Absolute extrema on closed bounded regions

WHY this works at all

WHY do we need both "closed" AND "bounded"?

• Drop bounded: on $D=\mathbb{R}^2$, $f=x$ has no max — it runs off to $+\infty$.
• Drop closed: on the open disk $x^2+y^2<1$, $f=x$ never reaches its sup $1$ (the boundary point is excluded).

Both conditions plug the "escape routes" — value can't run to infinity, and can't sneak toward a missing edge.

HOW: the algorithm

WHY no second-derivative test here? Because we have a finite list of candidate points and we just compare actual values. We never need to classify saddle vs. extremum — direct comparison wins.

Worked Example 1 — over a rectangle

Find the absolute extrema of $f(x,y)=x^2-2xy+2y$ on $D=\{0\le x\le 3,\ 0\le y\le 2\}$.

Step 1 — interior. $f_x = 2x-2y=0,\qquad f_y=-2x+2=0.$ Why this step? Interior extrema need a flat tangent plane, so set both partials to $0$. From $f_y$: $x=1$. From $f_x$: $y=x=1$. Point $(1,1)$ is inside $D$. ✓ $f(1,1)=1-2+2=1$.

Step 2 — boundary (4 edges). Why? The fence may hold the extremum even if interior is flat-free.

• Bottom $y=0,\ 0\le x\le 3$: $f=x^2$. Increasing → check ends: $f(0,0)=0$, $f(3,0)=9$.
• Top $y=2,\ 0\le x\le 3$: $f=x^2-4x+4=(x-2)^2$. Critical $x=2$: $f=0$. Ends: $f(0,2)=4,\ f(3,2)=1$.
• Left $x=0,\ 0\le y\le 2$: $f=2y$. Ends: $f(0,0)=0,\ f(0,2)=4$.
• Right $x=3,\ 0\le y\le 2$: $f=9-6y+2y=9-4y$. Decreasing → ends: $f(3,0)=9,\ f(3,2)=1$.

Step 3 — compare. Candidate values: $\{1,\ 0,\ 9,\ 0,\ 4,\ 0,\ 4,\ 9,\ 1\}$. $\boxed{\text{abs max}=9 \text{ at }(3,0);\qquad \text{abs min}=0 \text{ at }(0,0)\text{ and }(2,2).}$ Why this step? We just pick biggest and smallest from the finite candidate list.

Worked Example 2 — over a disk (use Lagrange / parametrize)

Find absolute extrema of $f(x,y)=x^2+y^2-x-y+1$ on the disk $x^2+y^2\le 1$.

Step 1 — interior. $f_x=2x-1=0\Rightarrow x=\tfrac12,\qquad f_y=2y-1=0\Rightarrow y=\tfrac12.$ Is $(\tfrac12,\tfrac12)$ inside? $\tfrac14+\tfrac14=\tfrac12<1$ ✓. $f(\tfrac12,\tfrac12)=\tfrac14+\tfrac14-\tfrac12-\tfrac12+1=\tfrac12.$

Step 2 — boundary $x^2+y^2=1$. Substitute $x^2+y^2=1$: $f = 1 - x - y + 1 = 2-(x+y).$ Why this step? The constraint kills the quadratic terms, leaving a clean linear thing. Parametrize $x=\cos t,\ y=\sin t$: $f=2-(\cos t+\sin t)=2-\sqrt2\sin(t+\tfrac\pi4)$. Since $\sin$ ranges in $[-1,1]$: $f_{\max}^{bdry}=2+\sqrt2\ \ (\text{at }x=y=-\tfrac1{\sqrt2}),\qquad f_{\min}^{bdry}=2-\sqrt2.$

Step 3 — compare. Values: interior $0.5$, boundary $2-\sqrt2\approx 0.586$, $2+\sqrt2\approx 3.414$. $\boxed{\text{abs max}=2+\sqrt2;\qquad \text{abs min}=\tfrac12 \text{ at }(\tfrac12,\tfrac12).}$

Common Mistakes

Flashcards

Connections

• Critical points and gradient — interior step solves $\nabla f=0$.
• Second derivative test (Hessian) — used for local classification, not here.
• Lagrange multipliers — alternative way to handle the boundary constraint.
• Extreme Value Theorem (1D) — the single-variable ancestor of this method.
• Closed and bounded sets (topology) — why the guarantee needs both.
• Parametrization of curves — turning a boundary into a 1-var problem.

Concept Map

Hinglish (regional understanding)