Difficulty: Intermediate | Prerequisites: Derivatives (power rule, product rule, chain rule), solving polynomial equations, natural logarithm properties.
Optimization is the reason most of Calculus I exists in the applied world. You have spent weeks learning to differentiate; this chapter is where that skill pays off. The core idea: a function's maximum or minimum occurs either at a point where the derivative equals zero, at a point where the derivative does not exist, or at an endpoint of a closed interval. Every optimization problem in this course reduces to three steps: build a function, find its critical points, and determine which critical point gives the best value. If you are comfortable taking derivatives and solving for x, you already have the mechanical tools. The new challenge is translating a word problem into a function to differentiate.
To find a maximum or minimum, take the derivative, set it to zero, and solve. If the domain is a closed interval, also check the endpoints. If there is a constraint (like x + y = 120), use it to eliminate one variable before differentiating.
Critical point
A value x = c in the domain of f where f'(c) = 0 or f'(c) does not exist. Think of it as any spot where the slope goes flat or the derivative breaks down.
Absolute (global) maximum / minimum
The single largest or smallest value a function achieves over its entire domain (or over a specified interval). In simple terms, this is the overall winner, not just a local peak or valley.
Local (relative) maximum / minimum
A value f(c) that is larger (or smaller) than all nearby values, but not necessarily the biggest (or smallest) everywhere. Think of it as the tallest hill in your neighbourhood, even if a taller mountain exists elsewhere.
Closed Interval Method
A procedure for finding absolute extrema on a closed interval [a, b]: find all critical points in (a, b), evaluate f at each critical point and at both endpoints, then compare. In simple terms, you make a shortlist of candidates and pick the winner.
Constraint
An equation that limits the variables in an optimization problem (e.g. x + y = 120). Think of it as a budget or a rule that ties your variables together.
Objective function
The function you are trying to maximize or minimize. In simple terms, this is the quantity you care about, the thing you want to make as large or as small as possible.
First Derivative Test
A method for classifying a critical point by checking whether f' changes sign around it. If f' changes from positive to negative, the critical point is a local max; if from negative to positive, a local min. Think of it as watching whether the curve goes up-then-down or down-then-up.
When a problem gives you two variables and a constraint linking them, the strategy is: use the constraint to eliminate one variable, then optimise a single-variable function.
Method
Identify the objective function (the quantity to maximise or minimise).
Identify the constraint (the equation linking the variables).
Solve the constraint for one variable and substitute into the objective function.
Differentiate, set the derivative to zero, and solve for the remaining variable.
Use the first derivative test or second derivative test to confirm whether you have a max or min.
Substitute back to find the other variable.
Worked Example (Problem 1)
Maximise Q = x²y subject to x + y = 120.
From the constraint: y = 120 – x.
Substitute: Q = x²(120 – x) = 120x² – x³.
Differentiate: Q' = 240x – 3x².
Set Q' = 0: 0 = –3x² + 240x = –3x(x – 80).
Critical points: x = 0 and x = 80.
First derivative test at x = 80: Q' is positive for x < 80 and negative for x > 80, so x = 80 is a maximum.
Back-substitute: y = 120 – 80 = 40.
Answer: the maximum value of Q occurs at x = 80, y = 40.
Why x = 0 is discarded: at x = 0, Q = 0, which is clearly not a maximum of a product that can be positive. The number line confirms Q' changes from negative to positive at x = 0 (a minimum), not a maximum.
When you need the absolute max and min of a continuous function on a closed interval [a, b], the Extreme Value Theorem guarantees they exist. The closed interval method finds them.
Method
Find all critical points of f in the open interval (a, b).
Evaluate f at each critical point.
Evaluate f at the endpoints a and b.
The largest value is the absolute maximum; the smallest is the absolute minimum.
No sign chart or derivative test is needed here. You are simply comparing a finite list of values.
Worked Example (Problem 2)
Find the absolute max and min of f(x) = 2x² – 4 ln(x) on [1/2, 2].
Differentiate: f'(x) = 4x – 4/x.
Set f'(x) = 0: 4x – 4/x = 0, so 4/x = 4x, giving 4x² = 4, hence x² = 1, so x = 1. (x = –1 is outside the domain since ln(x) requires x > 0.)
Evaluate at the critical point and endpoints:
f(1/2) = 2(1/4) – 4 ln(1/2) = 0.5 + 4 ln(2) ≈ 3.27
f(1) = 2(1) – 4 ln(1) = 2 – 0 = 2
f(2) = 2(4) – 4 ln(2) = 8 – 4 ln(2) ≈ 5.22
Compare: absolute minimum is f(1) = 2; absolute maximum is f(2) ≈ 5.22.
Key observation: the absolute max here occurs at an endpoint, not at a critical point. This is common, and it is precisely why you must always check endpoints on a closed interval.
These problems give a physical scenario (a box, a fence, a container) with a constraint on one quantity (volume, perimeter, cost) and ask you to optimise another (surface area, area, material). The method is the same as constrained optimization, but the setup requires translating geometry into algebra.
Method
Draw and label the shape. Identify all variables.
Write the objective function (the quantity to minimise or maximise).
Write the constraint equation.
Solve the constraint for one variable, substitute into the objective function.
Differentiate, set to zero, solve.
Verify it is a min or max (first or second derivative test).
Find remaining dimensions.
Worked Example (Problem 3)
Minimise the surface area of an open-top rectangular box with a square base and a fixed volume of 12 cubic units.
Variables: base side length x, height y.
Volume constraint: V = x²y = 12, so y = 12/x².
Surface area (open top, so one x² face missing): SA = 2x² + 4xy.
