ECON300 Problem Set 1
To find the best level of an activity, take the derivative of the benefit and cost functions to get marginal benefit (MB) and marginal cost (MC), then set MB = MC and solve. The solution is only valid if the net benefit there is higher than at both boundary points (zero activity and maximum activity). This is the workhorse method for every unconstrained optimisation problem in the course.
Total benefit, B(H)
The overall benefit gained from H units of an activity. In this problem set, B(H) = 654H − 40H². Think of it as the cumulative payoff from all the hours you hire the mechanic.
Total cost, C(H)
The overall cost incurred from H units of an activity. Here, C(H) = 110H + 120H². In simple terms, this is everything you pay, including the accelerating expense of additional hours.
Marginal benefit, MB(H)
The additional benefit from one more unit of activity. Formally, MB(H) = dB/dH. It tells you what the next hour is worth to you.
Marginal cost, MC(H)
The additional cost from one more unit of activity. Formally, MC(H) = dC/dH. It tells you what the next hour will cost you.
Net benefit, NB(H)
Total benefit minus total cost: NB(H) = B(H) − C(H). The quantity you are trying to maximise.
Best choice (interior solution)
The value of H where MB(H) = MC(H), provided it lies within the feasible range and yields higher net benefit than the endpoints.
Boundary choices
The extreme endpoints of the feasible range (here, H = 0 and H = 6). You must check these to confirm your interior solution is the true maximum.
Take the first derivative of each function with respect to H.
B(H) = 654H − 40H² → MB(H) = dB/dH = 654 − 80H
C(H) = 110H + 120H² → MC(H) = dC/dH = 110 + 240H
MB slopes downward (each additional hour is worth less). MC slopes upward (each additional hour costs more). This is the standard pattern in diminishing-returns problems.
Set the two derivatives equal and solve for H:
654 − 80H = 110 + 240H
654 − 110 = 240H + 80H
544 = 320H
H* = 1.7 hours
This is the interior candidate. At H = 1.7, the benefit from one more hour exactly equals the cost of one more hour, so there is no gain from adjusting.
The feasible range is 0 ≤ H ≤ 6. Calculate net benefit at each boundary and at the interior solution:
NB(0) = B(0) − C(0) = 0 − 0 = 0
NB(1.7) = B(1.7) − C(1.7) = (654 × 1.7 − 40 × 1.7²) − (110 × 1.7 + 120 × 1.7²) = (1111.8 − 115.6) − (187 + 346.8) = 996.2 − 533.8 = 462.4
NB(6) = B(6) − C(6) = (654 × 6 − 40 × 36) − (110 × 6 + 120 × 36) = (3924 − 1440) − (660 + 4320) = 2484 − 4980 = −2496
Since NB(1.7) = 462.4 is higher than both NB(0) = 0 and NB(6) = −2496, the interior solution H* = 1.7 is confirmed as the best choice.
Setting MB = MC finds a candidate, but it could be a minimum or a saddle point. The boundary check ensures you have the global maximum over the entire feasible range. In this problem, hiring for all six hours would leave you with a large negative net benefit, which confirms the interior solution.
General rule for unconstrained optimisation:
MB(H) = MC(H) \quad \Rightarrow \quad \frac{dB}{dH} = \frac{dC}{dH}For this problem:
654 - 80H = 110 + 240H \quad \Rightarrow \quad H^* = 1.7Net benefit at the optimum:
NB(H^*) = B(H^*) - C(H^*) = 462.4The MB curve is a downward-sloping line starting at 654 (when H = 0) and falling. The MC curve is an upward-sloping line starting at 110. They cross at H = 1.7.
Students often think the best choice is wherever MB is highest. It is not. The best choice is where MB = MC, because that is where the gap between total benefit and total cost is largest.
Some students set B(H) = C(H) instead of MB(H) = MC(H). Setting the totals equal finds the break-even point, not the optimum.
Forgetting to check boundary choices. Even if MB = MC gives a valid H, the boundaries could yield higher net benefit (not the case here, but it can happen when the interior solution falls outside the feasible range).
Confusing "best choice" with "maximum benefit." You want maximum net benefit, not maximum benefit. Benefit alone ignores costs.
⚠️ You will almost certainly be asked to derive MB and MC from given total functions. Know your power rule cold.
⚠️ The boundary check step is easy marks. Examiners include it to see whether you understand that MB = MC is necessary but not sufficient.
⚠️ Expect variations where the interior solution falls outside the feasible range, making a boundary the best choice.
⚠️ This method reappears in consumer theory (MRS = price ratio) and producer theory (MR = MC). The logic is identical, only the labels change.
True or False: The best choice always occurs where total benefit equals total cost. _______
If MB(H) = 200 − 10H and MC(H) = 50 + 5H, the optimal H is _______.
True or False: If the interior solution gives a negative net benefit, a boundary choice could be better. _______
The slope of the MB curve is determined by the _______ of the total benefit function.
True or False: You only need to check boundaries when the problem explicitly asks you to. _______
Answers: 1. False (you want MB = MC, not B = C). 2. H = 10 (set 200 − 10H = 50 + 5H). 3. True. 4. Second derivative (the coefficient on H²). 5. False (always check, even if not asked, to confirm the global maximum).
Q: Given B(H) = 654H − 40H² and C(H) = 110H + 120H², find the marginal benefit and marginal cost functions.
A: MB(H) = 654 − 80H. MC(H) = 110 + 240H. Take the first derivative of each with respect to H.
Q: Find the best choice of H and verify it is better than the boundaries (H = 0 and H = 6).
A: Set MB = MC: 654 − 80H = 110 + 240H → 544 = 320H → H* = 1.7. NB(0) = 0, NB(1.7) = 462.4, NB(6) = −2496. The interior solution H* = 1.7 is the best choice.
Q: Suppose instead that B(H) = 500H − 50H² and C(H) = 100H + 50H², with 0 ≤ H ≤ 5. What is the best choice?
A: MB = 500 − 100H, MC = 100 + 100H. Set equal: 500 − 100H = 100 + 100H → 400 = 200H → H* = 2. Check: NB(0) = 0, NB(2) = (1000 − 200) − (200 + 200) = 400, NB(5) = (2500 − 1250) − (500 + 1250) = −500. Best choice is H* = 2.
Q: Why might the boundary H = 0 be the best choice in some problems?
A: If the interior MB = MC solution yields a negative net benefit, or if MB < MC for all positive H (the first unit costs more than it is worth), then doing nothing (H = 0) is optimal.
This method is the foundation for constrained optimisation (Problem Set 2, Question 2), where you add a budget constraint and equalize marginal benefit per dollar across activities. The same MB = MC logic reappears in consumer choice (set MRS equal to the price ratio) and in firm theory (produce where MR = MC). If you are comfortable with this unconstrained version, the constrained version is a short step forward.
marginal analysis, marginal benefit, marginal cost, net benefit maximisation, unconstrained optimisation, interior solution, boundary solution, corner solution, first-order condition, MB = MC rule, derivative optimisation, ECON300, intermediate microeconomics, optimal choice, benefit-cost analysis