Source: Lecture notes, Advanced Calculus I, University of Central Florida
Tags: derivatives, applications of derivatives, extrema, critical points, optimization, rates of change, second derivative test, calculus, UCF, Advanced Calculus I
Difficulty: Intermediate Prerequisites: Basic differentiation rules (power rule, product rule, chain rule), understanding of what a derivative represents geometrically and as a rate of change.
This material covers what you can do with derivatives once you know how to compute them. It sits at the transition from "calculate f'(x)" to "use f'(x) to answer real questions." The three applications here, finding extrema, solving optimisation problems, and interpreting rates of change, form the backbone of nearly every applied calculus problem you will see on exams and in later courses. If you are comfortable differentiating polynomials and rational functions, you are ready for this.
Derivatives let you find where a function peaks or bottoms out (extrema), solve constrained max/min problems (optimisation), and describe how quantities change over time (rates of change). The core technique across all three is the same: set the derivative equal to zero, solve for the variable, then interpret the result.
Critical point
A value of x where f'(x) = 0 or f'(x) is undefined. These are the only candidates for local maxima or minima. Think of it as: the places where the function's slope is flat or the slope doesn't exist.
Extrema (local maximum / local minimum)
The highest or lowest values a function reaches in some neighbourhood. A local max is a hilltop; a local min is the bottom of a valley. They are "local" because the function may go higher or lower elsewhere.
Second derivative test
A method for classifying critical points. If f''(c) > 0 at a critical point c, the function is concave up there, so c is a local minimum. If f''(c) < 0, the function is concave down, so c is a local maximum. If f''(c) = 0, the test is inconclusive. In simple terms, the second derivative tells you whether the curve bends upward (bowl) or downward (arch) at that point.
Optimisation (constrained optimisation)
Finding the absolute largest or smallest value of a quantity subject to one or more constraints. Think of it as: you want the best possible outcome, but the rules of the problem limit your choices.
Constraint
An equation that restricts the variables in an optimisation problem. For example, "the volume must equal V" is a constraint. You use it to eliminate a variable so you can differentiate with respect to the remaining ones.
Rate of change
The derivative of a quantity with respect to time (or another variable). Velocity is the rate of change of position; acceleration is the rate of change of velocity. In simple terms, it answers "how fast is this thing changing right now?"
To find local maxima and minima of f(x), compute f'(x) and set it equal to zero. Also check where f'(x) is undefined.
The solutions are the critical points, the only places where a local extremum can occur.
Classify each critical point using the second derivative test:
f''(c) > 0 → local minimum (concave up, bowl shape)
f''(c) < 0 → local maximum (concave down, arch shape)
f''(c) = 0 → test is inconclusive; use the first derivative test instead
Worked example:
f(x) = x³ – 6x² + 9x + 2
f'(x) = 3x² – 12x + 9
Set f'(x) = 0: 3x² – 12x + 9 = 0, which factors as 3(x – 1)(x – 3) = 0
Critical points: x = 1 and x = 3
f''(x) = 6x – 12
f''(1) = 6(1) – 12 = –6 < 0, so x = 1 is a local maximum
f''(3) = 6(3) – 12 = 6 > 0, so x = 3 is a local minimum
(Note: the original lecture source labels x = 1 as a local minimum and x = 3 as a local maximum. The second derivative values show the opposite. The working above is correct.)
The general approach:
Identify what you want to maximise or minimise (the objective function)
Write down the constraint(s)
Use the constraint to eliminate one variable from the objective function
Differentiate the objective with respect to the remaining variable(s), set equal to zero, and solve
Check that your solution satisfies the constraint and is indeed a max or min
Worked example: minimising surface area of a box with fixed volume
Given a rectangular box with dimensions x, y, z:
Constraint: xyz = V (fixed volume)
Objective: minimise A = 2xy + 2xz + 2yz (surface area)
Step-by-step:
Use the constraint to eliminate z: z = V/(xy)
Substitute into the surface area formula: A = 2xy + 2V/x + 2V/y
Take partial derivatives and set each to zero:
∂A/∂x = 2y – 2V/x² = 0, giving x = √(V/y)
∂A/∂y = 2x – 2V/y² = 0, giving y = √(V/x)
Substituting back: z = V/(xy)
By symmetry, the minimum surface area occurs when x = y = z, i.e. the box is a cube with side length ∛V
If s(t) is the position of an object at time t, then:
v(t) = s'(t) is the velocity (first derivative of position)
a(t) = v'(t) = s''(t) is the acceleration (second derivative of position)
Setting v(t) = 0 finds the moment when the object changes direction (or reaches a peak/trough in position)
Worked example: ball thrown vertically
Position: s(t) = –½gt² + v₀t, where g is gravitational acceleration and v₀ is initial velocity.
