Difficulty: Intermediate to Advanced | Prerequisites: All earlier sections of Chapter 3 (extrema, MVT, derivatives and graphs). For antiderivatives, you need solid fluency with differentiation rules in reverse.
Optimisation is the pay-off for all the max/min theory: real-world problems where you build a function from a word problem and find its best value. Newton's method is a numerical tool that uses tangent lines to zero in on roots of equations that cannot be solved algebraically. Antiderivatives reverse the differentiation process and are the bridge to integral calculus in Chapter 4. Together these three topics show that differentiation is not just an abstract operation; it solves design problems, finds approximate solutions, and undoes itself.
Optimisation: translate the word problem into a function of one variable, find its critical numbers, and confirm you have a max or min. Newton's method: guess a root, draw the tangent, take its x-intercept as the next guess, repeat. Antiderivatives: the reverse of differentiation, always with a "+ C" because the derivative of a constant is zero; initial conditions pin down C.
Optimisation problem
A word problem asking you to maximise or minimise a quantity subject to one or more constraints. Think of it as the applied version of finding absolute extrema.
Newton's method
An iterative algorithm for approximating roots of f(x) = 0. Starting from an initial guess x_1, each subsequent approximation is x_{n+1} = x_n - f(x_n) / f'(x_n). In simple terms, you follow the tangent line to the x-axis, then repeat.
Antiderivative
A function F such that F'(x) = f(x) on an interval I. In simple terms, it is the function whose derivative gives you back f.
General antiderivative
F(x) + C, where F is any one antiderivative and C is an arbitrary constant. The "+ C" accounts for the fact that infinitely many functions share the same derivative.
Differential equation
An equation involving derivatives of an unknown function. Example: f'(x) = 2x.
Initial value problem (IVP)
A differential equation paired with an initial condition such as f(c) = a. The initial condition pins down the particular value of C.
Position, velocity, acceleration
If s(t) is position, then v(t) = s'(t) is velocity and a(t) = v'(t) = s''(t) is acceleration. Going backwards: the antiderivative of acceleration gives velocity; the antiderivative of velocity gives position.
Read the problem in full. Identify what is given and what is to be maximised or minimised.
Draw a diagram and label it.
Introduce notation: assign a variable to the quantity you want to optimise and to other unknowns.
Write an equation for the quantity to be optimised.
Use the constraint(s) to eliminate variables, reducing to a function of one variable.
Find the absolute max or min using the closed interval method or the first/second derivative test.
A farmer has 300 ft of fencing and wants a rectangular area divided into two pens by a fence parallel to one side.
Let x = width, y = length. Total fencing: 2x + 3y = 300, so y = (300 - 2x) / 3.
Area A = xy = x(300 - 2x) / 3 = (300x - 2x^2) / 3.
A'(x) = (300 - 4x) / 3 = 0 gives x = 75. y = (300 - 150) / 3 = 50.
A''(x) = -4/3 < 0, so x = 75 gives a maximum.
Maximum area = 75 x 50 = 3750 sq ft.
Square base (side x), height h, volume = 5 cubic feet. Glass sides cost $3/sq ft, plastic base costs $2/sq ft, open top.
Constraint: x^2 * h = 5, so h = 5/x^2.
Cost = 2x^2 (base) + 3 * 4xh (four glass sides) = 2x^2 + 60/x.
C'(x) = 4x - 60/x^2 = 0 gives x^3 = 15, so x = 15^(1/3).
C''(x) = 4 + 120/x^3 > 0, confirming a minimum.
A window with a rectangle below and a semicircle on top, total outside perimeter 46 ft.
Let r = radius of the semicircle, h = height of the rectangle. Width of rectangle = 2r.
Perimeter: 2h + 2r + pir = 46, so h = (46 - 2r - pir) / 2.
Area = 2rh + (1/2)pi*r^2. Substitute h and differentiate to find the optimal r.
Choose an initial guess x_1 near the root you want.
Compute x_{n+1} = x_n - f(x_n) / f'(x_n).
