B. Derivatives and Their Applications · Applications of Derivatives: Motion, Rates and OptimizationOptimization Basics: Absolute Maximum and Minimum Values
B2.4About 65 minutes
Learning goals- I can tell the difference between local and absolute extremes.
- I can find the absolute maximum and minimum of a continuous function on a closed interval.
- I can find absolute extremes of polynomial, simple rational and exponential functions.
- I can justify an absolute extreme on an open interval with a single critical number.
Words to know
absolute (global) maximum
The largest value of a function over its whole domain or a given interval.
absolute (global) minimum
The smallest value of a function over its whole domain or a given interval.
critical number
A number c in the domain where f′(c) = 0 or f′(c) does not exist.
Extreme Value Theorem
A function continuous on a closed interval [a, b] has an absolute maximum and an absolute minimum on it.
closed interval
An interval that includes its endpoints, such as 0 ≤ x ≤ 5 or [0, 5].
The best possible
Optimization means finding the best value of something: the largest area, the least cost, the greatest profit, the shortest time. The Ontario curriculum (B2.4) asks you to solve optimization problems with polynomial, simple rational and exponential models, and these problems are among the most useful things calculus does. Engineers, economists, biologists and logistics planners solve them every day.
In Curve Sketching: Connecting f, f′ and f″ you found local maximums and minimums. A real problem wants the absolute (or global) maximum or minimum: the very largest or smallest value over all the allowed inputs. This lesson shows how to find it, and the next five lessons use it on real problems.
Where can the absolute extremes be?
Suppose f is continuous on a closed interval a ≤ x ≤ b. Then:
- f has an absolute maximum and an absolute minimum on the interval. (This is the Extreme Value Theorem; you will prove it in university.)
- Each one happens either at a critical number inside the interval (where f′(x) = 0 or f′(x) does not exist) or at an endpoint.
So you do not need to sketch the whole graph. You only need to check the candidates.
The closed-interval method
- Find f′(x).
- Find the critical numbers in the interval (a < x < b).
- Evaluate f at each critical number and at both endpoints a and b.
- The largest value is the absolute maximum; the smallest is the absolute minimum.
Worked exampleExample 1: A polynomial on an interval
Find the absolute maximum and minimum of f(x) = x³ − 3x² − 9x + 5 on −2 ≤ x ≤ 6.
- f′(x) = 3x² − 6x − 9 = 3(x² − 2x − 3) = 3(x − 3)(x + 1).
- Critical numbers: x = −1 and x = 3. Both lie inside the interval.
- Evaluate: f(−2) = −8 − 12 + 18 + 5 = 3; f(−1) = −1 − 3 + 9 + 5 = 10; f(3) = 27 − 27 − 27 + 5 = −22; f(6) = 216 − 108 − 54 + 5 = 59.
- Compare the four values: 3, 10, −22, 59.
Absolute maximum 59 at x = 6 (an endpoint). Absolute minimum −22 at x = 3. The local maximum at x = −1 is not the absolute maximum.
The four candidates for Example 1. The local maximum at x = −1 is beaten by the endpoint x = 6, which is why you must always check the endpoints.Worked exampleExample 2: A rational function
Find the absolute extremes of A(x) = x + 9/x on 1 ≤ x ≤ 10.
- Write A(x) = x + 9x⁻¹. Then A′(x) = 1 − 9x⁻² = 1 − 9/x².
- A′(x) = 0 when x² = 9, so x = 3 (x = −3 is outside the interval). A′ exists everywhere on the interval (x = 0 is not in it).
- Evaluate: A(1) = 1 + 9 = 10; A(3) = 3 + 3 = 6; A(10) = 10 + 0.9 = 10.9.
Absolute minimum 6 at x = 3; absolute maximum 10.9 at x = 10.
Worked exampleExample 3: An exponential model
The concentration of a medication in the blood is C(t) = 20te^(−0.5t) mg/L, t hours after a dose, for 0 ≤ t ≤ 8. When is the concentration highest, and what is it?
- Product rule: C′(t) = 20e^(−0.5t) + 20t(−0.5)e^(−0.5t) = 20e^(−0.5t)(1 − 0.5t).
- Since e^(−0.5t) > 0 always, C′(t) = 0 only when 1 − 0.5t = 0, so t = 2.
- Evaluate: C(0) = 0; C(2) = 40e^(−1) ≈ 14.7; C(8) = 160e^(−4) ≈ 2.93.
The concentration is highest at t = 2 h, about 14.7 mg/L.
