Difficulty: Intermediate to Advanced | Prerequisites: Ch. 4.5 (absolute extrema on closed intervals), derivative rules, critical numbers
Big Picture
Chapter 4.6 introduces two things. First, the second derivative test: a way to classify critical points as local maxima or minima without checking a sign chart. Second, applied optimization: real-world word problems where you build your own function, find its critical number, and confirm it gives the best outcome. This is where calculus meets business, because most of the exam problems involve maximising revenue or profit, or minimising cost. You need the closed interval method from Ch. 4.5 and all your derivative skills.
The second derivative test says: at a critical number c, if f''(c) < 0 it is a local max, if f''(c) > 0 it is a local min, and if f''(c) = 0 the test is inconclusive. For applied optimization, translate the word problem into a function of one variable, find the critical number, and use the second derivative test (or the closed interval method) to confirm it is the extremum you want.
Second derivative test
A method for classifying critical points. Evaluate f''(c) at a critical number c: negative means local max (concave down), positive means local min (concave up), zero means inconclusive.
Think of it as: the curve's curvature at the peak or valley tells you which one it is.
Local maximum (relative maximum)
A point where f(c) is greater than or equal to all nearby values. Not necessarily the highest point on the entire domain.
In simple terms, this means: a hilltop in the immediate neighbourhood.
Local minimum (relative minimum)
A point where f(c) is less than or equal to all nearby values.
In simple terms, this means: a valley floor in the immediate neighbourhood.
Concavity
A function is concave up where f''(x) > 0 (curves like a bowl) and concave down where f''(x) < 0 (curves like a dome).
Think of it as: concave up holds water, concave down spills it.
Revenue function R(x)
R(x) = x * p, where x is the number of units sold and p is the price per unit. If p depends on x (a demand equation), then R(x) = x * p(x).
Profit function P(x)
P(x) = R(x) - C(x). Revenue minus cost. Maximising profit is the most common business optimization question.
Constraint equation
A relationship between variables that lets you reduce a two-variable problem to one variable. For example, if x + y = 13, you can write y = 13 - x and substitute.
The Second Derivative Test
Find f'(x) and set it to zero to get critical numbers.
Compute f''(x).
At each critical number c: if f''(c) > 0, then (c, f(c)) is a local minimum. If f''(c) < 0, then (c, f(c)) is a local maximum. If f''(c) = 0, the test is inconclusive (try the first derivative test instead).
Worked Example: f(x) = 3x^3 - 2x^2 - 3
f'(x) = 9x^2 - 4x = x(9x - 4). Critical numbers: x = 0 and x = 4/9.
f''(x) = 18x - 4. f''(0) = -4 < 0, so (0, -3) is a local maximum. f''(4/9) = 18(4/9) - 4 = 4 > 0, so (4/9, -3.1) is a local minimum.
Worked Example: f(x) = x^4 - 3
f'(x) = 4x^3. Critical number: x = 0. f''(x) = 12x^2. f''(0) = 0, so the test is inconclusive.
When there is only one critical number on an open interval
If a function has exactly one critical number on an open interval and it is a local max (or min) by the second derivative test, then it is also the absolute max (or min) on that interval. This principle appears in Problems 9, 10, 11, and 12.
Worked Example: f(x) = 20 - 4x - 243/x^4 on (0, infinity)
f'(x) = -4 + 972/x^5. Set to zero: 972/x^5 = 4, so x^5 = 243, giving x = 3.
f''(x) = -4860/x^6. f''(3) = -4860/729 < 0, so (3, 5) is a local max.
Since x = 3 is the only critical number on (0, infinity), it is the absolute maximum. The absolute max value is 5.
Applied Optimization: The Problem-Solving Framework
Every word problem follows the same pattern:
Read the problem. Identify the quantity to maximise or minimise.
Assign variables and write the objective function (the thing you want to optimise).
If there are two variables, use the constraint to eliminate one.
