Mathematics Advanced • Year 11 • Module 4 • Lesson 14
Optimisation with Exponentials
Practise HSC-style writing on exponential optimisation, including a structured profit-maximisation problem with explicit marking criteria.
1. Short-answer questions
1.1 Find the maximum value of f(x) = 3 x e^(−x) for x ≥ 0. Show that you have found a maximum (not a minimum or inflection). 3 marks Band 3-4
1.2 Find the coordinates of the stationary point of y = x² e^(−x) (for x > 0) and classify it. 4 marks Band 4
1.3 Find the minimum value of f(x) = e^x + 4 e^(−x). State which test you used to classify it. 3 marks Band 4
Stuck on 1.3? f'(x) = e^x − 4 e^(−x); set equal to 0 and solve via e^(2x) = 4.2. Extended response
2.1 A start-up sells a subscription at price $x per month. Their revenue (in thousands of dollars per month) is modelled by R(x) = 100 x · e^(−0.05 x), for x > 0. The marketing team wants the optimal price.
(a) Differentiate R(x) using the product rule. Factor out the positive exponential.
(b) Find the optimal price x* algebraically and prove it is a maximum using either the sign-of-R' test or the second-derivative test, naming which.
(c) Compute the maximum revenue R(x*), giving the exact value (in terms of e) and a decimal value rounded to the nearest hundred dollars.
(d) The CEO wonders whether a small cut to $18 (slightly below x*) would gain much by attracting more customers. Compute R(18) and compare it to R(x*) as a percentage. State whether the cut is materially worthwhile. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — applies product rule with u = 100x, v = e^(−0.05 x), getting u' = 100, v' = −0.05 e^(−0.05 x).
• 1 mark — factors to R'(x) = 100 e^(−0.05 x)(1 − 0.05 x).
Part (b) — 2 marks
• 1 mark — notes e^(−0.05 x) > 0 and solves 1 − 0.05 x = 0 ⇒ x* = 20.
• 1 mark — names a test (sign of R' OR R'' < 0 at x = 20) and concludes maximum.
Part (c) — 2 marks
• 1 mark — exact value R(20) = 100 · 20 · e^(−1) = 2000 / e (in thousands).
• 1 mark — decimal: 2000 / e ≈ 735.76, rounded to the nearest hundred dollars: ≈ $735 800 (or $735.8 thousand).
Part (d) — 1 mark
• 1 mark — computes R(18) ≈ 1800 · e^(−0.9) ≈ 732.0, expresses as ≈ 99.5% of R(x*), and concludes the cut is not materially worthwhile (a tenth of a percent of revenue lost).
Your response:
Stuck on (d)? Compute the percentage gap: 1 − R(18)/R(20).How did this worksheet feel?
What I'll revisit before next class:
1.1 — Max of f(x) = 3 x e^(−x), x ≥ 0 (3 marks)
Sample response. f'(x) = 3 e^(−x) + 3x(−e^(−x)) = 3 e^(−x)(1 − x). Since e^(−x) > 0, f'(x) = 0 iff x = 1. For 0 ≤ x < 1: 1 − x > 0, so f' > 0 (increasing). For x > 1: 1 − x < 0, so f' < 0 (decreasing). Sign + then −, so x = 1 is a maximum. Maximum value: f(1) = 3 · 1 · e^(−1) = 3/e ≈ 1.104.
Marking notes. 1 mark — correct f'(x) with the e^(−x) factor isolated. 1 mark — stationary point x = 1 with the "e^(−x) > 0" remark. 1 mark — classification using sign test AND max value 3/e (exact). Numerical answer alone caps at 2/3.
1.2 — Stationary point of y = x² e^(−x), x > 0 (4 marks)
Sample response. dy/dx = 2x e^(−x) + x²(−e^(−x)) = e^(−x) · x · (2 − x). Since e^(−x) > 0 and we want x > 0, set 2 − x = 0 ⇒ x = 2. y(2) = 4 e^(−2) = 4/e². To classify: for 0 < x < 2, (2 − x) > 0 ⇒ y' > 0 (increasing). For x > 2, (2 − x) < 0 ⇒ y' < 0 (decreasing). Hence the stationary point at (2, 4/e²) is a local maximum.
