Absolute Minimums and Maximums

Main idea

On a closed, bounded region, a continuous function always reaches a highest and a lowest value. They can only hide in two places: at critical points inside the region, or somewhere on its edge.

\text{candidates} = \{\text{critical points inside } D\} \cup \{\text{extrema of } f \text{ on the boundary of } D\}

Explanation

An absolute maximum is simply the single highest value a function reaches anywhere in a region, and the absolute minimum is the single lowest. ("Absolute" as opposed to "relative": it has to beat every point in the region, not just its neighbors.) It's what you need when there are real limits, like finding the hottest spot on a metal plate of a given size. The candidate list in the formula looks formal, but it's just two places to look: critical points inside (level spots in the middle) and the extremes along the boundary (the edge of the region). Check both lists, compare, and you're done.

Remember the closed interval method from Calculus I? To find the biggest value of $f(x)$ on $[a, b]$ , you checked the critical points inside and the two endpoints, then compared. This is the same method in two dimensions. The only difference is that the "endpoints" are now a whole boundary curve, which takes more work to check.

Why are those the only places? If the highest point is in the interior, the ground is level there, so it's a critical point from the last lesson. If it's not in the interior, it's on the edge. The theorem that a max and min exist at all needs the region to be closed (it includes its edge) and bounded (it doesn't run off to infinity), the same way the interval had to be closed.

Steps:

Find the critical points inside the region and evaluate $f$ there.
On each piece of the boundary, write $f$ as a function of one variable, by substituting the boundary's equation or a parametrization. That turns it into a Calculus I problem: find its extrema, including the corners where pieces meet.
Compare all the values. The biggest is the absolute max and the smallest is the absolute min. You don't need the second derivative test here, since you're comparing everything directly.

See it

A circular plate x² + y² ≤ 4, heated to T = 20 + 2x + 4y − x² − y², with a dot circling its edge. The only critical point, (1, 2), is off the plate, so the hottest and coldest spots have to be on the edge. As you can see, the edge temperature 16 + 4 cos θ + 8 sin θ swings between about 7.1 and 24.9, peaking in the direction of (1, 2).

Worked examples

Example 1

Find the absolute extrema of $f(x, y) = x^2 + y^2 - 2x$ on the disk $x^2 + y^2 \le 4$ .

1
Inside: $\nabla f = \langle 2x - 2, 2y\rangle = \vec 0$ at $(1, 0)$ , which is inside the disk, and $f = -1$ there.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find the absolute maximum of $f = x^2 + y^2 - 2x$ on $x^2 + y^2 \le 4$ .

Problem 2+5 XP

Find the absolute minimum on the same disk.

Problem 3+5 XP

Find the absolute maximum of $f = xy$ on the square $0 \le x, y \le 1$ .

Problem 4+5 XP

Find the absolute maximum of $f = x + y$ on the disk $x^2 + y^2 \le 2$ .

Problem 5+5 XP

Find the absolute minimum of $f = x^2 - y^2$ on the square $-1 \le x, y \le 1$ .

Problem 6+5 XP

Noise in a rectangular park, $0 \le x \le 4$ , $0 \le y \le 3$ (hundreds of meters), is $N(x, y) = x^2 + y^2 - 2x - 4y + 10$ decibels above a baseline. What's the lowest noise level in the park?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find the absolute maximum of $f(x, y) = x^2y$ on the triangle with vertices $(0, 0)$ , $(3, 0)$ and $(0, 3)$ .

Where it's used

Choosing the best design within physical limits, such as the strongest beam that fits a space or the best mix of ingredients within a budget, is optimization on a bounded region.

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Relative Minimums and Maximums Lagrange Multipliers