Difficulty: Intermediate to Advanced | Prerequisites: Differentiation, critical points, second derivative test, implicit differentiation
This is where calculus meets the real world in MAT 210. Optimisation problems ask you to maximise or minimise a quantity (area, revenue, cost) subject to a constraint. Elasticity of demand measures how sensitive quantity is to price. Related rates use implicit differentiation with respect to time to connect changing quantities. These topics draw on everything covered earlier in the course.
Optimisation: write the objective function, use the constraint to reduce it to one variable, differentiate, set equal to zero, and verify with the second derivative. Elasticity E < 1 means inelastic, E > 1 means elastic. Related rates: relate the quantities with an equation, differentiate with respect to time, plug in known values, and solve for the unknown rate. Absolute extrema on a closed interval: evaluate f at critical points and endpoints, then compare.
Optimisation (applied max/min)
The process of finding the maximum or minimum value of a quantity subject to a constraint. In simple terms, it is finding the best possible value (biggest area, highest profit, lowest cost) given a limitation.
Objective function
The function you want to maximise or minimise (for example, area, revenue, or cost). Think of it as the thing you are trying to make as large or small as possible.
Constraint
A condition that limits the variables (for example, a fixed perimeter or a fixed budget). It lets you eliminate one variable so the objective function depends on just one.
Elasticity of demand (E)
A measure of how responsive quantity demanded is to a change in price. E = |p/q · dq/dp|. When E < 1, demand is inelastic (quantity changes less than price). When E > 1, demand is elastic (quantity changes more than price). When E = 1, demand is unit elastic.
Related rates
Problems where two or more quantities change with time, linked by an equation. You differentiate the equation with respect to t and solve for the unknown rate. Think of it as applying implicit differentiation where the independent variable is time.
Absolute extrema (on a closed interval)
The largest and smallest values a function takes on a closed interval [a, b]. Found by evaluating f at all critical points inside the interval and at both endpoints, then comparing.
Identify the objective function (what to maximise or minimise) and the constraint equation.
Use the constraint to express the objective in terms of one variable.
Differentiate, set equal to zero, and solve for the critical point.
Confirm with the second derivative test.
Example (track problem): A running track has a rectangular field with two semicircular ends. Perimeter = 2L + pi · W = 400. Area = L · W. Substituting L = (400 - pi · W) / 2 = 200 - pi · W / 2, then A(W) = 200W - pi · W^2 / 2. A'(W) = 200 - pi · W. Setting A'(W) = 0 gives W = 200/pi (roughly 63.66 m). Then L = 100 m. A''(W) = -pi < 0 confirms a maximum.
Revenue R(p) is a function of price. Find R'(p), set it to zero, and solve.
Example: R(p) = 750p - 10p^2. R'(p) = 750 - 20p. Setting to zero: p = 37.5.
From the demand equation p = f(q), solve for q as a function of p.
Compute dq/dp.
Apply the formula E = |p/q · dq/dp|.
Interpret: E < 1 is inelastic, E > 1 is elastic.
Example: p = 90 - 10q, so q = 9 - 0.1p. dq/dp = -0.1. At p = 30: q = 6. E = |30/6 · (-0.1)| = |5 · (-0.1)| = 0.5. Since 0.5 < 1, demand is inelastic.
Write the equation relating the quantities.
Differentiate both sides with respect to time t.
Substitute known values and rates, then solve for the unknown rate.
Example (balloon): V = (4/3) pi r^3. dV/dt = 4 pi r^2 (dr/dt). Given dV/dt = 100 and r = 50: 100 = 4 pi (2500)(dr/dt), so dr/dt = 1/(100 pi) cm/s.
Example (ladder): x^2 + y^2 = 13^2 = 169. Differentiating: 2x(dx/dt) + 2y(dy/dt) = 0. Given dx/dt = 5 and x = 12: y = 5. Then 2(12)(5) + 2(5)(dy/dt) = 0, so 120 + 10(dy/dt) = 0, giving dy/dt = -12 ft/s. The top slides down at 12 ft/s.
Find all critical points of f in (a, b).
Evaluate f at each critical point and at x = a and x = b.
The largest value is the absolute maximum; the smallest is the absolute minimum.