Marking notes. 1 mark — correct derivative factored as e^(−x) · x · (2 − x). 1 mark — both stationary x-values noted (x = 0 ruled out by domain or treated as endpoint); selects x = 2. 1 mark — correct coordinates (2, 4/e²) in exact form. 1 mark — classified with named test. Missing classification caps at 3/4.
1.3 — Min of f(x) = e^x + 4 e^(−x) (3 marks)
Sample response. f'(x) = e^x − 4 e^(−x). Set = 0: e^x = 4 e^(−x) ⇒ e^(2x) = 4 ⇒ 2x = ln 4 ⇒ x = (ln 4)/2 = ln 2. f(ln 2) = e^(ln 2) + 4 e^(−ln 2) = 2 + 4 · (1/2) = 2 + 2 = 4. Second-derivative test: f''(x) = e^x + 4 e^(−x), always positive, so the stationary point is a minimum.
Marking notes. 1 mark — correct derivative and reduction to e^(2x) = 4. 1 mark — exact x = ln 2 (not just x ≈ 0.693). 1 mark — exact min value 4 AND named test (second-derivative, with f''(x) > 0 stated).
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). Let u = 100x and v = e^(−0.05 x). Then u' = 100 and v' = −0.05 e^(−0.05 x). [1 mark]
By the product rule, R'(x) = 100 · e^(−0.05 x) + 100x · (−0.05 e^(−0.05 x)) = 100 e^(−0.05 x) · (1 − 0.05 x). [1 mark — factored]
Part (b). The factor 100 e^(−0.05 x) is strictly positive for all real x, so R'(x) = 0 if and only if 1 − 0.05 x = 0, i.e. x* = 20. [1 mark]
Using the second-derivative test: differentiate R'(x). With g(x) = 100 e^(−0.05 x)(1 − 0.05 x), g'(x) = 100 e^(−0.05 x) · ( −0.05(1 − 0.05x) − 0.05 ) = 100 e^(−0.05 x) · ( 0.0025 x − 0.1 ). At x = 20: g'(20) = 100 e^(−1) · (0.05 − 0.1) = −5/e < 0. So R''(20) < 0, confirming x* = 20 is a local maximum. [1 mark — test named and conclusion]
Part (c). R(20) = 100 · 20 · e^(−1) = 2000 / e (in thousands of dollars). [1 mark — exact]
Decimal: 2000 / 2.71828 ≈ 735.76 thousand ≈ $735 800 (to the nearest hundred dollars). [1 mark — rounded]
Part (d). R(18) = 100 · 18 · e^(−0.9) = 1800 · e^(−0.9) ≈ 1800 · 0.4066 ≈ 731.9 thousand.
Percentage of R(x*): R(18)/R(20) = 731.9 / 735.76 ≈ 0.9948, i.e. ≈ 99.5%. So R(18) is only about 0.5% less than R(20) — the revenue is not materially worth sacrificing the $2 cut to reach $18 (a half-percent gap). [1 mark]
Total: 7/7.
Band descriptors for marker.
Band 3: Computes R'(x) correctly but does not factor out e^(−0.05 x); finds x = 20 without justifying that the exponential is never zero; no classification test named; ≈ 2-3 marks.
Band 4: Correctly factors derivative; finds x* = 20 and states "max" with hand-waving; computes R(x*) as a decimal but no exact form; no part (d); ≈ 4-5 marks.
Band 5: Full algebra, exact value 2000/e quoted; classification test named (sign test or f''); part (d) attempted with rough comparison; ≈ 5-6 marks.
Band 6: Full algebra, "iff" reasoning in (b), exact + decimal values in (c), percentage comparison and clear business-style conclusion in (d). 7/7.