Repeat until the desired accuracy is reached.
The convergence is typically quadratic: the number of correct digits roughly doubles with each iteration.
f'(x) = 6x^5 + 3.
x_2 = 0 - f(0)/f'(0) = 0 - 2/3 = -2/3.
x_3 = -2/3 - f(-2/3) / f'(-2/3). Compute f(-2/3) = (64/729) + (-2) + 2 = 64/729. f'(-2/3) = 6(-32/243) + 3 = -192/243 + 3 = -192/243 + 729/243 = 537/243. x_3 = -2/3 - (64/729) / (537/243).
Horizontal tangent. If f'(x_n) = 0, the formula divides by zero and the method breaks.
Cycling. The iterates can bounce back and forth without converging.
Wrong root. The method may converge to a different root from the one you intended.
All three issues can usually be resolved by choosing a better initial guess x_1.
To approximate the cube root of 29, solve x^3 - 29 = 0. Let f(x) = x^3 - 29, f'(x) = 3x^2. Start with x_1 = 3 (since 3^3 = 27 is close). x_2 = 3 - (27 - 29)/(27) = 3 + 2/27 = 83/27.
To approximate a critical number of g(x), apply Newton's method to g'(x) = 0 instead. Set f(x) = g'(x) and use f'(x) = g''(x) in the iteration formula.
F is an antiderivative of f on interval I if F'(x) = f(x) for all x in I. By the MVT corollary, any two antiderivatives of the same function differ by a constant, so the most general antiderivative is F(x) + C.
Antiderivative of x^n (n != -1) is x^(n+1) / (n+1) + C.
Antiderivative of cos(x) is sin(x) + C.
Antiderivative of sin(x) is -cos(x) + C.
Antiderivative of sec^2(x) is tan(x) + C.
Antiderivative of 1/x^2 = x^(-2) is -1/x + C.
Sums and constant multiples carry through: the antiderivative of af(x) + bg(x) is aF(x) + bG(x) + C.
Many functions need algebraic simplification first.
f(x) = sqrt(x)(6 + 7x) = 6x^(1/2) + 7x^(3/2). Antiderivative: 6 * (2/3)x^(3/2) + 7 * (2/5)x^(5/2) + C = 4x^(3/2) + (14/5)x^(5/2) + C.
h(t) = (3 - t + t^2) / sqrt(t) = 3t^(-1/2) - t^(1/2) + t^(3/2). Antiderivative: 6t^(1/2) - (2/3)t^(3/2) + (2/5)t^(5/2) + C.
Given f'(x) and f(c) = a, find f(x):
Antidifferentiate f'(x) to get F(x) + C.
Plug in the initial condition: F(c) + C = a. Solve for C.
Write the particular solution.
Example: f'(x) = 1 + 3sqrt(x), f(4) = 25. F(x) = x + 2x^(3/2) + C. F(4) = 4 + 2(8) + C = 20 + C = 25, so C = 5. f(x) = x + 2x^(3/2) + 5.
Example: v(t) = 7 - 10t, s(0) = 1. s(t) = 7t - 5t^2 + C. s(0) = C = 1. So s(t) = 7t - 5t^2 + 1.
Example: a(t) = t^2 - 4t + 6, s(1) = 20, s(0) = 0. First find v(t) = t^3/3 - 2t^2 + 6t + C_1. Then s(t) = t^4/12 - (2/3)t^3 + 3t^2 + C_1*t + C_2. Use s(0) = 0 to get C_2 = 0. Use s(1) = 20 to find C_1.
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \quad \text{(Newton's Method)}\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)\int \cos x\,dx = \sin x + C \qquad \int \sin x\,dx = -\cos x + C\int \sec^2 x\,dx = \tan x + Cs(t) = \int v(t)\,dt, \qquad v(t) = \int a(t)\,dtStudents forget that optimisation requires confirming the critical number is a max or min (not just finding where the derivative is zero). Always justify with the first derivative test, second derivative test, or closed interval method.