Note: 2x² accounts for top and bottom if the box is closed, or just the base plus internal structure depending on the specific formulation. The homework uses SA = 2x² + 4xy.
Substitute: SA = 2x² + 4x(12/x²) = 2x² + 48/x.
Differentiate: SA' = 4x – 48x⁻² = 4x – 48/x².
Set SA' = 0: 4x = 48/x², so 4x³ = 48, giving x³ = 12, hence x = ∛12 ≈ 2.289.
Back-substitute: y = 12/(∛12)² = 12/∛144 = ∛12 ≈ 2.289.
The optimal box has x = y = ∛12. When the base is square and the height equals the base side length, you use the least material.
Why this makes intuitive sense: a cube (or near-cube) minimises surface area for a given volume. The closer a box is to a cube, the less material it wastes.
\frac{d}{dx}[x^n] = n x^{n-1}\frac{d}{dx}[\ln x] = \frac{1}{x}\frac{d}{dx}[f \cdot g] = f'g + fg' \quad \text{(product rule)}Surface area of a rectangular box with square base (side x, height y):
Open top: SA = x² + 4xy
Closed: SA = 2x² + 4xy
Volume of a rectangular box with square base: V = x²y
Critical point condition: set f'(x) = 0 and solve, then also check where f'(x) is undefined.
Manufacturers use exactly this type of surface-area minimisation to design cans, boxes, and containers that use the least material for a given capacity. Shipping companies optimise box dimensions to minimise cardboard cost. Engineers designing heat exchangers or fuel tanks solve the same constrained optimization problems with more complex geometries.
Students often assume the critical point is automatically the answer. It is not. On a closed interval, the absolute max or min can occur at an endpoint. Always check endpoints.
Students sometimes forget to verify whether a critical point is a max or a min. Finding f'(c) = 0 tells you the slope is zero; it does not tell you which type of extremum you have. Use a sign chart, the first derivative test, or the second derivative test.
When given a constraint like x + y = 120, students occasionally differentiate without substituting first, leaving two variables in the derivative. You must reduce to a single variable before differentiating.
Students often discard x = 0 without explanation in problems where the domain is (0, ∞). State why: at x = 0 the objective function gives a trivial or undefined value.
⚠️ Optimization word problems appear on nearly every Calculus I final. Expect at least one constrained optimization problem and one closed-interval extrema problem.
⚠️ The closed interval method is mechanical and high-reward on exams: if you can differentiate and evaluate, you will get the marks. Do not skip endpoint checks.
⚠️ Professors frequently test whether students can set up the objective function from a word problem. The calculus is often straightforward; the translation from English to algebra is where marks are lost.
⚠️ Watch for domain restrictions. If a variable represents a physical length, it must be positive. If x + y = 120, then x must be between 0 and 120.
True or false: if f'(c) = 0, then f(c) is guaranteed to be a local maximum or minimum.
False. f'(c) = 0 means c is a critical point, but it could be an inflection point (e.g. f(x) = x³ at x = 0).
Fill in the blank: to find absolute extrema on a closed interval, evaluate f at all ______ in the interval and at both ______.
Critical points; endpoints.
True or false: when optimising Q = x²y subject to x + y = 120, you should differentiate Q with respect to both x and y.
False. You eliminate one variable using the constraint first, then differentiate with respect to the remaining variable only.
Fill in the blank: the derivative of ln(x) is ______.
1/x.
True or false: if a function is continuous on [a, b], it must attain both an absolute max and an absolute min on that interval.
True. This is the Extreme Value Theorem.
Q: Two positive numbers x and y satisfy x + y = 100. What values of x and y maximise the product P = xy?
A: Substitute y = 100 – x, so P = x(100 – x) = 100x – x². Differentiate: P' = 100 – 2x. Set to zero: x = 50. Then y = 50. The maximum product is 2500.
Q: Find the absolute maximum and minimum of g(x) = x³ – 3x + 1 on [–2, 2].
A: g'(x) = 3x² – 3 = 3(x² – 1) = 0, so x = 1 and x = –1. Evaluate: g(–2) = –8 + 6 + 1 = –1, g(–1) = –1 + 3 + 1 = 3, g(1) = 1 – 3 + 1 = –1, g(2) = 8 – 6 + 1 = 3. Absolute max = 3 (at x = –1 and x = 2); absolute min = –1 (at x = –2 and x = 1).
Q: A farmer has 200 metres of fencing and wants to enclose a rectangular area against a straight river (no fence needed on the river side). What dimensions maximise the enclosed area?
A: Let x = side perpendicular to the river, y = side parallel to the river. Constraint: 2x + y = 200, so y = 200 – 2x. Area = xy = x(200 – 2x) = 200x – 2x². Differentiate: A' = 200 – 4x = 0, so x = 50. Then y = 100. Maximum area is 5000 m².
Q: Find the critical points of h(x) = x⁴ – 4x³ + 6 and classify each.
A: h'(x) = 4x³ – 12x² = 4x²(x – 3). Critical points at x = 0 and x = 3. At x = 3, h' changes from negative to positive, so x = 3 is a local minimum. At x = 0, h' does not change sign (negative on both sides), so x = 0 is neither a max nor a min (it is an inflection-like critical point).
This material connects directly to related rates (Chapter 3), which also requires translating a word problem into a calculus setup, though there you differentiate with respect to time rather than optimising. The second derivative test used to confirm maxima and minima ties into concavity and curve sketching from the same chapter. In Calculus II, these optimization ideas extend to multivariable functions where Lagrange multipliers replace the substitution method used here.
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