Average Cost, Marginal Cost and an Area Under the Curve
A small workshop produces a single product in batches of one hundred units. The manager models the total cost of producing x hundred units per day as C(x) = x³ − 12x² + 60x, measured in hundreds of yen, for 0 < x ≤ 10; the factory cannot run above 1000 units per day. Before the annual pricing review she wants three things: the daily output that minimises the average cost per unit A(x) = C(x)/x; the relationship between marginal cost C′(x) and average cost at that output; and the total excess of the average cost over its minimum value as output runs from very small values up to the optimal level, which she wants reported as an area under the graph of A. A colleague warns her that setting A′(x) = 0 is not by itself enough, because the feasible interval has endpoints, and that an algebraic solution of C′(x) = A(x) may turn out not to be admissible. Production is treated as continuous for the purposes of the model.
1. Find the output minimising average cost and the minimum average cost, justifying that it is a minimum on the whole interval.
Read the worked answer
Divide first: A(x) = C(x)/x = x² − 12x + 60 for 0 < x ≤ 10.
Stationary point: A′(x) = 2x − 12 = 0 gives x = 6. Since A″(x) = 2 > 0, the parabola opens upward and x = 6 is a local (indeed global on ℝ) minimum. Value: A(6) = 36 − 72 + 60 = 24.
Now check the interval, as the colleague insisted. As x → 0⁺, A(x) → 60; at the right endpoint A(10) = 100 − 120 + 60 = 40. Both exceed 24, so on (0, 10] the minimum is 24 at x = 6.
Interpretation: produce 600 units per day; the average cost is 24 hundred yen per hundred units, i.e. about ¥24 per unit. Note that A is decreasing on (0, 6) and increasing on (6, 10], which is the monotonicity statement behind the minimum.
2. Show the relation between marginal cost and average cost at the optimal output, and solve C′(x) = A(x) in general.
Read the worked answer
Differentiate the cost: C′(x) = 3x² − 24x + 60. At x = 6, C′(6) = 108 − 144 + 60 = 24, which equals A(6) = 24.
In general,
C′(x) = A(x) ⇔ 3x² − 24x + 60 = x² − 12x + 60 ⇔ 2x² − 12x = 0 ⇔ 2x(x − 6) = 0,
so x = 0 or x = 6. The root x = 0 must be discarded: the domain is 0 < x ≤ 10 and A(x) is not even defined at x = 0. Hence the only admissible solution is x = 6, the average-cost minimiser.
This is not a coincidence. From A = C/x, the quotient rule gives A′ = (xC′ − C)/x² = (C′ − A)/x, so A′ = 0 exactly when C′ = A. The algebraic solution set therefore always contains the stationary points of A, and the judgement is to discard the ones outside the domain or where A is undefined.
3. Compute the total excess of average cost over its minimum, from output 0 to the optimal output, as an area.
Read the worked answer
The excess is A(x) − 24, and
A(x) − 24 = x² − 12x + 36 = (x − 6)²,
a perfect square, so the graph of A lies on or above the line y = 24 for all x and touches it only at x = 6. No sign correction is needed.
Excess = ∫₀⁶ (x − 6)² dx = [ (x − 6)³ / 3 ]₀⁶ = 0/3 − (−6)³/3 = 0 + 216/3 = 72.
Equivalently, substituting u = x − 6 gives ∫₋₆⁰ u² du = 72. The figure 72 is in units of (hundreds of yen per hundred units) × (hundreds of units); the numerical value is what matters here, and the point of the exercise is that an area between a curve and a horizontal line is computed as ∫ (upper − lower) with the integrand kept non-negative.