Optimisation and Word Problems: Maxima Minima Guide

Optimization word problems ask for the "best" choice: the largest area, the smallest cost, the shortest time. They all follow the same strategy — translate the story into a function of one variable, then use derivatives to find its maximum or minimum. The calculus part is routine; the real skill is setting up the right function.

This matters because optimization is calculus's most visible real-world payoff. Fencing a field with limited wire, designing a can that uses the least metal, pricing a product for maximum revenue — every one of these is a maxima-minima problem in disguise, and the method below solves them all.

Key Concepts

• Objective function: the quantity to maximize or minimize, written as a function of a single variable (e.g. area A(x)).
• Constraint: the given condition (fixed perimeter, fixed volume) used to eliminate extra variables so only one remains.
• Domain: the physically meaningful range of the variable (lengths must be positive, for instance) — the optimum must lie in it.
• Verification: use the second derivative test or endpoint analysis to confirm the critical point is the desired maximum or minimum.

How It Works

1. Step 1: Read the problem, draw a diagram, and name the variables.
2. Step 2: Write the objective function and the constraint equation.
3. Step 3: Use the constraint to eliminate all but one variable from the objective.
4. Step 4: Differentiate, set the derivative to zero, and solve for the critical point.
5. Step 5: Verify it is a maximum (or minimum) with the second derivative test, and check the answer makes physical sense.

Worked Example

A farmer has 40 metres of fencing to enclose a rectangular field. What dimensions give the maximum area?

1. Step 1: Let the sides be x and y metres.
2. Step 2: Objective: area A = xy. Constraint: perimeter 2x + 2y = 40.
3. Step 3: From the constraint, y = 20 − x, so A(x) = x(20 − x) = 20x − x², with domain 0 smaller than x smaller than 20.
4. Step 4: A'(x) = 20 − 2x. Setting to zero: 20 − 2x = 0 gives x = 10, and then y = 20 − 10 = 10.
5. Step 5: A''(x) = −2, which is smaller than 0, confirming a maximum. The field should be 10 m by 10 m — a square — with maximum area 100 m².

Common Misconceptions

• "The answer is just the critical point." You must verify max vs min and sanity-check the result — a negative length or an absurd value means the setup went wrong.
• "The constraint can be ignored after setup." It also restricts the domain; always check that the critical point lies within the physically allowed range.
• "A square is always the answer." Only for the problems where symmetry applies — different constraints (like a fence against a river needing only three sides) give different shapes.

Real-World Applications

• Manufacturing: designing cylindrical cans that minimize metal for a fixed volume (the classic surface-area problem).
• Business: choosing the price that maximizes revenue when demand falls as price rises.
• Logistics: finding the warehouse location that minimizes total transport distance to customers.
• Physics: light takes the path of least time (Fermat's principle), which is an optimization problem solved by minimizing a time function.

To recap: name variables, write the objective and constraint, eliminate down to one variable, differentiate and solve, then verify. The setup is the hard part — the calculus is the same every time.