Visual walkthrough — Absolute extrema on closed bounded regions

Our terrain for the whole page: $f(x,y)=x^2-2xy+2y\text{over the rectangle }D=\{0\le x\le 3,0\le y\le 2\}.$

Let me name every symbol before we use it.

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Step 1 — See the terrain and the fence

WHAT. Before any algebra, look at what we're standing on. The rectangle $D$ is a flat floor plan; above it sits a curved surface whose height is $f$.

WHY. The Extreme Value Theorem's 2D cousin promises that a continuous height on a closed, bounded field must hit a genuine highest and lowest value somewhere. Our $f$ is a polynomial, so it's smooth everywhere — no gaps, no cliffs — so the promise holds. We are allowed to go hunting; a treasure is guaranteed to exist.

PICTURE. The floor is the rectangle; the coloured surface is the height. Notice it dips and rises — our job is to find the extreme dips and rises without checking infinitely many points.

The whole strategy: the top/bottom can only hide in two kinds of place — a flat dimple inside the fence, or somewhere on the fence. So we search those two, and nowhere else.

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Step 2 — Hunt the flat spots inside (the gradient)

WHAT. Inside the field, at a highest or lowest point the ground must be locally flat — the surface has a level tangent plane there. "Flat in both directions" means the two partial slopes are zero.

WHY the gradient? If the gradient arrow were not zero at a spot, you could take a tiny step along it and climb higher — so that spot could not have been the summit. Flat interior extrema therefore force $\nabla f=0$. (See Critical points and gradient.)

For our $f=x^2-2xy+2y$:

$$\underset{\text{east–west slope}}{\underbrace{f_x}}=\underset{\text{from }{x}^2}{\underbrace{2x}}-\underset{\text{from }-2xy}{\underbrace{2y}}=0,\underset{\text{north–south slope}}{\underbrace{f_y}}=\underset{\text{from }-2xy}{\underbrace{-2x}}+\underset{\text{from }2y}{\underbrace{2}}=0.$$

Solve. The second equation gives $-2x+2=0\Rightarrow x=1$. Put that in the first: $2(1)-2y=0\Rightarrow y=1$. One flat spot: $(1,1)$.

Is it inside the fence? $0<1<3$ and $0<1<2$ — yes, strictly inside. ✓ Its height: $f(1,1)=1^2-2(1)(1)+2(1)=1-2+2=1.$

PICTURE. The gradient arrows all flow away from the flat point; right at $(1,1)$ the arrow shrinks to nothing.

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Step 3 — Walk the bottom edge $y=0$

WHAT. Now we walk the fence. On the bottom edge, $y$ is stuck at $0$, and only $x$ is free to move from $0$ to $3$. Freezing one variable turns the surface into a plain 1-variable curve (see Parametrization of curves).

WHY. The boundary is a whole curve of points where the interior "flat" rule doesn't apply — the fence can pin an extreme even when the ground there is sloped. So we must reduce $f$ on each edge to a single-variable function and use ordinary 1-D max/min (from Extreme Value Theorem (1D)).

Substitute $y=0$:

$$g(x)=f(x,0)=x^2\underset{=0}{\underbrace{-2x\cdot 0}}\underset{=0}{\underbrace{+2\cdot 0}}=x^2,0\le x\le 3.$$

$g(x)=x^2$ only ever increases on $[0,3]$ (its slope $2x$ is positive). So no interior critical point on this edge — the extremes sit at the two endpoints: $g(0)=f(0,0)=0,g(3)=f(3,0)=9.$

PICTURE. The bottom edge lifted into a rising parabola; the max is the right corner.

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Step 4 — Walk the top edge $y=2$

WHAT. Freeze $y=2$; slide $x$ from $0$ to $3$.

WHY. Same reason — reduce to 1-D. This edge is more interesting because the reduced curve actually has a low point inside the interval.

Substitute $y=2$:

$$h(x)=f(x,2)=x^2\underset{-4x}{\underbrace{-2x\cdot 2}}\underset{+4}{\underbrace{+2\cdot 2}}=x^2-4x+4=(x-2{)}^2,0\le x\le 3.$$

The tidy form $(x-2{)}^2$ says: a bowl bottoming out at $x=2$, height $0$. Check that it's an interior critical point of the edge: derivative $2(x-2)=0\Rightarrow x=2$, and $0<2<3$ ✓. Then the endpoints: $h(2)=f(2,2)=0,h(0)=f(0,2)=4,h(3)=f(3,2)=1.$

PICTURE. The top edge lifted into a bowl; its lowest touch is at $(2,2)$, height $0$.

