Peaks and Valleys: A Journey Through the Extreme Value Theorem

The Extreme Value Theorem guarantees that a function continuous on a closed, bounded interval reaches both a highest and a lowest value somewhere on it. Below: the statement, why each condition is needed, the closed interval method for finding those values, and a worked optimization example.

What the Extreme Value Theorem says

If a function \( f \) is continuous on a closed, bounded interval \( [a, b] \), then \( f \) attains both an absolute maximum and an absolute minimum somewhere on that interval. There is a point \( c \) in \( [a, b] \) where \( f(c) \ge f(x) \) for every \( x \) in the interval, and a point \( d \) where \( f(d) \le f(x) \) for every \( x \).

The theorem is an existence result. It promises the highest and lowest values are reached; it does not tell you where. Finding them is the work of the closed interval method below.

Why both conditions matter

The interval must be closed. On \( (0, 1) \), the function \( f(x) = x \) gets arbitrarily close to 1 without ever reaching it. There is no maximum, only a supremum. Include the endpoint and the maximum exists.

The interval must be bounded. On \( [0, \infty) \), \( f(x) = x \) has a minimum at 0 but no maximum at all.

The function must be continuous. Define \( f(x) = \frac{1}{x} \) for \( x \) in \( (0, 1] \) and \( f(0) = 0 \). The interval is closed and bounded, but the jump at 0 means the values climb without limit and no maximum is attained.

All three requirements are sufficient rather than necessary. A discontinuous function can still happen to have a maximum; the theorem simply stops guaranteeing one.

The closed interval method

1. Confirm \( f \) is continuous on \( [a, b] \).
2. Find the critical points inside the interval: values where \( f'(x) = 0 \) or where \( f'(x) \) does not exist.
3. Evaluate \( f \) at each critical point and at both endpoints \( a \) and \( b \).
4. The largest of those values is the absolute maximum and the smallest is the absolute minimum.

The endpoints are the step most often skipped. A function can rise steadily across the whole interval, in which case its maximum sits at \( b \) and there is no critical point at all.

A worked application: solar panel angle

Consider the optimization of a solar panel’s efficiency. The goal is to maximize the panel’s energy output. By using the Extreme Value Theorem, engineers can determine the panel’s angle and orientation that yield the highest energy capture. This involves analyzing the function that relates the angle of the solar panel to the amount of absorbed solar energy. The theorem helps identify the optimal angle, the maximum point of the function, within a given range (e.g., 0 to 90 degrees), ensuring the solar panel operates at peak efficiency throughout the day.

Let’s create a simplified numerical example for optimizing a solar panel’s efficiency. Suppose the energy output \( E \) of a solar panel, measured in kilowatt-hours, varies with its angle ( \theta ) (measured in degrees from the ground) according to the function:

\( E(\theta) = -2\theta^2 + 90\theta \)

This function might represent a simplified model where the energy output is maximal at a certain angle and decreases as the angle moves away from this optimal point. We’re interested in finding the angle that maximizes \( E \) within the practical range of \( 0^\circ \) to \( 45^\circ \).

1. Find the Critical Points: We first find the derivative of \( E(\theta) \) and set it to zero to find critical points. The derivative is:
   \( E'(\theta) = -4\theta + 90 \) Setting this equal to zero gives:
   \( -4\theta + 90 = 0 \)
   \( \theta = \frac{90}{4} = 22.5^\circ \)
2. Evaluate at Endpoints and Critical Point: We evaluate \( E \) at \( 0^\circ \), \( 22.5^\circ \), and \( 45^\circ \) to find which gives the maximum output. \( E(0) = -2 \times 0^2 + 90 \times 0 = 0 \) kilowatt-hours
   \( E(22.5) = -2 \times 22.5^2 + 90 \times 22.5=1012.5 \) kilowatt-hours
   \( E(45) = -2 \times 45^2 + 90 \times 45=0 \) kilowatt-hours

Reading the solar panel result

Following the closed interval method on \( E(\theta) = -2\theta^2 + 90\theta \) over \( [0, 90] \):

• \( E'(\theta) = -4\theta + 90 \), which is zero at \( \theta = 22.5 \) degrees. That is the only critical point, and it lies inside the interval.
• \( E(22.5) = -2(506.25) + 2025 = 1012.5 \) kilowatt-hours.
• \( E(0) = 0 \) and \( E(90) = -16200 + 8100 = -8100 \).

So the absolute maximum is 1012.5 kWh at 22.5 degrees, and the absolute minimum on the modeled range is at \( \theta = 90 \). The negative value there shows that a quadratic model only holds over the range it was fitted to: a panel standing vertically produces little energy, not negative energy.

Extreme Value Theorem or First Derivative Test?

They answer different questions. The First Derivative Test classifies a critical point as a local maximum or minimum by looking at the sign of \( f’ \) on either side of it. The Extreme Value Theorem is about absolute extremes across a whole closed interval, and the endpoints compete for the title even though they are not critical points.

A local maximum need not be the absolute maximum, and the absolute maximum need not be a local one. On \( [0, 3] \) the function \( f(x) = x \) has its absolute maximum at \( x = 3 \), an endpoint, where there is no local maximum in the usual sense.

Practice questions on the Extreme Value Theorem

1. Find the absolute maximum and minimum of \( f(x) = x^3 – 3x \) on \( [-2, 2] \).
2. Does \( f(x) = \tan x \) attain a maximum on \( [0, \pi] \)? Explain.
3. Find the absolute extremes of \( f(x) = x^{2/3} \) on \( [-1, 8] \). Where is the derivative undefined?
4. A fence 40 m long encloses a rectangular pen against a wall, so only three sides are fenced. What dimensions give the largest area?

Answers

1. \( f'(x) = 3x^2 – 3 = 0 \) gives \( x = \pm 1 \). Values: \( f(-2) = -2 \), \( f(-1) = 2 \), \( f(1) = -2 \), \( f(2) = 2 \). Absolute maximum 2, at \( x = -1 \) and \( x = 2 \); absolute minimum \( -2 \), at \( x = -2 \) and \( x = 1 \).
2. No. \( \tan x \) is undefined at \( x = \frac{\pi}{2} \), which lies inside the interval, so the function is not continuous on \( [0, \pi] \) and the theorem does not apply. The values grow without bound near \( \frac{\pi}{2} \).
3. \( f'(x) = \frac{2}{3}x^{-1/3} \) is undefined at \( x = 0 \), making 0 a critical point. Values: \( f(-1) = 1 \), \( f(0) = 0 \), \( f(8) = 4 \). Absolute maximum 4 at \( x = 8 \); absolute minimum 0 at \( x = 0 \).
4. With width \( w \) and the two perpendicular sides \( \frac{40 – w}{2} \) each, area \( A(w) = w \cdot \frac{40 – w}{2} \) on \( [0, 40] \). \( A'(w) = 20 – w = 0 \) at \( w = 20 \), giving \( A = 200 \) m², against \( A(0) = A(40) = 0 \). The pen is 20 m along the wall and 10 m deep.

The Ultimate Algebra Bundle 2026 From Pre-Algebra to Algebra II

Download

$109.99 Original price was: $109.99.$54.99Current price is: $54.99.

Comparing these values, the maximum energy output of \( 1012.5 \) kilowatt-hours occurs at \( 22.5^\circ \). Therefore, in this simplified model, angling the solar panel at \( 22.5^\circ \) from the ground optimizes its energy capture, demonstrating an application of the Extreme Value Theorem in determining the most efficient operational angle.

The Ultimate Elementary School Math Bundle Grade 3-5

Download

$109.99 Original price was: $109.99.$54.99Current price is: $54.99.