This set has twelve original questions, ordered from easier to harder, covering all five lessons in stationary points and optimisation. Questions 1 to 5 practise finding and classifying points, 6 to 9 add constraints, endpoints and domains, and 10 to 12 mix everything.
Attempt each question on paper before opening the answer. Write the constraint, the objective function and the domain as separate lines. Mark the ones you got wrong and use the routing list at the end.
Questions
1. Find the stationary point of y = x² + 6x − 1 and state whether it is a maximum or minimum.
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dy/dx = 2x + 6 = 0 gives x = −3. Then y = 9 − 18 − 1 = −10. d²y/dx² = 2 > 0, so it is a minimum.
(−3, −10), a minimum
2. Find and classify the stationary points of y = x³ − 12x + 5.
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dy/dx = 3x² − 12 = 3(x − 2)(x + 2), so x = 2 or x = −2. d²y/dx² = 6x.
At x = 2: y = 8 − 24 + 5 = −11, and d²y/dx² = 12 > 0, so a minimum.
At x = −2: y = −8 + 24 + 5 = 21, and d²y/dx² = −12 < 0, so a maximum.
(2, −11) is a minimum and (−2, 21) is a maximum
3. Show that y = x³ − 3x² + 3x has one stationary point and that it is a stationary point of inflection.
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dy/dx = 3x² − 6x + 3 = 3(x − 1)². This is zero only at x = 1, where y = 1 − 3 + 3 = 1.
Since 3(x − 1)² is never negative, the gradient is positive on both sides of x = 1. The sign does not change.
(1, 1) is the only stationary point and it is an inflection
4. Find and classify the stationary points of y = x⁴ − 4x³.
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dy/dx = 4x³ − 12x² = 4x²(x − 3), so x = 0 or x = 3.
At x = 3: y = 81 − 108 = −27. d²y/dx² = 12x² − 24x = 108 − 72 = 36 > 0, so a minimum.
At x = 0: y = 0 and d²y/dx² = 0, so test the gradient. At x = −1, dy/dx = 4(1)(−4) = −16 (negative). At x = 1, dy/dx = 4(1)(−2) = −8 (negative). No sign change, so an inflection.
(3, −27) is a minimum and (0, 0) is a stationary point of inflection
5. Find the stationary point of y = 3x + 12/x² for x > 0 and show it is a minimum.
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y = 3x + 12x⁻², so dy/dx = 3 − 24x⁻³ = 3 − 24/x³. Setting to zero gives x³ = 8, so x = 2. Then y = 6 + 3 = 9.
d²y/dx² = 72x⁻⁴ = 72/x⁴. At x = 2 it is 72/16 = 4.5 > 0.
(2, 9) is a minimum
6. A rectangle has perimeter 50 cm. Find its greatest area.
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Let the sides be x and y. Constraint: x + y = 25, so y = 25 − x. Objective: A = x(25 − x) = 25x − x².
dA/dx = 25 − 2x = 0 gives x = 12.5, and y = 12.5. d²A/dx² = −2 < 0, so a maximum.
Greatest area = 12.5 × 12.5 = 156.25 cm²
7. Find the greatest and least values of y = 2x³ − 9x² + 12x for 0 ≤ x ≤ 3.
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dy/dx = 6x² − 18x + 12 = 6(x − 1)(x − 2), so x = 1 or x = 2, both inside the interval.
y(1) = 2 − 9 + 12 = 5. y(2) = 16 − 36 + 24 = 4. Endpoints: y(0) = 0 and y(3) = 54 − 81 + 36 = 9.
Greatest 9 (at x = 3), least 0 (at x = 0)
8. A farmer has 40 m of fencing for three sides of a pen against a wall that is 12 m long. Find the largest area.
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Let the two sides be x m and the side along the wall be y m. Constraint: 2x + y = 40, so y = 40 − 2x. Objective: A = 40x − 2x².
Domain: y ≤ 12 gives x ≥ 14, and y > 0 gives x < 20. So 14 ≤ x ≤ 20.
dA/dx = 40 − 4x = 0 gives x = 10, which is outside the domain. Compare endpoints: A(14) = 560 − 392 = 168 and A(20) = 800 − 800 = 0.
Largest area = 168 m², with x = 14 m and y = 12 m
9. A ball is thrown so that its height is h = 1.5 + 12t − 4t² metres after t seconds. Find the maximum height and the time it occurs.
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dh/dt = 12 − 8t = 0 gives t = 1.5. h = 1.5 + 18 − 9 = 10.5. d²h/dt² = −8 < 0, so a maximum.
Maximum height 10.5 m, after 1.5 seconds
10. An open-topped box has a square base of side x cm and a volume of 32 cm³. Its surface area is S = x² + 4xh, where h is the height. Find the value of x that minimises S, and the least S.
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Constraint: x²h = 32, so h = 32/x². Then S = x² + 4x(32/x²) = x² + 128/x.
dS/dx = 2x − 128/x² = 0 gives x³ = 64, so x = 4. Then h = 32/16 = 2 and S = 16 + 32 = 48.
d²S/dx² = 2 + 256/x³ = 2 + 4 = 6 > 0, so a minimum.
x = 4 cm, least surface area 48 cm²
11. The cost of a journey is C = 2v + 800/v ringgit, where v km/h is the speed and v > 0. Find the speed that gives the least cost and the cost.
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dC/dv = 2 − 800/v² = 0 gives v² = 400, so v = 20 (v > 0). C = 40 + 40 = 80.
d²C/dv² = 1600/v³ = 1600/8000 = 0.2 > 0, so a minimum.
Speed 20 km/h, least cost RM80
12. Two positive numbers x and y have sum 20. Find the values that make x²y as large as possible, and that largest value to 1 decimal place.
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y = 20 − x, so F = x²(20 − x) = 20x² − x³ for 0 < x < 20.
dF/dx = 40x − 3x² = x(40 − 3x) = 0 gives x = 40/3 (x = 0 is not allowed). Then y = 20 − 40/3 = 20/3.
d²F/dx² = 40 − 6x = 40 − 80 = −40 < 0, so a maximum. F = (40/3)² × (20/3) = 1600/9 × 20/3 = 32000/27 ≈ 1185.2.
x = 13.3 (40/3), y = 6.7 (20/3), largest value ≈ 1185.2
If you got these wrong
- Questions 1 to 3, or the y-coordinate was missing: go back to finding stationary points from a derivative.
- Questions 2 to 5, or the wrong classification: revisit classifying stationary behaviour, especially the gradient test when d²y/dx² = 0.
- Questions 6, 10 and 12, or you could not start: practise forming an objective function from a constraint.
- Question 7, or a greatest value that looked too small: review checking endpoints.
- Questions 8 and 9, or answers without units: work through interpreting an optimum with units and a domain.
- Derivative slips in questions 5 and 11: revise differentiation techniques.
Record each slip in the mistake log and retest queue, and use the non-calculator working trainer or the quadratic structure explorer for extra checking. If one error type keeps returning, Additional Mathematics tuition is one way to work on it with a teacher.