Mathematics Advanced • Year 11 • Module 4 • Lesson 14
Optimisation with Exponentials
Apply exponential differentiation and stationary-point analysis to profit, dose, area and design problems in real contexts.
Problem 1 — Maximising revenue
A streaming service prices a subscription at $x per month. Based on past sign-up data, monthly revenue (in thousands of dollars) is modelled by
R(x) = 50 x · e^(−x/20), for x > 0
Set up: What are we solving for?
(i) Use the product rule to find R'(x) and factor out e^(−x/20). 2 marks
(ii) Solve R'(x) = 0 and state the optimal price x* (in dollars). 2 marks
(iii) Compute the maximum monthly revenue at x = x*, and verify the second-derivative test gives R''(x*) < 0. 3 marks
Stuck? Revisit lesson § Worked example 2 — profit function.Problem 2 — Drug absorption curve
After taking a tablet, the plasma concentration of a drug (in mg/L) follows
C(t) = 4 t · e^(−0.5 t), for t ≥ 0 (t in hours)
Set up: What are we solving for?
(i) Differentiate C(t) using the product rule and factor out e^(−0.5 t). 2 marks
(ii) Find the time t* at which the peak concentration occurs. 2 marks
(iii) Compute the peak concentration C(t*), and confirm by computing C at t* − 1 and t* + 1 that you have found a maximum. 3 marks
Problem 3 — Cable sag (cosh function)
The shape of a hanging cable suspended from two equal-height points is described, near the centre, by the cosh function
y(x) = (e^x + e^(−x)) / 2
Set up: What are we solving for?
(i) Find y'(x) and y''(x), simplifying as far as you can. 2 marks
(ii) Show that y has a single stationary point at x = 0 and use y''(0) to classify it. 3 marks
(iii) State the minimum value of y, and explain in one sentence why the minimum represents the lowest point of the hanging cable (the sag). 2 marks
Stuck? cosh(x) = (e^x + e^(−x))/2 is even and always ≥ 1.Problem 4 — Average cost minimisation
Manufacturing x units has total cost (in dollars)
C(x) = 50 + 10 e^(0.1 x), for x > 0
Set up: What are we solving for?
(i) Write the average cost function Ā(x) = C(x) / x. 1 mark
(ii) Differentiate Ā(x) using the quotient rule. Simplify and set Ā'(x) = 0 to get a single equation in x. 3 marks
(iii) The equation in (ii) cannot be solved exactly with elementary functions. Solve it numerically (a simple iteration or "guess and refine"): show that the minimum-cost x lies between x = 5 and x = 12 by evaluating Ā' at each, then narrow to a one-unit window. 3 marks
Problem 5 — Tent design with an exponential side profile
A pop-up tent has a side profile described by y = h · e^(−x²/L²), with peak height h at x = 0 and width parameter L. The area under the curve (i.e. cross-section enclosed by the profile and the x-axis from x = −L to x = L) is approximately A ≈ 1.5 h L (you don't need to derive this).
Set up: What are we solving for?
(i) Fix the cross-section area at A = 6 m². Express h as a function of L. 1 mark
(ii) The fabric needed scales with peak height plus width: M(L) = 2 h(L) + 2 L. Substitute h from (i) and differentiate M(L). 3 marks
(iii) Find the L that minimises M, and the corresponding h. Comment in one sentence on the trade-off. 3 marks
Stuck? Once h = 4/L, M(L) = 8/L + 2L. Min where dM/dL = 0.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Revenue
Set up. We differentiate the revenue model, find the optimal price, and confirm a maximum.
(i) Product rule: R'(x) = 50 e^(−x/20) + 50 x · (−1/20) e^(−x/20) = 50 e^(−x/20) · (1 − x/20).
(ii) e^(−x/20) > 0, so R'(x) = 0 iff 1 − x/20 = 0 iff x* = 20 (dollars per month).
(iii) R(20) = 50 · 20 · e^(−1) = 1000/e ≈ $367.88 thousand. R''(x) = 50 e^(−x/20)(−2/20 + x/400) = 50 e^(−x/20)(x − 40)/400. At x* = 20: R''(20) = 50 e^(−1)(−20)/400 ≈ −9.2 < 0 ✓ confirms maximum.
Problem 2 — Drug absorption
Set up. We differentiate C(t), find the peak time, and confirm with neighbouring values.
(i) C'(t) = 4 e^(−0.5 t) + 4 t · (−0.5 e^(−0.5 t)) = 4 e^(−0.5 t)(1 − 0.5 t) = 2 e^(−0.5 t)(2 − t).
(ii) e^(−0.5 t) > 0, so C'(t) = 0 iff 2 − t = 0 iff t* = 2 hours.
(iii) C(2) = 4 · 2 · e^(−1) = 8/e ≈ 2.943 mg/L. C(1) = 4 · e^(−0.5) ≈ 2.426. C(3) = 12 · e^(−1.5) ≈ 2.678. Both lower than C(2) ✓ confirms maximum at t = 2.
Problem 3 — Cable sag (cosh)
Set up. We differentiate the cosh function, find its stationary point, and interpret the minimum.
(i) y'(x) = (e^x − e^(−x))/2 = sinh(x). y''(x) = (e^x + e^(−x))/2 = cosh(x) = y(x).
(ii) y'(x) = 0 iff e^x = e^(−x) iff e^(2x) = 1 iff x = 0. y''(0) = (1 + 1)/2 = 1 > 0, so x = 0 is a minimum.
(iii) y(0) = (1 + 1)/2 = 1. Physically, the cable hangs symmetrically; its lowest point sits directly below the midspan (at x = 0), which is the minimum of the cosh profile.
Problem 4 — Average cost
Set up. We minimise Ā(x) = C(x)/x and locate the minimum numerically.
(i) Ā(x) = (50 + 10 e^(0.1 x)) / x = 50/x + 10 e^(0.1 x)/x.
(ii) Quotient rule: Ā'(x) = [C'(x) · x − C(x)] / x² with C'(x) = e^(0.1 x). Setting Ā'(x) = 0 gives x · C'(x) = C(x), i.e. x · e^(0.1 x) = 50 + 10 e^(0.1 x), or equivalently (x − 10) e^(0.1 x) = 50.
(iii) Evaluate (x − 10) e^(0.1 x): at x = 5: (−5)(1.6487) ≈ −8.24 < 50; at x = 12: (2)(3.32) ≈ 6.64 < 50; at x = 14: (4)(4.055) ≈ 16.2; at x = 16: (6)(4.953) ≈ 29.7; at x = 18: (8)(6.05) ≈ 48.4; at x = 19: (9)(6.69) ≈ 60.2. So the root lies between x = 18 and x = 19, i.e. x ≈ 18-19 units for minimum average cost. (Refinement: x ≈ 18.1.)
Problem 5 — Tent design
Set up. We use the area constraint to eliminate h, then minimise M(L).
(i) A = 1.5 h L = 6 ⇒ h = 4/L.
(ii) M(L) = 2 · (4/L) + 2 L = 8/L + 2 L. dM/dL = −8/L² + 2.
(iii) Set dM/dL = 0: 2 = 8/L² ⇒ L² = 4 ⇒ L = 2 m (take positive). h = 4/2 = 2 m. (M''(L) = 16/L³ > 0 at L = 2, confirming a minimum.) M(2) = 4 + 4 = 8 m of fabric. Trade-off: a tall, narrow tent uses lots of fabric for the peak; a short, wide tent uses lots for the width — the optimum balances them, with peak height equal to half-width.