Week 2 — Drills

Exam format. Show all work. Core items are for everyone; stretch items for whoever finishes early. Solutions at the end, worked the way an examiner wants to see them.

A. Reading a function

D1 (core). For each function, state the domain as an interval and say what the restriction means for the firm. (a) f(q) = √(q − 20) — profit per bike above a minimum run. (b) g(q) = ln(q − 5) — a log revenue that only starts once five units are sold. (c) h(q) = 120 / (q − 40) — average fixed cost per bike above capacity 40.

D2 (core). Arno Bikes faces demand p(q) = 1600 − 4q. (a) Write revenue R(q) and simplify. (b) Write profit π(q) using C(q) = 60,000 + 400q. (c) Evaluate π(100) and π(200). Which quantity is better, and by how much?

D3 (stretch). A second firm has R(q) = 30 ln(q + 1) (€ thousand) and C(q) = 0.4q² + 3q + 25. (a) State the domain of R for q ≥ 0. (b) Compute π(5) and π(10) to two decimals.

B. Limits

D4 (core). Evaluate, showing the factoring step: (a) lim (q→10) (q² − 100)/(q − 10) (b) lim (q→3) (q² − 5q + 6)/(q − 3)

D5 (core — exam form). Evaluate the one-sided limit lim (q→1⁺) (q − 1)²/(q − q²). Hint: factor the denominator first. Note the sign of each factor as q approaches 1 from above.

D6 (core). Evaluate the limits at infinity and say what each means for the firm: (a) lim (q→∞) (1200q − 4q²)/(q² ) (b) lim (q→∞) (60,000 + 400q)/q — average cost per bike as the run gets long.

D7 (stretch). lim (q→−∞) (3q³ − 2q + 1)/(5q³ + q²).

C. The derivative

D8 (core). Differentiate, term by term: (a) π(q) = −4q² + 1200q − 60,000 (b) C(q) = 0.4q² + 3q + 25 (c) R(q) = 30 ln(q + 1) (d) f(q) = 5q³ − 2q² + 7q − 1

D8b (core — one inside another). Differentiate, naming the inside each time: (a) ln(q + 1) (b) ln(5q) (c) (3q − 4)⁵ (d) e^(0.08t) (e) 20 ln(2q + 7)

D9 (core — the point of the week). For π(q) = −4q² + 1200q − 60,000: (a) Compute π'(80) and say, in one sentence a manager would use, what it means. (b) Compute π'(200) and do the same. (c) Solve π'(q) = 0. Compare with the vertex you found in week 1.

D10 (stretch). Using the definition, f'(a) = lim (h→0) [f(a+h) − f(a)]/h, show that the derivative of f(q) = 3q² at q = a is 6a.

D. Reversing the derivative

D11 (core — exam form). Find f(q) given f'(q) = 6q² − 2q and f(1) = 4.

D12 (core). Find the total cost function given marginal cost MC(q) = 0.8q + 3 and fixed cost C(0) = 25.

D13 (core — exam form). Integrate term by term; split the fraction first where there is one. (a) ∫ (6q² − 4q + 1) dq (b) ∫ (3q³ + 4q² − 2)/q dq

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Solutions

D1. (a) Need q − 20 ≥ 0 → [20, +∞). Below 20 bikes the formula has no meaning: the firm does not operate under a 20-bike run. (b) Need q − 5 > 0 (strict: ln 0 is undefined) → (5, +∞). Revenue only exists past the fifth unit. (c) Need q − 40 ≠ 0 → (−∞, 40) ∪ (40, +∞); for a quantity, [0, 40) ∪ (40, +∞). At exactly capacity the expression blows up — the model breaks there.

D2. (a) R(q) = (1600 − 4q)q = 1600q − 4q². (b) π(q) = 1600q − 4q² − 60,000 − 400q = −4q² + 1200q − 60,000. (c) π(100) = −40,000 + 120,000 − 60,000 = €20,000. π(200) = −160,000 + 240,000 − 60,000 = €20,000. Equal. Two different quantities, the same profit — the parabola is symmetric about q = 150. Say this out loud: “more” is not the question, “where on the hill” is.

