Optimisation problems ask you to find the input that makes a function as large or as small as possible. In business calculus, this usually means finding the price, quantity, or dimensions that maximise revenue or profit, or minimise cost. The method is always the same: build the function, take its derivative, set the derivative to zero, and verify with the second derivative test.
Difficulty: Intermediate
Prerequisites: Critical points and the second derivative test, basic integration, comfort translating word problems into algebraic expressions.
This is where all the calculus you have learnt pays off. Optimisation problems are the reason business students take calculus: they model real decisions about pricing, production volume, and resource allocation. If you can set up and solve these problems, you can answer questions like "How many units should we produce to make the most profit?" or "What rental rate brings in the most income?" Every optimisation problem follows the same skeleton (build a function, differentiate, solve, verify), so once you have the pattern, the variety of word problems stops being intimidating.
Revenue function R(x)
Total revenue from selling x units, calculated as price per unit times number of units: R(x) = p · x. When price depends on quantity (a demand curve), substitute the demand equation for p.
Cost function C(x)
Total cost of producing x units. Typically includes a fixed cost (constant term) and variable costs that depend on x.
Profit function P(x)
Revenue minus cost: P(x) = R(x) - C(x). The quantity that maximises profit is where P'(x) = 0 and P''(x) < 0.
Price-demand equation
A relationship between the selling price p and the number of units x that consumers will buy at that price. In simple terms, it tells you how demand drops as you raise the price.
Marginal cost
The derivative of the total cost function, C'(x). It represents the cost of producing one additional unit. Think of it as the answer to "how much more does the next unit cost?"
Marginal revenue
The derivative of the total revenue function, R'(x). It represents the additional revenue from selling one more unit.
Fixed costs
Costs that do not change with production volume (rent, salaries, insurance). In the cost function, these are the constant term. When you integrate a marginal cost function, fixed costs become the constant of integration.
Constraint
A limit on the resources available (total fencing, total budget, maximum capacity). In a constrained optimisation problem, you use the constraint to eliminate one variable so the function depends on only one variable.
The standard setup: a company sells x units at price p, with a price-demand equation and a cost function given.
Worked example from the source: a company manufactures smartphones. p = 500 - 0.4x and C(x) = 20,000 + 20x.
Revenue: R(x) = x · p = x(500 - 0.4x) = 500x - 0.4x^2.
To maximise revenue, take R'(x) = 500 - 0.8x. Set to zero: x = 625. So 625 phones maximise revenue.
The price at maximum revenue: p = 500 - 0.4(625) = 500 - 250 = 250.
Profit: P(x) = R(x) - C(x) = 500x - 0.4x^2 - 20,000 - 20x = -0.4x^2 + 480x - 20,000.
P'(x) = -0.8x + 480. Set to zero: x = 600. So 600 phones maximise profit.
Verify: P''(x) = -0.8 < 0, confirming a maximum.
Note: the quantity that maximises revenue (625) is different from the quantity that maximises profit (600). This is a common exam distinction.
Another classic format: a car rental agency rents 200 cars per day at $30 per day. For each $1 increase in rate, 5 fewer cars are rented.
Let x = number of $1 increases. New rate = 30 + x. Cars rented = 200 - 5x.
Income: I(x) = (30 + x)(200 - 5x) = 6000 - 150x + 200x - 5x^2 = 6000 + 50x - 5x^2.
I'(x) = 50 - 10x. Set to zero: x = 5.
Optimal rate = 30 + 5 = $35 per day. Cars rented = 200 - 25 = 175.
Maximum income = 35 × 175 = $6,125.
When a problem gives you the marginal cost C'(x) and the fixed costs, you integrate to recover the total cost function.
Integrate C'(x) to get C(x) + a constant.
Use the fixed cost (the value of C(0)) as the initial condition to solve for the constant.
This is a direct application of particular antiderivatives.
These problems give you a fixed amount of material (fencing, wall length) and ask you to find dimensions that maximise or minimise area.
General approach:
Identify the quantity to optimise (usually area) and write it as a function of two variables.
Use the constraint (total fencing available) to eliminate one variable.
You now have a single-variable function. Differentiate, set to zero, solve.
Verify with the second derivative test.
Example: maximise the area of a rectangular field with P metres of fencing.
A = l × w, subject to 2l + 2w = P.
From the constraint: w = (P - 2l)/2.