Students try to use Newton's method without checking whether f'(x_n) is close to zero. A near-zero derivative means the tangent is nearly horizontal and the next guess can shoot off to a completely wrong region.
Students omit the "+ C" in antiderivatives. Every indefinite antiderivative has it. Losing C means losing the family of solutions.
In initial value problems, students sometimes find C but then forget to write the final answer with C substituted in.
Optimisation word problems are high-value exam questions. The setup (drawing the diagram, writing the equations) is where most marks are lost.
Newton's method problems typically ask you to compute x_2 or x_3, not to simplify. Show the formula and substitute.
Antiderivative problems range from routine (power rule) to tricky (rewrite the function first, then antidifferentiate). Expect both types.
IVPs with two conditions (second-order, e.g. given a(t), s(0), s(1)) require two rounds of antidifferentiation and solving for two constants.
True or false: In an optimisation problem, you always need a constraint equation.
Answer: True (when the objective function has more than one variable, a constraint reduces it to one variable).
Fill in the blank: Newton's method iteration formula is x_{n+1} = x_n - ______.
Answer: f(x_n) / f'(x_n).
True or false: The antiderivative of cos(x) is -sin(x) + C.
Answer: False. The antiderivative of cos(x) is sin(x) + C.
Fill in the blank: If a(t) is acceleration, then the antiderivative of a(t) gives ______.
Answer: velocity, v(t).
Q: The sum of two positive numbers is 16. What is the smallest possible value of the sum of their squares?
A: Let the numbers be x and 16 - x. Minimise f(x) = x^2 + (16 - x)^2 = 2x^2 - 32x + 256. f'(x) = 4x - 32 = 0 gives x = 8. f''(x) = 4 > 0, confirming a minimum. f(8) = 64 + 64 = 128.
Q: Use Newton's method to approximate the positive solution to x^2 = sin(x). Take x_1 = pi/3. Find x_2.
A: Let f(x) = x^2 - sin(x). f'(x) = 2x - cos(x). x_2 = pi/3 - f(pi/3) / f'(pi/3). f(pi/3) = pi^2/9 - sin(pi/3) = pi^2/9 - sqrt(3)/2. f'(pi/3) = 2pi/3 - cos(pi/3) = 2pi/3 - 1/2. Substitute and simplify (no need to reduce to a decimal).
Q: Find the most general antiderivative of g(x) = 8x - 3sec^2(x).
A: Antiderivative of 8x is 4x^2. Antiderivative of 3sec^2(x) is 3tan(x). So the answer is 4x^2 - 3tan(x) + C.
Q: Solve the IVP: dy/dx = 6/x^6 + 9x^3, y(1) = 14.
A: dy/dx = 6x^(-6) + 9x^3. Antiderivative: y = 6 * x^(-5)/(-5) + 9 * x^4/4 + C = -6/(5x^5) + (9/4)x^4 + C. y(1) = -6/5 + 9/4 + C = 14. Solve for C: C = 14 + 6/5 - 9/4 = 14 + 24/20 - 45/20 = 14 - 21/20 = 259/20.
Q: Find the position of a particle given a(t) = t^2 - 4t + 6, s(0) = 0, s(1) = 20.
A: v(t) = t^3/3 - 2t^2 + 6t + C_1. s(t) = t^4/12 - (2/3)t^3 + 3t^2 + C_1*t + C_2. s(0) = C_2 = 0. s(1) = 1/12 - 2/3 + 3 + C_1 = 20. Solve: C_1 = 20 - 1/12 + 2/3 - 3 = 20 - 1/12 + 8/12 - 36/12 = 20 - 29/12 = 211/12.
Optimisation problems are the applied pay-off of the absolute extrema theory from Section 1. Newton's method connects to linearisation from Chapter 2 (the tangent line approximation). Antiderivatives are the conceptual reverse of differentiation from Chapter 2, and they lead directly into the definite integral and the Fundamental Theorem of Calculus in Chapter 4.
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