D3. (a) q + 1 > 0 → q > −1, so for a quantity the whole of [0, +∞). (b) π(5) = 30 ln 6 − (0.4·25 + 15 + 25) = 53.75 − 50 = €3.75k. π(10) = 30 ln 11 − (40 + 30 + 25) = 71.94 − 95 = −€23.06k. The second firm is past its best scale at 10.

D4. (a) (q² − 100)/(q − 10) = (q − 10)(q + 10)/(q − 10) = q + 10 → **20**. (b) (q² − 5q + 6)/(q − 3) = (q − 3)(q − 2)/(q − 3) = q − 2 → **1**.

D5. Factor: (q − 1)² / (q(1 − q)). Now 1 − q = −(q − 1), so the expression is (q − 1)² / (−q(q − 1)) = −(q − 1)/q. As q → 1⁺, (q − 1) → 0⁺ and q → 1, so the value → −0 = **0**. The exam wants the factoring and the sign discussion written out, not just the answer.

D6. (a) Divide numerator and denominator by q²: (1200/q − 4)/1 → **−4**. Far out, profit falls like −4q²: the quadratic term dominates and the firm is deep in loss. (b) (60,000 + 400q)/q = 60,000/q + 400 → **400**. Average cost per bike settles at the variable cost: fixed cost spreads to nothing. This is economies of scale in one line.

D7. Divide by q³: (3 − 2/q² + 1/q³)/(5 + 1/q) → **3/5**.

D8. (a) π'(q) = −8q + 1200 (b) C'(q) = 0.8q + 3 (c) R'(q) = 30/(q + 1) (d) f'(q) = 15q² − 4q + 7

D8b. Outside first, inside untouched, then times the inside’s derivative. (a) inside q + 1, derivative 1 → 1/(q+1). The multiplication by 1 is why this one looks free. (b) inside 5q, derivative 5 → (1/5q)·5 = **1/q**. (Also visible from ln 5q = ln 5 + ln q, and ln 5 is a constant.) (c) outside (·)⁵, inside 3q − 4, derivative 3 → 5(3q−4)⁴·3 = **15(3q−4)⁴**. (d) inside 0.08t, derivative 0.08 → 0.08 e^(0.08t). (e) inside 2q + 7, derivative 2 → 20 · (1/(2q+7)) · 2 = **40/(2q+7)**.

D9. (a) π'(80) = −640 + 1200 = **+560**. At 80 bikes a month, one more bike adds about €560 to profit. Make more. (b) π'(200) = −1600 + 1200 = **−400**. At 200 bikes, one more bike costs about €400 of profit. Make fewer. (c) −8q + 1200 = 0 → q = 150, exactly the vertex −b/2a from week 1. The formula worked because the curve was a parabola; the derivative works whatever the curve.

D10. [3(a+h)² − 3a²]/h = [3a² + 6ah + 3h² − 3a²]/h = (6ah + 3h²)/h = 6a + 3h. As h → 0 this → 6a. Divide before letting h go: that is the whole trick.

D11. Antiderivative term by term: f(q) = 2q³ − q² + c. Then f(1) = 2 − 1 + c = 4 → c = 3, so f(q) = 2q³ − q² + 3. Never forget the constant: the initial condition is what pins it down, and the exam always gives you one.

D12. C(q) = 0.4q² + 3q + c, and C(0) = c = 25, so C(q) = 0.4q² + 3q + 25. The constant of integration is the fixed cost. That is not a coincidence; it is the economics.

D13. (a) Power up and divide, term by term: 2q³ − 2q² + q + c. Check: differentiate and you get back 6q² − 4q + 1. (b) Divide each term by q first: 3q² + 4q − 2/q. Then integrate each piece: ∫ 3q² dq = q³, ∫ 4q dq = 2q², ∫ −2/q dq = −2 ln|q|, so q³ + 2q² − 2 ln|q| + c. The 1/q term is the only one the power rule cannot do (it would divide by zero): it is the log.