Difficulty: Intermediate | Prerequisites: Curve sketching study notes, first and second derivative tests
Optimisation is where calculus earns its keep. The problems in this section ask you to find the largest or smallest value a function can take, either on a closed interval or in an applied setting with real-world constraints. This is the payoff for all the derivative mechanics you have learnt so far. If you are comfortable finding critical points and applying the second derivative test, the new material here is mostly about translating a word problem into a function and a constraint.
Absolute extrema problems ask for the highest or lowest value a function reaches on a given domain. On a closed interval, compare function values at every critical point and both endpoints. On an open interval with only one critical point, the second derivative test alone decides. Optimisation word problems add one layer: translate a real scenario into an objective function and a constraint, reduce to one variable, then apply the same tools.
Absolute maximum
The largest value f(c) attains on its entire domain (or on a specified interval). There can be more than one point that achieves the same maximum value.
In simple terms, it is the highest point on the graph over the region you are looking at.
Absolute minimum
The smallest value f(c) attains on its entire domain (or on a specified interval).
Think of it as the lowest point on the graph over the region.
Closed interval method (Max-Min #1)
Used when the domain is a closed interval [a, b]. Find all critical values in (a, b), evaluate f at each critical value and at both endpoints, then compare. The largest is the absolute max, the smallest is the absolute min.
In simple terms, check every candidate and pick the winner.
One-critical-point method (Max-Min #2)
Used when the domain is an open interval and there is exactly one critical point. Apply the second derivative test at that point: if f''(c) > 0 it is an absolute minimum, if f''(c) < 0 it is an absolute maximum.
Think of it as: only one candidate, so if it is a local extremum it must also be the global one.
Objective function
The quantity you want to maximise or minimise in a word problem (area, volume, revenue, cost, distance).
In simple terms, it is "the thing you are solving for."
Constraint equation
A relationship that limits the variables in the problem (total fencing available, fixed perimeter, given surface area). You use it to eliminate one variable so the objective function depends on only one variable.
Think of it as the budget or the limit that ties the variables together.
f'(x) = 4x³ – 4x = 4x(x² – 1). Setting f'(x) = 0 gives x = –1, 0, 1.
The candidate list is all critical values plus the endpoints: {–2, –1, 0, 1, 2}.
Evaluate f at each: f(–2) = 13, f(–1) = 4, f(0) = 5, f(1) = 4, f(2) = 13.
Absolute maxima at (–2, 13) and (2, 13). Absolute minima at (–1, 4) and (1, 4).
The method works because a continuous function on a closed interval is guaranteed (by the Extreme Value Theorem) to attain its absolute max and min somewhere on that interval.
f'(x) = 2 – 72/x². Setting f'(x) = 0 gives x² = 36, so x = 6 (discard x = –6, outside the interval).
Only one critical point: x = 6, f(6) = 24.
The interval (0, ∞) is open, so there are no endpoints to evaluate. With only one CP, use Max-Min #2.
f''(x) = 144/x³. f''(6) = 144/216 > 0, confirming a local (and therefore absolute) minimum at (6, 24).
Every optimisation problem follows a four-step pattern.
Step i: Identify the quantity to optimise. State what you are maximising or minimising (area, volume, revenue).
Step ii: Write the objective function. Express the quantity in terms of the problem's variables.
Step iii: Use the constraint to reduce to one variable. Substitute so the objective function depends on a single variable.
Step iv: Differentiate, set equal to zero, solve. Then confirm whether the result is a max or min using the second derivative test or context.
Objective: maximise A = x · y.
Constraint: 3x + y = 240, so y = 240 – 3x.
A(x) = x(240 – 3x) = 240x – 3x².
A'(x) = 240 – 6x. Setting to zero gives x = 40, y = 120.
Maximum area = 40 × 120 = 4800 yd².
After cutting squares of side x, the base is (20 – 2x) × (20 – 2x) and the height is x.
V(x) = (20 – 2x)² · x = 4x³ – 80x² + 400x.
V'(x) = 12x² – 160x + 400 = 4(3x – 10)(x – 10). Setting to zero gives x = 10/3 (x = 10 makes the box vanish).
Maximum volume = (40/3)² · (10/3) = 16000/27 in³.
R = (number of customers)(ticket price). Let x = price increase from the base price.
R(x) = (100 – 10x)(3 + x) = –10x² + 70x + 300.
R'(x) = –20x + 70. Setting to zero gives x = 3.50.
R''(x) = –20 < 0, so this is a maximum. Optimum ticket price = 3.00 + 3.50 = $6.50.
f'(x) = –1/x² + 8x. Setting f'(x) = 0 gives 8x = 1/x², so 8x³ = 1, x = 1/2.
f''(x) = 2/x³ + 8. f''(1/2) = 16 + 8 = 24 > 0, confirming a minimum.
