Absolute extreme values on [a, b]
The highest point of the road between two towns could be a hilltop in the middle or simply one of the towns. We must check both the hilltops and the end points.
A continuous function on a closed interval [a, b] always has an absolute maximum and an absolute minimum. Find the critical points inside the interval, compute f at these and at a and b; the largest value is the absolute maximum and the smallest is the absolute minimum.
Formula: Compare f(a), f(b) and f(c) for each critical point c in (a, b).
Rule 1: Evaluate and compare — Make a short table of x and f(x).
- Easy: f(x) = x² − 4x on [0, 5]: f(0) = 0, f(2) = −4, f(5) = 5, so the absolute maximum is 5 and the absolute minimum is −4.
- Medium: f(x) = x³ − 3x on [−2, 3]: f(−2) = −2, f(−1) = 2, f(1) = −2, f(3) = 18: the absolute maximum is 18 and the absolute minimum is −2.
- Exam-type: f(x) = 2x³ − 15x² + 36x on [1, 5]: values 23, 28, 27, 55 at 1, 2, 3, 5: maximum 55, minimum 23.
Rule 2: A line or a monotonic function — The extremes occur at the end points.
- Easy: f(x) = 2x + 3 on [1, 4]: minimum 5, maximum 11.
- Medium: f(x) = x³ on [−2, 2]: minimum −8, maximum 8.
- Exam-type: f(x) = sin x on [0, π]: maximum 1 at π/2 and minimum 0 at both ends.
Do not forget the end points: the extremes may be there.
Looking only at critical points and ignoring the end points.
Candidates = critical points inside + both end points.
Practice questions
- 1Find the absolute extremes of f(x) = x² − 6x on [0, 5].
- 2Find the absolute extremes of f(x) = 2x + 3 on [1, 4].
- 3Find the absolute extremes of f(x) = x³ on [−2, 2].
- 4Find the absolute extremes of f(x) = x² + 2x on [−3, 1].
- 5Find the absolute extremes of f(x) = x³ − 3x + 1 on [0, 2].
- 6Find the absolute extremes of f(x) = sin x on [0, π].
- 7Find the absolute extremes of f(x) = x + 1/x on [1/2, 3].
- 8Find the absolute extremes of f(x) = x³ − 6x² + 9x + 1 on [0, 4].
Show answers (try them yourself first)
- 1Critical point 3: f(0) = 0, f(3) = −9, f(5) = −5; maximum 0, minimum −9
- 2Increasing: minimum f(1) = 5, maximum f(4) = 11
- 3Minimum −8, maximum 8
- 4Critical point −1: f(−3) = 3, f(−1) = −1, f(1) = 3; maximum 3, minimum −1
- 5Critical point 1: f(0) = 1, f(1) = −1, f(2) = 3; maximum 3, minimum −1
- 6Maximum 1 at π/2; minimum 0 at 0 and π
- 7Critical point 1: f(1/2) = 5/2, f(1) = 2, f(3) = 10/3; maximum 10/3, minimum 2
- 8Critical points 1 and 3: f(0) = 1, f(1) = 5, f(3) = 1, f(4) = 5; maximum 5, minimum 1
Optimisation problems on a closed interval
A farmer with a fixed length of fence wants the biggest field. This is a maximum problem on an interval of possible side lengths.
Write the quantity to be maximised or minimised in terms of one variable, find the interval of allowed values, then use the critical points and the end points.
Formula: Quantity as a function of one variable, then f′ = 0
Rule 1: Set up and solve — Express, differentiate, test.
- Easy: 40 m of fence: A = x (20 − x) on [0, 20] is largest at x = 10, area 100 m² (the end points give 0).
- Medium: A wire of 24 cm bent into a rectangle: maximum area 36 cm², a 6 cm square.
- Exam-type: A box from a 12 × 12 sheet by cutting squares of side x from the corners: V = x (12 − 2x)², V′ = (12 − 2x)(12 − 6x), so x = 2 and V = 128 cubic units.
Rule 2: Interpret — Give units and check that the answer is in the interval.
- Easy: Profit P(x) = −x² + 12x − 20: P′ = −2x + 12 = 0 at x = 6, P = 16.
- Medium: A ball thrown up: h = 20t − 5t²: h′ = 20 − 10t = 0 at t = 2, h = 20 m.
- Exam-type: Cost C(x) = x² − 10x + 60: least at x = 5 with C = 35.
Always compare with the end points when the interval is closed.
Giving x as the answer when the question asks for the maximum value.
Write, differentiate, solve, compare, answer with units.
Practice questions
- 1Two numbers add to 14. Find the largest product.
- 2A farmer has 60 m of fence for a rectangular field. Find the largest area.
- 3Two positive numbers add to 10. Find the least sum of their squares.
- 4A rectangle has area 64 m². Find the least perimeter.
- 5Profit in thousands is P(x) = −x² + 12x − 20. Find the maximum profit.
- 6A ball thrown upwards has height h = 20t − 5t² metres. Find the greatest height.
- 7A box is made from a 12 × 12 sheet by cutting squares of side x from the corners. Find x for maximum volume and the volume.
- 8Find the maximum of x² (12 − x) for 0 ≤ x ≤ 12.
Show answers (try them yourself first)
- 1Parts 7 and 7 give 49
- 2x (30 − x) is largest at x = 15: 225 m²
- 3x² + (10 − x)² is least at x = 5: 50
- 4Sides x and 64/x: the perimeter is least at x = 8: 32 m
- 5P′ = −2x + 12 = 0 at x = 6, so the maximum profit is 16 (thousand)
- 6h′ = 20 − 10t = 0 at t = 2, so h = 40 − 20 = 20 m
- 7V = x (12 − 2x)² is largest at x = 2, V = 2 × 64 = 128 cubic units
- 8f′ = 24x − 3x² = 3x (8 − x): f(0) = 0, f(8) = 256, f(12) = 0; the maximum is 256 at x = 8