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4.4.14 · D5Multivariable Calculus

Question bank — Absolute extrema on closed bounded regions

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Every reveal below is a full sentence of reasoning — never a bare "true" or "false". The point is why.

Setup you need before the traps

Before the questions make sense, three words must be pinned down precisely.

Figure — Absolute extrema on closed bounded regions
Figure — Absolute extrema on closed bounded regions

True or false — justify

A continuous function on any bounded set attains an absolute max.
False — bounded is not enough; the set must also be closed. On the open disk , approaches but never reaches it because the boundary point is missing. See Closed and bounded sets (topology).
A continuous function on any closed set attains an absolute max.
False — closed is not enough; it must also be bounded. On the closed set , runs off to with no maximum.
If everywhere in the interior of , then has no absolute extrema on .
False — it only means the extrema are not in the interior; they still exist and must live on the boundary, guaranteed by the Extreme Value Theorem (which holds in every dimension for closed bounded sets).
The absolute maximum of on is always an interior critical point.
False — it can sit on the boundary. In the restated Worked Example 1 above, the max value occurred at the corner , not at the interior critical point .
Once you find the single interior critical point, you can stop.
False — you must still walk the whole boundary and compare, because a boundary point may beat the interior candidate; the interior critical point is just one entry in the candidate list.
If is differentiable everywhere on , then all candidate points come from solving .
False — differentiability rules out "corner/kink" candidates in the interior, but the boundary still supplies candidates (its own critical points, plus endpoints and corners).
The Hessian second-derivative test is needed to confirm which candidate is the absolute max.
False — for absolute extrema you simply compare numerical values of at a finite candidate list; the Second derivative test (Hessian) only classifies local interior behaviour and can even mislabel a boundary winner.
Every point where inside is an absolute extremum of on .
False — a critical point may be a saddle or a mere local feature; it is only a candidate. Its status is decided by comparison, not by being critical.
On a disk boundary, substituting into finishes the boundary analysis.
False — substitution simplifies but a remaining freedom (the angle) still varies; you must optimise over that one variable, e.g. via from Parametrization of curves.
A function can attain its absolute minimum at two different points.
True — nothing forbids ties. In the restated Worked Example 1 the minimum value was attained at both and .
If has a hole (a non-simply-connected region), you only need to walk the outer fence.
False — the boundary of a holed region has two parts, the outer fence and the inner rim around the hole. Both are part of the boundary of a closed , so extrema can hide on the inner rim; you must walk every boundary component.

Spot the error

" has interior critical point with , so the absolute minimum is ." Where is the mistake?
The interior value was never compared against boundary values; the true minimum is (on the boundary), so calling the minimum skips Step 3, comparison entirely.
"On the edge , has critical point giving , so the edge's extreme values are just ." What's wrong?
Only the interior critical point of the edge was checked; the endpoints and ( and ) were dropped, and one of them is the edge's maximum.
" on : on the boundary is constant, so the boundary contributes nothing." Is that reasoning safe?
The conclusion (boundary value is constant ) is correct here, but the phrase "contributes nothing" is dangerous — that constant is still a candidate and turns out to be the absolute max; you must include it in the comparison.
"The gradient is zero at a saddle point, so a saddle can never be an absolute extremum." Correct?
Wrong reasoning — a saddle is not a local extremum, but on a restricted region it could still tie the largest or smallest value seen; regardless, we don't classify, we compare values.
"I parametrised the boundary as for and found the extremes." What did they miss?
Half the circle — a full boundary needs ; stopping at silently discards the lower semicircle where an extremum may hide.
"On the square I found each edge's extremes but treated the four corners as belonging to no edge, so I ignored them." Fix?
Each corner is the shared endpoint of two edges and is a mandatory candidate; ignoring corners can miss the answer, as the max at the corner in Worked Example 1 shows.
"My region is an annulus (a disk with a smaller disk removed). I optimised on the interior and on the outer circle." What's missing?
The inner circle bounding the removed hole is also part of the boundary; skipping it can miss an extremum that lives on the rim of the hole.

Why questions

Why does an interior extremum force ?
If you could step in the direction to increase (or to decrease it), contradicting that the point was a max (or min); see Critical points and gradient.
Why do we reduce the boundary to a single-variable problem instead of using on it?
On the boundary is constrained to a curve, so its unconstrained gradient generally isn't zero there; parametrising collapses it to one free variable where ordinary 1-variable calculus applies, or you use Lagrange multipliers.
Why do we compare actual values rather than classify each candidate?
We have a finite candidate list, and the absolute max/min is simply the largest/smallest of those actual numbers — classification (saddle vs. extremum) answers a different, local question.
Why must be closed and bounded, not just one?
Bounded blocks the value from escaping to ; closed blocks it from sneaking toward a missing boundary point — both escape routes must be sealed for the guarantee to hold.
Why is continuity essential to the Extreme Value Theorem?
A jump or spike lets approach a value it never attains (e.g. a removed point at the peak); continuity forbids such gaps so the supremum is actually reached.
Why doesn't the recipe need the Hessian even though it appears in local-extrema chapters?
The Hessian answers "is this critical point a local max/min/saddle?"; the absolute problem instead asks "which candidate has the largest value?", and that is settled by direct comparison.
Why does a holed region make the boundary step longer?
Because "boundary" means all the edges of at once — a hole adds a whole extra inner curve, so you must parametrise and optimise on each separate boundary component, then include every result in the comparison.

Edge cases

If is constant on all of , what are its absolute extrema?
Every point is simultaneously an absolute max and an absolute min; the extreme value is that single constant, attained everywhere.
What if the interior critical point lies outside ?
You discard it — only interior critical points inside are candidates; a solution of off the region is irrelevant to this problem.
What if the interior critical point lies exactly on the boundary of ?
It is automatically captured by the boundary analysis, so you handle it there; it is still a valid candidate, just counted as a boundary point.
Can a single point (a degenerate region ) have absolute extrema?
Yes — a one-point set is closed and bounded, and is trivially both the max and the min; the theorem still applies.
What if has a corner (non-differentiable point) inside ?
Then may fail to exist there, so that point is added to the candidate list as a non-differentiable interior candidate, alongside points where .
If two adjacent boundary edges meet at a corner where is smooth, must you still list the corner?
Yes — smoothness of doesn't remove the fact that each edge is a closed interval whose endpoint (the corner) can hold an extreme value.
For an annulus , how many boundary curves must you walk?
Two — the outer circle and the inner circle ; both bound the region, so each must be parametrised and searched.
Recall One-line survival checklist

Interior critical points inside ➜ every boundary component (outer fence and any inner rims) reduced to 1-var with all endpoints/corners ➜ compare every value; never classify, never skip a fence, never drop a corner.

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