Difficulty: Intermediate | Prerequisites: Critical values and the First Derivative Test (see companion study notes).
When a function is continuous on a closed interval [c, d], the Extreme Value Theorem guarantees it has both an absolute maximum and an absolute minimum somewhere on that interval. To find them, you evaluate the function at every critical value inside the interval and at both endpoints, then compare. The largest value is the absolute max; the smallest is the absolute min.
Tags: absolute maximum, absolute minimum, Extreme Value Theorem, EVT, closed interval, bounded interval, endpoints, global max, global min
Absolute (global) maximum
The highest function value over the entire domain or over a specified closed interval. There is no point anywhere on the interval where f(x) is larger.
Think of it as: the tallest peak on the entire map, not just the tallest hill in your neighbourhood.
Absolute (global) minimum
The lowest function value over the entire domain or over a specified closed interval.
Think of it as: the deepest valley on the entire map.
Closed interval
An interval [c, d] that includes both of its endpoints. Written with square brackets.
In simple terms, the function's domain has definite start and end values, and those values count.
Bounded interval
An interval with finite length. A closed and bounded interval has real-number endpoints on both sides (no infinity).
Extreme Value Theorem (EVT)
If y = f(x) is continuous over a closed and bounded interval [c, d], then f is guaranteed to have both an absolute maximum point and an absolute minimum point on that interval.
In simple terms, if the function has no breaks and the interval has definite ends, the function must reach a highest value and a lowest value somewhere on that interval. No exceptions.
Endpoint
The boundary values c and d of a closed interval [c, d]. When finding absolute extrema, you must evaluate f at these points in addition to the critical values.
Think of it as: the edges of the playing field. The highest or lowest point might be right on the boundary.
The First Derivative Test (covered in the companion notes) tells you about relative extrema, the local peaks and valleys. But many real problems ask for the absolute best or worst outcome over a fixed range: the pricing strategy that maximises revenue, the production level that minimises cost. The Extreme Value Theorem guarantees those absolute extrema exist whenever the function is continuous on a closed interval, and it gives you a clean procedure to find them.
If y = f(x) is continuous over a closed and bounded interval [c, d], then f is guaranteed to have both an absolute maximum point and an absolute minimum point.
The critical detail: the absolute extremum might occur at a critical value inside the interval, or it might occur at one of the endpoints. You have to check all candidates.
Find f'(x)
Solve f'(x) = 0 to get all critical values
Discard any critical values that fall outside [c, d]
Evaluate f at each remaining critical value and at both endpoints c and d
The largest value is the absolute maximum; the smallest is the absolute minimum
A hotel manager charges $90 per room and rents 100 rooms per night (revenue = $9,000). For every $1 she reduces the price, she rents 2 additional rooms.
Building the model:
Let x = the dollar amount she reduces the price.
Price per room: p = 90 - x
Number of rooms rented: q = 100 + 2x
Revenue: R(x) = p × q = (90 - x)(100 + 2x)
Expanding:
R(x) = 9,000 + 180x - 100x - 2x²
R(x) = 9,000 + 80x - 2x²
Domain:
She can reduce the price by at most $90 (to $0 per room) and at least $0 (no reduction). So the domain is the closed interval [0, 90].
Step 1: Differentiate.
R'(x) = 80 - 4x
Step 2: Set equal to zero.
80 - 4x = 0, so x = 20
Step 3: x = 20 is inside [0, 90], so it stays.
Step 4: Evaluate R at the critical value and both endpoints.
R(0) = 9,000 + 0 - 0 = 9,000
R(20) = 9,000 + 80(20) - 2(20²) = 9,000 + 1,600 - 800 = 9,800
R(90) = 9,000 + 80(90) - 2(90²) = 9,000 + 7,200 - 16,200 = 0
Step 5: Compare.
Absolute maximum: R(20) = 9,800 at x = 20
Absolute minimum: R(90) = 0 at x = 90
Interpretation:
The manager should reduce the price by $20 (charging $70 per room, renting 140 rooms) for a maximum nightly revenue of $9,800. The First Derivative Test confirms this independently: R'(10) = 40 > 0 and R'(30) = -40 < 0, so the sign changes from positive to negative, confirming x = 20 is a relative maximum. Since R(20) = 9,800 is also larger than both endpoint values, it is the absolute maximum.
If she were to reduce the price all the way to $0, she would rent 280 rooms but earn nothing.
Revenue model:
R = p × q (price per unit times quantity sold)
Hotel revenue function:
R(x) = (90 - x)(100 + 2x) = 9,000 + 80x - 2x²
where x is the dollar reduction from the base price of $90.
Derivative of the revenue function:
R'(x) = 80 - 4x
Absolute extrema checklist (on a closed interval [c, d]):
Compute f'(x)
Solve f'(x) = 0 for critical values inside [c, d]
Make a list: f(c), f(each critical value), f(d)
Largest = absolute max, smallest = absolute min
The hotel revenue example is a textbook case of pricing optimisation. The manager faces a trade-off: lower prices attract more guests, but each dollar off the rate costs margin. Calculus pinpoints the exact sweet spot.