Take the derivative, set it to zero, and solve for the critical number.
Verify with the second derivative test (or by checking endpoints if the domain is closed).
Answer the question that was asked (price, number of units, dimensions, profit, etc.).
Type 1: Max product with a sum constraint
Problem 14: x + y = 13, maximise xy. Substitute y = 13 - x. Then P(x) = x(13 - x) = 13x - x^2. P'(x) = 13 - 2x = 0 gives x = 6.5, y = 6.5.
Type 2: Revenue and profit maximisation
Problem 15: p = 500 - 0.4x and C(x) = 20,000 + 20x.
Revenue: R(x) = xp = 500x - 0.4x^2. R'(x) = 500 - 0.8x = 0 gives x = 625 phones. Max revenue price: p = 500 - 0.4(625) = $250. Max revenue: R(625) = $156,250.
Profit: P(x) = R(x) - C(x) = 480x - 0.4x^2 - 20,000. P'(x) = 480 - 0.8x = 0 gives x = 600 phones. Max profit: P(600) = $124,000.
Type 3: Rate increase / decrease problems
Problem 16: 200 cars at $30/day; each $1 increase loses 5 cars. Let x = number of $1 increases. Cars rented: 200 - 5x. Rate: 30 + x. Income: I(x) = (30 + x)(200 - 5x) = 6000 + 50x - 5x^2. I'(x) = 50 - 10x = 0 gives x = 5. Rate = $35. Max income = $6,125.
Problem 17: 1,600 cups at $2.40; each $0.05 decrease adds 50 cups. Let x = number of $0.05 decreases. Revenue: R(x) = (2.40 - 0.05x)(1600 + 50x). Expanding and differentiating gives the optimal price of $2.00.
Type 4: Fencing and area problems
Problem 21 (river fencing): 1800 ft of fencing, three sides (one side is the river). If the two widths are w and the length is l, then 2w + l = 1800, so l = 1800 - 2w. Area: A(w) = w(1800 - 2w) = 1800w - 2w^2. A'(w) = 1800 - 4w = 0 gives w = 450. Then l = 900. Dimensions: 450 x 900 ft.
Problem 24 (playground divided in two): 480 ft of fencing, rectangle split by one internal fence parallel to the width. Constraint: 2l + 3w = 480. Area: A(w) = w(240 - 3w/2). Differentiate to find w = 80, l = 120. Max area = 9,600 ft^2.
Type 5: Minimum fencing / cost for a given area
Problem 18: Area = 5,000 sq yards, fence three sides. Let width = w, length = l. Then wl = 5000, so l = 5000/w. Fencing = 2w + l = 2w + 5000/w. Differentiate and solve: w = 50, l = 100. Min fencing = 200 yards.
Problem 19: Area = 450 sq ft, three sides cost $3/ft, fourth side costs $9/ft. Cost = 3(2w + l) + 9l = 6w + 12l. With wl = 450, substitute l = 450/w. Differentiate to get w = 30, l = 15. Dimensions: 15 ft by 30 ft.
Type 6: Pricing to maximise profit (unit cost given)
Problem 22: Cost $5/unit, demand = 100(20 - x), sell at $x. Profit = (x - 5)(100)(20 - x) = 100(x - 5)(20 - x). Differentiate: 100(25 - 2x) = 0 gives x = 12.50.
Problem 20: Flashlights at $6, 3,000/month; each $1 increase loses 1,000 units; cost $4/unit. Let x = price increase. Price = 6 + x, units = 3000 - 1000x. Profit = (6 + x - 4)(3000 - 1000x) = (2 + x)(3000 - 1000x). Differentiate to get optimal price = $6.50.
Problem 23: 60 houses at $50,000 profit each, each extra house drops profit by $500/house. Let x = extra houses. Total profit = (60 + x)(50000 - 500x). Differentiate to find x = 20, so build 80 houses.