Velocity: v(t) = s'(t) = –gt + v₀
Maximum height occurs when v(t) = 0:
0 = –gt + v₀
t = v₀/g
Acceleration: a(t) = v'(t) = –g (constant, directed downward)
Formula | Meaning |
|---|---|
f'(c) = 0 or undefined | Condition for c to be a critical point |
f''(c) > 0 → local min | Second derivative test (concave up) |
f''(c) < 0 → local max | Second derivative test (concave down) |
s(t) = –½gt² + v₀t | Position of a vertically thrown object |
v(t) = –gt + v₀ | Velocity (first derivative of position) |
t = v₀/g | Time to reach maximum height |
A = 2xy + 2V/x + 2V/y | Surface area of box after eliminating z |
Optimisation with derivatives is how engineers design packaging that uses the least material for a given volume, and how economists find the production level that maximises profit. Rates-of-change problems appear constantly in physics and engineering: every time you describe how a rocket accelerates, how a drug concentration decays in the bloodstream, or how a population grows, you are working with derivatives as rates of change.
Students often assume every point where f'(x) = 0 is an extremum. It is not: f(x) = x³ has f'(0) = 0 but x = 0 is an inflection point, not a max or min. Always classify with the second derivative test or the first derivative test.
In optimisation, students sometimes forget to use the constraint to reduce the number of variables before differentiating. If you are differentiating a function of two variables without using partial derivatives or substitution, something has gone wrong.
The second derivative test is inconclusive when f''(c) = 0. Students sometimes treat this as "no extremum," but it could still be one. You need the first derivative test in that case.
With rates of change, students confuse velocity and speed. Velocity can be negative (indicating direction); speed is the absolute value of velocity.
⚠️ Critical point identification and classification (first and second derivative tests) appear on nearly every calculus exam. Know both tests and when to use each.
⚠️ Optimisation problems are a favourite for longer exam questions. Practice the full workflow: define variables, write the constraint, eliminate a variable, differentiate, solve, interpret.
⚠️ Rates-of-change problems often appear as word problems. The key step is translating the physical description into a function and its derivative.
⚠️ Watch the algebra when setting derivatives to zero, especially with rational expressions (multiply through by the denominator to clear fractions).
True or false: If f'(c) = 0, then f has a local extremum at x = c.
False. f'(c) = 0 is necessary but not sufficient. The point could be an inflection point.
Fill in the blank: If f''(c) > 0 at a critical point, then c is a local ______.
Minimum.
True or false: The velocity of a ball thrown upward is zero at its maximum height.
True. Setting v(t) = 0 gives the time of maximum height.
Fill in the blank: In a constrained optimisation problem, you use the ______ to eliminate a variable before differentiating.
Constraint.
Q: Given f(x) = x³ – 6x² + 9x + 2, find the critical points and classify each as a local max or local min using the second derivative test.
A: f'(x) = 3x² – 12x + 9 = 3(x – 1)(x – 3) = 0, so critical points are x = 1 and x = 3. f''(x) = 6x – 12. f''(1) = –6 < 0, so x = 1 is a local maximum. f''(3) = 6 > 0, so x = 3 is a local minimum.
Q: A rectangular box must have volume V. Express the surface area A in terms of x, y, and V only (eliminating z), and write the equation you would solve to find the value of x that minimises A.
A: Since z = V/(xy), substituting gives A = 2xy + 2V/x + 2V/y. To minimise with respect to x, set ∂A/∂x = 2y – 2V/x² = 0 and solve for x.
Q: A ball is thrown straight up with initial velocity v₀. Write expressions for its position, velocity, and acceleration as functions of time. At what time does it reach its maximum height?
A: s(t) = –½gt² + v₀t, v(t) = –gt + v₀, a(t) = –g. Maximum height occurs when v(t) = 0, giving t = v₀/g.
Q: What should you do when the second derivative test gives f''(c) = 0?
A: The test is inconclusive. Use the first derivative test instead: check the sign of f'(x) on either side of c. If f' changes from positive to negative, c is a local max; from negative to positive, a local min; no sign change means no extremum.
This material connects directly to curve sketching, where you combine critical points, concavity, and behaviour at infinity to draw a full graph of a function. It also lays the groundwork for related rates problems, which use the chain rule to connect rates of change of multiple linked quantities. In multivariable calculus, these same ideas extend to functions of several variables using gradients and Lagrange multipliers.
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