Example: f(x) = x^3 - 6x^2 + 9x + 1 on [0, 5]. f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3). Critical points x = 1 and x = 3. f(0) = 1, f(1) = 5, f(3) = 1, f(5) = 21. Absolute max = 21 (at x = 5), absolute min = 1 (at x = 0 and x = 3).
E = \left| \frac{p}{q} \cdot \frac{dq}{dp} \right|V = \frac{4}{3}\pi r^3 \quad \Rightarrow \quad \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}x^2 + y^2 = L^2 \quad \Rightarrow \quad 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0R'(p) = 0 \quad \text{gives the revenue-maximising price}Students often skip the second derivative test in optimisation, leaving the answer unverified. The exam awards marks for confirming that the critical point is a maximum (or minimum).
Students confuse elastic and inelastic. E < 1 means inelastic (demand is relatively unresponsive to price changes). E > 1 means elastic. A common error is reversing the two.
In related rates, students sometimes forget to differentiate with respect to time, treating the equation as static. Every variable that changes with time gets a d/dt.
Students forget to check the endpoints when finding absolute extrema on a closed interval. The maximum or minimum can occur at an endpoint, not just at a critical point.
⚠️ Optimisation (the track problem or a similar geometry question) is a high-mark long-answer question. Show every step: constraint, substitution, derivative, critical point, second derivative test, and final answer with units.
⚠️ Absolute extrema on a closed interval is an 8-mark question. Evaluate f at all critical points and endpoints. Do not stop after finding the critical points.
⚠️ Related rates (ladder or balloon) is a long-answer staple. Write the relationship equation, differentiate with respect to t, substitute known values, solve, and include units in the final answer. The negative sign in the ladder problem indicates the direction (sliding down).
⚠️ Elasticity of demand is tested as both a multiple-choice conceptual question (which range is inelastic?) and a short-answer calculation. Know the formula and the interpretation.
⚠️ Revenue maximisation (setting R'(p) = 0) is a quick multiple-choice question. Be able to differentiate and solve in under a minute.
True or False: E < 1 means demand is elastic. (False. E < 1 means inelastic.)
Fill in the blank: To find the absolute maximum of f(x) on [a, b], evaluate f at all ______ and both ______. (critical points; endpoints)
True or False: In a related-rates problem, you differentiate the equation with respect to x. (False. You differentiate with respect to time, t.)
Fill in the blank: The revenue-maximising price is found by setting ______ = 0. (R'(p))
True or False: If A''(W) < 0 at a critical point in an optimisation problem, that critical point is a maximum. (True.)
Q: A revenue function is R(p) = 750p - 10p^2. At what price is revenue maximised?
A: R'(p) = 750 - 20p = 0, so p = 37.5.
Q: For what value range of E is demand considered inelastic?
A: 0 <= E < 1.
Q: Determine the elasticity of demand when p = 30, given p = 90 - 10q.
A: q = 9 - 0.1p = 6. dq/dp = -0.1. E = |30/6 · (-0.1)| = 0.5. Demand is inelastic.
Q: A 13-foot ladder leans against a wall. The base slides away at 5 ft/s. When the base is 12 ft from the wall, how fast is the top sliding down?
A: x^2 + y^2 = 169. At x = 12, y = 5. Differentiating: 2(12)(5) + 2(5)(dy/dt) = 0. dy/dt = -12 ft/s.
Q: Find the absolute max and min of f(x) = x^3 - 6x^2 + 9x + 1 on [0, 5].
A: Critical points x = 1, x = 3. f(0) = 1, f(1) = 5, f(3) = 1, f(5) = 21. Absolute max = 21, absolute min = 1.
Q: A running track with perimeter 400 m has a rectangular field with semicircular ends. Find L and W that maximise the field area.
A: Constraint: 2L + pi · W = 400. A(W) = 200W - pi · W^2 / 2. A'(W) = 200 - pi · W = 0 gives W = 200/pi. L = 100 m.
Optimisation problems combine the constraint-substitution algebra of the problem setup with the critical-point and second-derivative techniques from the derivatives notes. Related rates depend on implicit differentiation. Elasticity connects to revenue maximisation: when demand is inelastic, raising the price increases revenue; when elastic, lowering the price increases revenue. Absolute extrema on closed intervals use the same critical-point machinery but add the endpoint check.
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