The concentration model C(t) = 20te^(−0.5t) on 0 ≤ t ≤ 8. The highest point has a horizontal tangent; the endpoints give smaller values.When the interval is not closed
Many real problems have an open interval, such as x > 0. Then there are no endpoints to check, and an absolute extreme might not exist. A useful test:
If f is continuous on an interval and has only one critical number c there, and f has a local minimum at c (by the first or second derivative test), then f(c) is the absolute minimum on that interval. (The same is true for a maximum.)
Why? If the graph went down again somewhere else, it would have to turn around, which would create a second critical number.
Worked exampleExample 4: One critical number on an open interval
Find the absolute minimum of f(x) = x + 16/x for x > 0.
- f′(x) = 1 − 16/x². f′ = 0 when x² = 16, so x = 4 (the only critical number for x > 0).
- First derivative test: f′(2) = 1 − 4 = −3 < 0 and f′(8) = 1 − 0.25 = 0.75 > 0, so f changes from decreasing to increasing: a local minimum at x = 4.
- Since x = 4 is the only critical number on x > 0, the local minimum is the absolute minimum. f(4) = 4 + 4 = 8.
The absolute minimum is 8 at x = 4. There is no absolute maximum (f grows without bound as x → 0⁺ or x → ∞).
Always check
Write the domain before you differentiate, throw out critical numbers outside it, and check endpoints when there are any. Most errors in optimization come from skipping one of these.
Just for funThe Extreme Value Theorem promises that a continuous function on a closed interval has a maximum and a minimum. It does not promise to tell you where they are. It is the math version of "your keys are definitely somewhere in this house".
Hands-on activityHighs and lows of a day
You need: an outdoor thermometer or a weather app with an hourly forecast, graph paper, and a pencil.
- Record the temperature every hour from 6 a.m. to 9 p.m. (or copy the hourly forecast for one day).
- Plot temperature against time and join the points with a smooth curve.
- Mark the absolute maximum and absolute minimum on your interval. For each, decide: is it at an endpoint (6 a.m. or 9 p.m.) or at an interior point? If it is interior, is the curve flat there (horizontal tangent)?
- Repeat with a shorter interval, say 9 a.m. to 1 p.m.
What to look for: on a typical day the minimum is near sunrise and the maximum in mid-afternoon, so on the long interval the maximum is an interior point with a flat top. On the short interval (temperature still rising) both extremes are usually at the endpoints. That is exactly why the closed-interval method checks both.
Your turn
1Find the absolute extremes of f(x) = x² − 6x + 1 on 0 ≤ x ≤ 5.
f′ = 2x − 6 = 0 at x = 3. f(0) = 1, f(3) = −8, f(5) = −4. Max 1 at x = 0; min −8 at x = 3.
2Find the absolute extremes of g(x) = 2x³ − 9x² + 12x on 0 ≤ x ≤ 3.
g′ = 6(x − 1)(x − 2). g(0) = 0, g(1) = 5, g(2) = 4, g(3) = 9. Max 9 at x = 3; min 0 at x = 0.
3Find the absolute extremes of h(x) = x⁴ − 8x² on −1 ≤ x ≤ 3.
h′ = 4x(x − 2)(x + 2); in the interval, x = 0 and x = 2. h(−1) = −7, h(0) = 0, h(2) = −16, h(3) = 9. Max 9 at x = 3; min −16 at x = 2.
4Find the absolute extremes of f(x) = x + 4/x on 1 ≤ x ≤ 5.
f′ = 1 − 4/x² = 0 at x = 2. f(1) = 5, f(2) = 4, f(5) = 5.8. Max 5.8 at x = 5; min 4 at x = 2.
5Find the absolute extremes of f(t) = 10te^(−t) on 0 ≤ t ≤ 4.
f′ = 10e^(−t)(1 − t) = 0 at t = 1. f(0) = 0, f(1) = 10/e ≈ 3.68, f(4) = 40e^(−4) ≈ 0.73. Max ≈ 3.68 at t = 1; min 0 at t = 0.
6A spring day's temperature is T(t) = −0.5t² + 4t + 10 °C, t hours after 8 a.m., for 0 ≤ t ≤ 10. Find the highest and lowest temperatures.
T′ = −t + 4 = 0 at t = 4. T(0) = 10, T(4) = 18, T(10) = 0. Highest 18 °C at noon (t = 4); lowest 0 °C at 6 p.m. (t = 10).
7Explain why f(x) = 1/x has no absolute maximum on −1 ≤ x ≤ 1, even though the interval is closed.
f is not continuous on the interval (it is undefined at x = 0, and f → ∞ as x → 0⁺). The closed-interval guarantee needs continuity.
More practice: warm-up
1For a continuous function on [a, b], where can the absolute extremes occur?
At critical numbers inside the interval or at the endpoints.
2What is a critical number?
