Difficulty: Intermediate | Prerequisites: Differentiation Rules, Applied Derivatives notes
This unit pulls together everything you have learned about derivatives and puts it to work on the questions businesses care about most: when is a function increasing or decreasing, where are the peaks and valleys, should we raise the price or lower it, and what dimensions minimise material? Curve sketching gives you the visual intuition; optimisation gives you the numerical answers. Elasticity connects price changes to demand changes. If you can take derivatives and set them to zero, you are ready.
Critical points (where y' = 0 or changes sign) give you local maxima and minima. Inflection points (where y'' changes sign) mark shifts in concavity. Elasticity of demand measures how sensitive quantity demanded is to price. Optimisation problems reduce to writing a function, differentiating, setting the derivative to zero, and solving.
Critical point
A value of x where f'(x) = 0 or f'(x) is undefined. These are candidates for local maxima and minima. Think of it as: "Where the function pauses before changing direction."
Local maximum / local minimum
A point where the function reaches a peak (max) or a valley (min) relative to its neighbours. Found at critical points where f' changes sign.
Inflection point
A point where the concavity changes, from concave up to concave down or vice versa. Found where f'' changes sign.
Concave up / concave down
Concave up means f'' > 0 (the curve bends upward, like a cup). Concave down means f'' < 0 (the curve bends downward, like a cap).
Absolute maximum / absolute minimum
The largest or smallest value of f(x) on a closed interval [a, b]. Found by evaluating f at every critical point in [a, b] and at both endpoints, then comparing.
Elasticity of demand, E(p)
Measures the responsiveness of quantity demanded to a change in price. Formula: E(p) = -[p · q'(p)] / q(p). If E > 1, demand is elastic (sensitive to price). If E < 1, demand is inelastic. If E = 1, unit elastic.
Revenue maximisation
Revenue = price x quantity = x · p(x). To maximise, take the derivative of R(x), set it to zero, and solve.
y' > 0: the function is increasing.
y' < 0: the function is decreasing.
y'' > 0: the function is concave up (cup shape).
y'' < 0: the function is concave down (cap shape).
Set y' = 0 (or identify where y' changes sign).
These are your candidates for local max and local min.
Example from the course: given y' > 0 on (2, 4) and y' < 0 on (-∞, 2) and (4, ∞), critical points are x = 2 and x = 4. x = 2 is a local min (y' goes from negative to positive), x = 4 is a local max (y' goes from positive to negative).
Inflection points occur where y'' changes sign.
Example: y'' > 0 on (1, 3) and x < 0; y'' < 0 on (0, 1) and x > 3. Concavity changes at x = 1 and x = 3, so those are inflection points.
Find all critical points in [a, b].
Evaluate f at each critical point and at both endpoints.
The largest value is the absolute max; the smallest is the absolute min.
Example: y = x/9 + 1/x on [1, 18]. y' = 1/9 - x⁻² = 0 gives x = 3. Evaluate: y(1) = 10/9, y(3) = 2/3, y(18) = 19/9. Absolute min at x = 3, absolute max at x = 18.
E(p) = -[p · q'(p)] / q(p)
q(p) is the demand function (quantity as a function of price).
q'(p) is its derivative with respect to price. Use the derivative of the demand function, not the derivative evaluated at a specific price.
E > 1: elastic. A 1% increase in price causes more than a 1% decrease in quantity demanded. Revenue decreases if price goes up.
E = 1: unit elastic. Revenue is at its maximum.
E < 1: inelastic. A 1% increase in price causes less than a 1% decrease in quantity demanded. Revenue increases if price goes up.
q(p) = 600 - 3p, 0 ≤ p ≤ 200. q'(p) = -3 (a constant).
At p = 150: q(150) = 600 - 450 = 150.
E(150) = -[150 · (-3)] / 150 = 450/150 = 3.
Since 3 > 1, demand is elastic at $150. A price increase would reduce revenue.
A hotel charges $1600/night when 90 rooms are rented. Each extra room over 90 reduces the price by $10.