At $90 per room, she rents 100 rooms ($9,000 revenue). At $70 per room (a $20 reduction), she rents 140 rooms ($9,800 revenue). That extra $800 per night comes from finding the vertex of the revenue parabola using the derivative.
This same structure appears in any "price vs quantity" problem: ticket sales for events, subscription pricing, wholesale discount schedules. The approach is always the same: write revenue as a function of one variable, differentiate, set equal to zero, check endpoints if the domain is bounded.
Students often confuse relative extrema with absolute extrema. A relative maximum is the highest point in its immediate neighbourhood. An absolute maximum is the highest point on the entire interval (or domain). Every absolute extremum on a closed interval is either a relative extremum or an endpoint value, but not every relative extremum is an absolute one.
Students frequently forget to check the endpoints. On a closed interval, the absolute max or min can occur at x = c or x = d, not just at a critical value. Skipping endpoints is the single most common procedural error on these problems.
Students sometimes apply the Extreme Value Theorem to functions that are not continuous on the interval, or to intervals that are not closed (e.g. open intervals like (0, 90)). The theorem requires both conditions: continuity and a closed, bounded interval.
In applied problems, students sometimes find the critical value but forget to state the corresponding function value and interpret it. The question "what is the maximum revenue?" requires both the x-value (how much to reduce the price) and the R-value (the revenue itself).
⚠️ The Extreme Value Theorem and the closed-interval method are heavily tested. Expect a problem that gives you a function on [c, d] and asks for the absolute max and min. The procedure is mechanical: differentiate, solve, evaluate at critical values and endpoints, compare.
⚠️ Applied optimisation problems (revenue, cost, profit, area, volume) are among the most common exam questions in this unit. You will need to build the function from a word problem, identify the practical domain, and interpret the result in context.
⚠️ Know when the EVT applies and when it does not. If the interval is open, or the function has a discontinuity on the interval, the theorem does not guarantee extrema exist.
⚠️ Always state both pieces of the answer: the x-value where the extremum occurs and the function value at that point. Partial answers lose marks.
True or false: the Extreme Value Theorem applies to any function on any interval.
Answer: False. It requires (a) a continuous function on (b) a closed and bounded interval.
Fill in the blank: to find absolute extrema on [c, d], you must evaluate f at all ____________ inside the interval and at both ____________.
Answer: critical values; endpoints.
True or false: an absolute maximum on a closed interval must occur at a critical value, never at an endpoint.
Answer: False. The absolute maximum can occur at an endpoint.
In the hotel example, the revenue function is R(x) = 9,000 + 80x - 2x². What is R'(x)?
Answer: R'(x) = 80 - 4x.
True or false: if a function has only one critical value on a closed interval and the First Derivative Test confirms it is a relative maximum, you still need to check the endpoints to confirm it is the absolute maximum.
Answer: True. The endpoint values must be compared to the critical-point value.
Q: State the Extreme Value Theorem in your own words.
A: If a function is continuous on a closed, bounded interval [c, d], it must achieve both a highest value (absolute maximum) and a lowest value (absolute minimum) somewhere on that interval.
Q: A hotel charges $90 per room and rents 100 rooms. For each $1 reduction in price, 2 additional rooms are rented. Write the revenue function R(x) where x is the price reduction.
A: Price = 90 - x, rooms = 100 + 2x, so R(x) = (90 - x)(100 + 2x) = 9,000 + 80x - 2x².
Q: Find the critical value of R(x) = 9,000 + 80x - 2x² and determine the maximum revenue.
A: R'(x) = 80 - 4x. Setting R'(x) = 0 gives x = 20. R(20) = 9,000 + 1,600 - 800 = 9,800. Maximum revenue is $9,800.
Q: What is the practical domain for the hotel revenue problem, and why?
A: [0, 90]. The manager can reduce the price by at most $90 (bringing it to $0) and by at least $0 (no reduction). Both endpoints are included, making it a closed interval.
Q: Evaluate R at both endpoints of [0, 90] and confirm which point is the absolute maximum and which is the absolute minimum.
A: R(0) = 9,000. R(90) = 0. R(20) = 9,800. The absolute maximum is 9,800 at x = 20. The absolute minimum is 0 at x = 90.
This connects directly to the First Derivative Test (companion study notes). The First Derivative Test classifies relative extrema; the Extreme Value Theorem procedure finds absolute extrema by adding endpoint checks.
The Second Derivative Test, covered in the next MAT210 lesson, provides another route to classifying critical points. It uses concavity (f''(x)) instead of sign charts. For the hotel example, f''(x) = -4 < 0 everywhere, confirming the parabola opens downward and x = 20 is a maximum.
In later courses, these techniques extend to multivariable optimization (functions of two or more variables), where critical points are found using partial derivatives and classified using the Hessian matrix.
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