Revenue: R(x) = x * p(x), where p(x) is the demand (price) equation
Profit: P(x) = R(x) - C(x)
Second derivative test: f''(c) > 0 means local min; f''(c) < 0 means local max; f''(c) = 0 means inconclusive
Rectangle area: A = l * w
Rectangle perimeter (three-sided, one open side): P = 2w + l
Rectangle perimeter (divided in two by internal fence): P = 2l + 3w
Students often forget that the second derivative test can be inconclusive. When f''(c) = 0, the test tells you nothing, and you need to fall back on the first derivative test or other reasoning.
Confusing revenue with profit. Revenue is what comes in (price times quantity). Profit is what remains after subtracting costs. Maximising revenue and maximising profit usually give different answers.
In rate-increase problems, students sometimes set up the variable as the new price rather than the number of increases. Either works, but mixing them up mid-problem causes errors.
Forgetting to answer the specific question asked. If the problem says "at what price," you need to go back and find the price after you have found x.
Applied optimization problems make up a large portion of Exam 4. Expect at least two or three word problems.
The exam typically includes one revenue/profit problem and one geometry (fencing/area) problem.
When there is one critical number on an open interval, state clearly that this makes it an absolute extremum. The exam often tests whether you understand this principle.
Always show the second derivative test work. Simply finding the critical number without confirming it is a max or min will cost marks.
True or False: If f''(c) > 0, then c is a local maximum. (False, it is a local minimum.)
Fill in the blank: Profit equals ______ minus ______. (Revenue, Cost)
True or False: If there is only one critical number on an open interval and the second derivative test shows it is a local max, it is also the absolute max. (True)
Fill in the blank: In a fencing problem where one side borders a river, the constraint equation has ______ sides of fencing, not four. (three)
True or False: Maximising revenue and maximising profit always give the same number of units. (False)
Q: A car rental agency rents 210 cars per day at $30/day. For each $1 increase, 5 fewer cars are rented. What rate maximises revenue, and what is the maximum income?
A: Let x = number of $1 increases. Revenue R(x) = (30 + x)(210 - 5x) = 6300 + 60x - 5x^2. R'(x) = 60 - 10x = 0, so x = 6. Rate = $36. Max income = (36)(180) = $6,480.
Q: A farmer has 1800 ft of fencing and wants to fence a rectangular field along a river (no fence on the river side). What dimensions give the largest area?
A: Let w = width. Then length = 1800 - 2w. Area A(w) = w(1800 - 2w). A'(w) = 1800 - 4w = 0, so w = 450. Length = 900. Dimensions: 450 ft by 900 ft.
Q: Use the second derivative test to find the local extrema of f(x) = 8x^4 - 4x^2 on (0, infinity).
A: f'(x) = 32x^3 - 8x = 8x(4x^2 - 1). On (0, infinity), the critical number is x = 1/2. f''(x) = 96x^2 - 8. f''(1/2) = 96(1/4) - 8 = 16 > 0, so (1/2, -1/2) is a local minimum. Since it is the only critical number on (0, infinity), -1/2 is the absolute minimum.
Q: Find two numbers whose sum is 13 and whose product is maximum.
A: Let x and y be the numbers. y = 13 - x. Product P(x) = x(13 - x) = 13x - x^2. P'(x) = 13 - 2x = 0, so x = 6.5, y = 6.5.
The second derivative test builds on Ch. 4.5 (critical numbers and absolute extrema on closed intervals) and extends it to open intervals. The optimization word problems here use the same derivative-based approach that reappears in Ch. 5.1 when you integrate marginal cost or marginal revenue functions to recover the original cost or revenue function. The revenue and cost relationships (R = xp, P = R - C) will also appear in definite integral applications in Ch. 5.5.
second derivative test, local extrema, relative max, relative min, concavity, applied optimization, revenue maximization, profit maximization, fencing problems, area optimization, constraint equation, demand equation, MAC 2233, FIU, Ch 4.6, calculus for business, word problems