Mathematics Advanced • Year 11 • Module 3 • Lesson 8
Stationary Points
Apply stationary-point analysis to motion, profit, geometry and curve-sketching problems.
Problem 1 — Tennis-ball peak height (kinematics)
A tennis ball thrown from a balcony has height (in metres) above the court
h(t) = −5t² + 15t + 4, 0 ≤ t ≤ 3.5 s
Set up: What are we solving for?
(i) Find h′(t) and the time at which h′(t) = 0 (i.e. the stationary point). 2 marks
(ii) Use the second derivative to confirm this stationary point is a maximum, and find the maximum height. 3 marks
(iii) Compare the maximum height to the value of h at the endpoints t = 0 and t = 3.5 s. Why does the question ask for the stationary point rather than just inspecting the endpoints? 2 marks
Stuck? Revisit lesson § Worked Example 1 (find, then classify).Problem 2 — Profit curve for a Sydney pop-up cafe (business)
A pop-up cafe's daily profit, when x staff are rostered, is modelled by
P(x) = −2x² + 40x − 100 dollars, 1 ≤ x ≤ 15
Set up: What are we solving for?
(i) Find the stationary point of P(x). 2 marks
(ii) Use P″(x) to classify the stationary point, and state the staffing level x* that maximises profit and the corresponding daily profit P(x*). 3 marks
(iii) Suppose the cafe owner can only roster integer numbers of staff (you cannot half-employ someone). Show that the integer staffing level matching the stationary point is the same as the continuous optimum, and verify by computing P(x* − 1) and P(x* + 1). 2 marks
Problem 3 — Open-top box volume (geometry)
A square of side 12 cm has equal squares of side x cm cut from each corner; the resulting cross is folded into an open-top box of volume
V(x) = x(12 − 2x)², 0 < x < 6
Set up: What are we solving for?
(i) Expand V(x) so that it is a polynomial in x, then find V′(x). 3 marks
(ii) Solve V′(x) = 0 and reject any solution outside the domain 0 < x < 6. 2 marks
(iii) Use V″(x) to classify the remaining stationary point and state the maximum volume of the box. 2 marks
Stuck on (i)? (12 − 2x)² = 144 − 48x + 4x², so V(x) = 144x − 48x² + 4x³.Problem 4 — Sketch from stationary points (curve sketching)
Consider the curve y = x³ − 6x² + 9x.
Set up: What are we solving for?
(i) Find the y-intercept and the x-intercepts. 2 marks
(ii) Find and classify all stationary points. 3 marks
(iii) Sketch the curve, labelling intercepts and stationary points with their coordinates. 2 marks
(Sketch on grid below or attach a sheet.)
Problem 5 — Build a cubic from its features (work-backwards)
The curve y = ax³ + bx² + cx + d has a local maximum at (1, 4) and a local minimum at (3, 0).
Set up: What are we solving for?
(i) Write down the two equations coming from "the curve passes through (1, 4) and (3, 0)". 2 marks
(ii) Write down the two equations coming from "the gradient is 0 at x = 1 and x = 3". 2 marks
(iii) Solve the four equations to find a, b, c, d. Verify your cubic satisfies the original two stationary-point conditions. 3 marks
Stuck? Subtract pairs of equations to eliminate d first, then c.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Tennis ball
Set up. We are finding the peak height of a projectile by locating the stationary point of h(t).
(i) h′(t) = −10t + 15 = 0 ⇒ t = 1.5 s.
(ii) h″(t) = −10 < 0 (everywhere), so the stationary point is a maximum. h(1.5) = −5(2.25) + 22.5 + 4 = −11.25 + 22.5 + 4 = 15.25 m.
(iii) h(0) = 4 m (height of balcony); h(3.5) = −5(12.25) + 52.5 + 4 = −61.25 + 56.5 = −4.75 m (the ball is below court level at t = 3.5 s — actually it would have hit the ground earlier). The maximum (15.25 m) is much higher than either endpoint. The stationary-point analysis is necessary because a quadratic does not attain its maximum at an endpoint — it attains it at the turning point in the interior, which inspection of endpoints would miss.