Substitute: A(l) = l(P - 2l)/2 = Pl/2 - l^2.
A'(l) = P/2 - 2l = 0 gives l = P/4. Then w = P/4.
The optimal rectangle is a square, with side length P/4.
Variation: if one side is along an existing wall (only three sides need fencing), the constraint changes to l + 2w = P, and the optimal dimensions are different.
R(x) = x \cdot p(x)P(x) = R(x) - C(x)Profit is maximised where:
P'(x) = R'(x) - C'(x) = 0 \quad \Longrightarrow \quad R'(x) = C'(x)In words: profit is maximised when marginal revenue equals marginal cost.
Area of a rectangle:
A = l \times wPerimeter constraint (all four sides fenced):
2l + 2w = PPerimeter constraint (one side along a wall):
l + 2w = PEvery business that sets a price or chooses a production level is solving an optimisation problem, whether they use calculus explicitly or not. Airlines adjust ticket prices to maximise revenue per flight (a direct application of price-demand optimisation). Manufacturers decide batch sizes based on marginal cost analysis. Farmers and builders solve fencing and area problems when laying out fields or floor plans with a fixed budget.
Confusing the quantity that maximises revenue with the quantity that maximises profit. They are usually different. Revenue ignores costs; profit subtracts them.
Forgetting to define the variable clearly. In the car rental problem, x is not the rate; it is the number of dollar increases. The rate is 30 + x. Getting this wrong cascades through the entire solution.
Setting up the constraint equation incorrectly in fencing problems. Count the sides that actually need fencing. If one side is along a wall, that side costs no fencing.
Neglecting to verify with the second derivative test. Finding where the derivative is zero is not enough; you need to confirm whether the critical point is a maximum or minimum.
⚠️ The smartphone-style problem (given p and C(x), find max revenue and max profit) is a staple. Expect to see one or two of these on any exam.
⚠️ "At what price should phones be sold to maximise revenue?" requires you to find x first, then substitute back into the price-demand equation. Do not confuse the optimal quantity with the optimal price.
⚠️ The fencing/area problem is a common constrained optimisation question. Know the two variants: all sides fenced vs. one side along a wall.
⚠️ Profit is maximised where marginal revenue equals marginal cost (R'(x) = C'(x)). This is a frequently tested conceptual point.
Fill in the blank: R(x) = x · ______. (p, the price per unit)
True or false: Profit is maximised at the same quantity as revenue. (False, usually different)
Fill in the blank: Profit is maximised where marginal revenue equals ______. (marginal cost)
True or false: In a fencing problem with one side along a wall, the constraint is 2l + 2w = P. (False; it is l + 2w = P)
Fill in the blank: P(x) = R(x) - ______. (C(x))
Q: A company's price-demand equation is p = 500 - 0.4x and its cost function is C(x) = 20,000 + 20x. How many units maximise profit?
A: P(x) = R(x) - C(x) = 500x - 0.4x^2 - 20,000 - 20x = -0.4x^2 + 480x - 20,000. P'(x) = -0.8x + 480 = 0 gives x = 600 units.
Q: A car rental agency rents 200 cars/day at $30/day. For each $1 increase, 5 fewer cars are rented. What daily rate maximises income?
A: Let x = number of $1 increases. I(x) = (30 + x)(200 - 5x). I'(x) = 50 - 10x = 0 gives x = 5. Optimal rate = $35/day.
Q: You have 120 metres of fencing and one side of the rectangle is along a river (no fence needed). What dimensions maximise the enclosed area?
A: Constraint: l + 2w = 120, so l = 120 - 2w. A(w) = w(120 - 2w) = 120w - 2w^2. A'(w) = 120 - 4w = 0 gives w = 30. Then l = 60. Maximum area = 1,800 m^2.
Q: If C'(x) = 10x + 200 and fixed costs are $5,000, what is C(x)?
A: Integrate: C(x) = 5x^2 + 200x + K. C(0) = 5,000, so K = 5,000. Answer: C(x) = 5x^2 + 200x + 5,000.
Optimisation depends directly on critical points and the second derivative test (doc 1). The cost-function-from-marginal-cost problems use integration (doc 2). These three topics form a chain: differentiation rules let you find critical points, critical points let you classify extrema, and integration lets you reconstruct total functions from marginal ones. Together they cover the core toolkit of applied business calculus.
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