A number c in the domain where f′(c) = 0 or f′(c) does not exist.
3f(x) = 3x + 2 on 0 ≤ x ≤ 4. Find the absolute max and min.
No critical numbers. Max 14 at x = 4; min 2 at x = 0.
4f(x) = x² on −1 ≤ x ≤ 2. Find the absolute max and min.
f′ = 2x = 0 at 0. f(−1) = 1, f(0) = 0, f(2) = 4. Max 4; min 0.
5Find the critical number of f(x) = x² − 10x.
More practice: on your own
1Find the absolute extremes of f(x) = x³ − 12x + 1 on −3 ≤ x ≤ 5.
f′ = 3x² − 12 = 0 at x = ±2. f(−3) = 10, f(−2) = 17, f(2) = −15, f(5) = 66. Max 66 at x = 5; min −15 at x = 2.
2A share price is modelled by S(t) = t³ − 9t² + 24t + 10 dollars, t months after purchase, 0 ≤ t ≤ 5. Find the highest and lowest prices.
S′ = 3(t − 2)(t − 4). S(0) = 10, S(2) = 30, S(4) = 26, S(5) = 30. Highest $30, at t = 2 and again at t = 5; lowest $10 at t = 0.
3Find the absolute minimum of f(x) = x + 25/x for x > 0, and justify that it is absolute.
f′ = 1 − 25/x² = 0 at x = 5 (only critical number). f′ < 0 for 0 < x < 5 and f′ > 0 for x > 5, so it is a local min, and being the only critical number it is absolute: f(5) = 10.
4Find the absolute extremes of f(x) = x²e^(−x) on 0 ≤ x ≤ 4.
f′ = 2xe^(−x) − x²e^(−x) = xe^(−x)(2 − x): critical numbers 0 and 2. f(0) = 0, f(2) = 4e^(−2) ≈ 0.541, f(4) = 16e^(−4) ≈ 0.293. Max ≈ 0.541 at x = 2; min 0 at x = 0.
5Find the absolute extremes of f(x) = sin x + cos x on 0 ≤ x ≤ π.
f′ = cos x − sin x = 0 when tan x = 1, x = π/4. f(0) = 1, f(π/4) = √2/2 + √2/2 = √2, f(π) = −1. Max √2 ≈ 1.41 at x = π/4; min −1 at x = π.
Challenge
1For f(x) = x³ − ax on 0 ≤ x ≤ 3, find the value of a for which the absolute minimum occurs at x = 1, and give the minimum value.
A minimum inside the interval needs f′(1) = 0: 3 − a = 0, so a = 3. Then f(0) = 0, f(1) = −2, f(3) = 18. The absolute minimum is −2 at x = 1. ✓
2f(x) = x³ on −1 ≤ x ≤ 1 has a critical number at x = 0. Is f(0) an absolute or local extreme? Explain.
f′ = 3x² = 0 at x = 0, but f′ ≥ 0 on both sides, so f is increasing through 0: no local extreme at all. Absolute max 1 at x = 1 and min −1 at x = −1 (endpoints). A critical number is only a candidate.
3A drug's concentration is C(t) = 20(e^(−0.2t) − e^(−t)) mg/L. Find when it is greatest and the greatest concentration.
C′ = 20(−0.2e^(−0.2t) + e^(−t)) = 0 when e^(−t) = 0.2e^(−0.2t), so e^(−0.8t) = 0.2 and t = (ln 5)/0.8 ≈ 2.01 h. C ≈ 20(0.6687 − 0.1337) ≈ 10.7 mg/L. (C′ > 0 before and < 0 after, and C → 0 as t → ∞, so this is the absolute max for t ≥ 0.)
Quiz
1. For a continuous function on a closed interval, which values must you compare to find the absolute maximum?
2. f(x) = x² − 4x on 0 ≤ x ≤ 5. What is the absolute maximum?
3. f(x) = x + 1/x on 0.5 ≤ x ≤ 4. Where is the absolute minimum?
4. f(t) = te^(−t). At which t is f′(t) = 0?
5. A function on x > 0 has exactly one critical number, at x = 3, and f′ changes from negative to positive there. What can you conclude?
Big idea (B2.4): optimization is finding absolute maximums and minimums. On a closed interval, the answer is always at a critical number or an endpoint, so you compare a short list of values.
How to help: ask your teen to list the candidates before calculating anything: "Which x-values are you going to test, and why?" A good answer names every critical number in the interval and both endpoints.
Common difficulties:
- Forgetting to check the endpoints.
- Keeping critical numbers that are outside the interval.
- Comparing values of f′ instead of values of f.
- Calling a local maximum the absolute maximum without comparing.