Problem 2 — Pop-up cafe profit
Set up. We are finding the staffing level that maximises a quadratic profit function.
(i) P′(x) = −4x + 40 = 0 ⇒ x = 10. y-coord: P(10) = −200 + 400 − 100 = 100.
(ii) P″(x) = −4 < 0 ⇒ local maximum. x* = 10 staff, daily profit P(10) = $100.
(iii) x* = 10 is already an integer, so no rounding is needed. Verify by adjacent integers: P(9) = −162 + 360 − 100 = 98; P(11) = −242 + 440 − 100 = 98. Both are below P(10) = 100, confirming x = 10 maximises integer profit.
Problem 3 — Open-top box volume
Set up. We are differentiating a constrained volume to find the cut size x that maximises capacity.
(i) V(x) = x(144 − 48x + 4x²) = 4x³ − 48x² + 144x. V′(x) = 12x² − 96x + 144 = 12(x² − 8x + 12) = 12(x − 2)(x − 6).
(ii) V′(x) = 0 ⇒ x = 2 or x = 6. Since x must satisfy 0 < x < 6, reject x = 6 (gives zero base length). Keep x = 2.
(iii) V″(x) = 24x − 96; V″(2) = 48 − 96 = −48 < 0 ⇒ maximum. V(2) = 4(8) − 48(4) + 144(2) = 32 − 192 + 288 = 128 cm³.
Problem 4 — Sketch y = x³ − 6x² + 9x
Set up. We are extracting all the features needed for a complete cubic sketch.
(i) y-intercept: y(0) = 0. x-intercepts: x³ − 6x² + 9x = x(x² − 6x + 9) = x(x − 3)² = 0 ⇒ x = 0 or x = 3. So intercepts at (0, 0) and (3, 0); (3, 0) is a touching point because of the repeated root.
(ii) y′ = 3x² − 12x + 9 = 3(x − 1)(x − 3) = 0 ⇒ x = 1 or x = 3. y(1) = 1 − 6 + 9 = 4; y(3) = 27 − 54 + 27 = 0. y″ = 6x − 12; y″(1) = −6 < 0 ⇒ local max at (1, 4); y″(3) = 6 > 0 ⇒ local min at (3, 0).
(iii) Sketch: cubic with positive leading coefficient. Rises from negative infinity through (0, 0) up to a local max at (1, 4), falls to a local min at (3, 0) where it touches the x-axis, then rises to positive infinity. The repeated root at x = 3 (which also happens to be the local min) means the curve is tangent to the x-axis there.
Problem 5 — Build a cubic
Set up. We have four conditions (two points, two zero gradients) and four unknowns (a, b, c, d).
(i) Through (1, 4): a + b + c + d = 4. Through (3, 0): 27a + 9b + 3c + d = 0.
(ii) y′ = 3ax² + 2bx + c. y′(1) = 0: 3a + 2b + c = 0. y′(3) = 0: 27a + 6b + c = 0.
(iii) From (ii), subtract: 24a + 4b = 0 ⇒ b = −6a. Sub into 3a + 2b + c = 0: c = −3a − 2b = −3a + 12a = 9a. From (i), eqn 1: a + (−6a) + 9a + d = 4 ⇒ 4a + d = 4 ⇒ d = 4 − 4a. From (i), eqn 2: 27a + 9(−6a) + 3(9a) + d = 0 ⇒ 27a − 54a + 27a + d = 0 ⇒ d = 0. Substituting back: 0 = 4 − 4a ⇒ a = 1. Hence b = −6, c = 9, d = 0. The cubic is y = x³ − 6x² + 9x, which is exactly the curve from Problem 4 — y(1) = 4 ✓ and y(3) = 0 ✓; y′(1) = 0 ✓ and y′(3) = 0 ✓; and from Problem 4 the second-derivative test confirms (1, 4) is the max and (3, 0) is the min.