Closed Interval Method
Find all critical values in (a, b).
Evaluate f at each critical value and at both endpoints a and b.
The largest value is the absolute max, the smallest is the absolute min.
One-CP Method (open interval, one critical point)
Find the single critical value c.
If f''(c) > 0, then f(c) is the absolute minimum. If f''(c) < 0, then f(c) is the absolute maximum.
Optimisation Setup (four steps)
Identify the quantity to optimise and name it.
Write the objective function in terms of the problem's variables.
Write the constraint equation and use it to eliminate one variable.
Differentiate the single-variable objective function, set equal to zero, solve, and confirm with the second derivative test.
Students often try to use the closed interval method on an open interval. If the interval does not include its endpoints (like (0, ∞)), you cannot evaluate f at the endpoints. You need the one-CP method or another approach.
Forgetting to check that a critical value actually lies inside the given interval. In the open-top box problem, x = 10 is a solution to V'(x) = 0, but it makes the box dimensions zero, so it is not a valid answer.
Mixing up "objective function" and "constraint equation." The objective is what you optimise. The constraint is the fixed total (fencing, material, budget) that lets you eliminate a variable. Writing the constraint as the thing to optimise leads nowhere.
Assuming the answer to a word problem is just the x-value. The question usually asks for the maximum area, optimal price, or minimum value. Plug back in to get the final answer in the correct units.
⚠️ The exam review has both closed-interval and open-interval problems. Know which method to use before you start computing.
⚠️ Word problems are almost guaranteed on Exam 2. Practise translating the scenario into the four-step optimisation setup.
⚠️ Watch for problems where one solution to f'(x) = 0 must be discarded because it falls outside the domain or produces a nonsensical answer (negative length, zero volume).
⚠️ Always state your final answer with units. "x = 40" is incomplete. "Maximum area = 4800 yd²" is what the question wants.
True or false: the closed interval method requires you to check the endpoints. Answer: True. Always evaluate f at both endpoints of a closed interval.
Fill in the blank: to use the one-CP method, the interval must be ______ and there must be exactly ______ critical point(s). Answer: Open; one.
True or false: if f''(c) > 0 at the only critical point on an open interval, then f(c) is an absolute maximum. Answer: False. f''(c) > 0 means concave up, so f(c) is an absolute minimum.
In an optimisation word problem, the constraint equation is used to ______. Answer: Eliminate one variable so the objective function depends on a single variable.
True or false: if a critical value makes a physical dimension zero or negative, you should still include it in your answer. Answer: False. Discard critical values that produce nonsensical results in the context of the problem.
Q: Find the absolute extrema of f(x) = x⁴ – 2x² + 5 on [–2, 2].
A: f'(x) = 4x³ – 4x = 4x(x² – 1). Critical values: x = –1, 0, 1. Evaluate f at {–2, –1, 0, 1, 2}: {13, 4, 5, 4, 13}. Absolute maxima at (–2, 13) and (2, 13). Absolute minima at (–1, 4) and (1, 4).
Q: Find the absolute minimum of f(x) = 2x + 72/x on (0, ∞).
A: f'(x) = 2 – 72/x². Setting to zero gives x² = 36, x = 6. Only one CP on an open interval, so use the second derivative test. f''(x) = 144/x³, f''(6) > 0, confirming an absolute minimum at (6, 24).
Q: A farmer has 240 ft of fencing and wants to enclose a rectangular area against a barn wall, using fencing on three sides. What dimensions maximise the area?
A: Let x = width (two sides of fencing), y = length (one side). Constraint: 3x + y = 240, so y = 240 – 3x. A(x) = x(240 – 3x). A'(x) = 240 – 6x = 0 gives x = 40, y = 120. Maximum area = 4800 yd².
Q: Squares are cut from the corners of a 20-inch square sheet of metal and the sides are folded up to make an open-top box. What size square maximises the volume?
A: V(x) = (20 – 2x)² · x. V'(x) = 4(3x – 10)(x – 10) = 0 gives x = 10/3 (x = 10 is rejected). Maximum volume = 16000/27 in³.
Q: A cinema charges $3 per ticket and draws 100 customers. For each $1 increase in price, 10 fewer customers attend. What ticket price maximises revenue?
A: Let x = dollar increase. R(x) = (100 – 10x)(3 + x) = –10x² + 70x + 300. R'(x) = –20x + 70 = 0 gives x = 3.50. Optimal price = $6.50. R''(x) = –20 < 0 confirms a maximum.
Absolute extrema build directly on curve sketching: you already know how to find and classify critical points, and this section just adds endpoints and constraints to the picture. Optimisation word problems are one of the most common exam formats for testing derivative skills, so this material ties together everything from the differentiation chapters.
The fencing and box problems are classics that reappear in multivariable calculus (Lagrange multipliers), where the constraint is handled differently but the idea is the same.
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