Price as a function of rooms: P(x) = 1600 - 10(x - 90) = 2500 - 10x.
Revenue: R(x) = x · (2500 - 10x) = 2500x - 10x².
R'(x) = 2500 - 20x = 0 gives x = 125 rooms.
Same hotel, but now there is a $100 disinfection cost per room.
Profit = Revenue - Cost = (2500x - 10x²) - 100x = 2400x - 10x².
Profit'(x) = 2400 - 20x = 0 gives x = 120 rooms.
Adding cost shifts the optimal number of rooms downward. The cost reduces the point where the marginal revenue equals the marginal cost.
2400 ft² of land, rectangular plot with a fence down the middle. Minimise total fencing.
Area constraint: xy = 2400, so y = 2400/x.
Fencing function: F = 2x + 3y = 2x + 3(2400/x) = 2x + 7200/x.
F'(x) = 2 - 7200x⁻² = 0. Solve: x² = 3600, x = 60 ft. Then y = 2400/60 = 40 ft.
The three-fence side (3y) comes from the fence down the middle creating three parallel lengths.
Students confuse critical points with inflection points. Critical points are where y' = 0 (potential max/min). Inflection points are where y'' changes sign (change of concavity). They are different things.
In elasticity, students sometimes use q'(p) evaluated at the specific price instead of the general derivative. Use the general derivative (e.g. -3 for q = 600 - 3p), not q'(150).
In absolute max/min problems, students forget to check the endpoints. Even if there is a critical point inside the interval, the absolute max or min could be at an endpoint.
In optimisation word problems, students sometimes maximise when they should minimise (or vice versa). Read the question: "maximise revenue" means set R'(x) = 0; "minimise fencing" means set F'(x) = 0.
⚠️ Curve sketching questions give you sign information about y' and y'' and ask you to draw the shape. Know what each combination of signs means.
⚠️ Absolute max/min on a closed interval: evaluate at critical points AND endpoints. This is the complete method.
⚠️ Elasticity: memorise E(p) = -[p · q'(p)] / q(p). Know what E > 1, E = 1, and E < 1 mean for revenue.
⚠️ Optimisation word problems follow a template: write the objective function, use the constraint to eliminate a variable, differentiate, set to zero, solve. Practise the hotel problem and the fencing problem until the process is automatic.
⚠️ When a cost is added to a revenue maximisation problem, the optimal answer shifts. Expect a two-part question: first maximise revenue, then maximise profit.
Q: Given y' > 0 on (2, 4) and y' < 0 everywhere else, where are the critical points and what type are they?
A: Critical points at x = 2 and x = 4. x = 2 is a local min (y' goes from - to +), x = 4 is a local max (y' goes from + to -).
Q: q(p) = 600 - 3p. Is demand elastic or inelastic at p = 150?
A: q'(p) = -3. q(150) = 150. E(150) = -[150(-3)]/150 = 3. Since 3 > 1, demand is elastic.
Q: A hotel charges $1600 for 90 rooms, reducing by $10 per extra room. How many rooms maximise revenue?
A: P(x) = 2500 - 10x. R(x) = 2500x - 10x². R'(x) = 2500 - 20x = 0. x = 125 rooms.
Q: Same hotel, but each room costs $100 to disinfect. How many rooms maximise profit?
A: Profit = 2500x - 10x² - 100x = 2400x - 10x². Profit' = 2400 - 20x = 0. x = 120 rooms.
Q: Find the absolute min of y = x/9 + 1/x on [1, 18].
A: y' = 1/9 - 1/x² = 0 gives x = 3. y(1) = 10/9, y(3) = 2/3, y(18) ≈ 2.06. Absolute min is 2/3 at x = 3.
Curve sketching combines the first and second derivative tests that you learned in differentiation. Optimisation is the business payoff of the entire derivatives unit. Elasticity connects to revenue and pricing strategy. In the next unit (integration), you will work with the area under curves, which builds on understanding where functions are positive or negative.
Related Terms / Search Tags
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