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Step 5 — Walk the left and right edges

WHAT. Two vertical edges left. On the left edge $x=0$, only $y$ moves; on the right edge $x=3$, only $y$ moves.

WHY. Same 1-D reduction; now the other variable is frozen.

Left, $x=0$:

$$f(0,y)=\underset{x^2}{\underbrace{0}}\underset{-2xy}{\underbrace{-0}}+2y=2y,0\le y\le 2.$$

A straight climbing line — extremes at the ends: $f(0,0)=0$, $f(0,2)=4$.

Right, $x=3$:

$$f(3,y)=\underset{x^2}{\underbrace{9}}\underset{-2xy}{\underbrace{-6y}}+2y=9-4y,0\le y\le 2.$$

A straight descending line (slope $-4<0$) — extremes at the ends: $f(3,0)=9$, $f(3,2)=1$.

PICTURE. Left edge rises, right edge falls — both are straight ramps, so only their endpoints matter.

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Step 6 — Compare every candidate

WHAT. Gather all heights we recorded and simply read off the biggest and smallest.

WHY. We now hold a finite list of suspects. No classifying, no Hessian — direct comparison decides everything. (Comparing beats classifying: the winner might be a boundary corner the Hessian can't even score.)

Where	Point	Height $f$
interior flat	$(1,1)$	$1$
bottom ends	$(0,0),(3,0)$	$0,9$
top	$(2,2),(0,2),(3,2)$	$0,4,1$
left end	$(0,2)$	$4$
right end	$(3,2)$	$1$

The largest height is $9$; the smallest is $0$ (achieved twice). $\boxed{\text{abs max}=9\text{ at }(3,0);\text{abs min}=0\text{ at }(0,0)\text{ and }(2,2).}$

PICTURE. The floor plan with every candidate dotted; the winner (max) and the two lowest (min) circled.

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Step 7 — The degenerate cases (so you're never surprised)

WHAT. What if the field or function is unusual? Three quick pictures of the edge behaviours.

WHY. The contract: every scenario must be shown, so you never hit one you weren't warned about.

• Flat spot lands outside the fence. If solving $\nabla f=0$ gave, say, $(5,5)$, you'd discard it — it's not in $D$. The extrema then live entirely on the boundary.
• No interior flat spot at all. Some $f$ have $\nabla f\ne 0$ everywhere (e.g. $f=x$). Then Step 2 yields nothing and all candidates come from the fence.
• Ties. Our min $0$ happens at two points, $(0,0)$ and $(2,2)$. That's allowed — "the" minimum is a value; it can be attained in several places.

PICTURE. Left: an outside critical point crossed out. Middle: a tilted plane with no flat spot. Right: two separate spots at the same lowest height.

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The one-picture summary

Everything above in a single frame: the field, the one interior flat spot, the four edges each reduced to a 1-D curve, the star on the winner ($9$) and dots on the tied minima ($0$).

Recall Feynman: the whole walk in plain words

We had a bumpy sheet stretched over a rectangular fenced yard, and we wanted its highest and lowest points. First we found where the sheet is flat inside the yard by asking "which way is uphill?" — the gradient — and finding the one spot where uphill points nowhere: $(1,1)$, height $1$. Then we walked all four sides of the fence. On each side one coordinate is frozen, so the sheet becomes a simple 1-D curve: the bottom was a rising parabola (peaks at the corner $(3,0)$, height $9$), the top was a bowl (touches $0$ at $(2,2)$), the left was a rising ramp, the right a falling ramp — ramps only matter at their ends. We wrote down the height at every flat spot and every corner and every edge-dip. Finally we just looked at the list and pointed: highest is $9$ at $(3,0)$; lowest is $0$, and it happens at two places, $(0,0)$ and $(2,2)$. No fancy tests — just search inside, walk the fence, and compare.

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Connections

• Critical points and gradient — Step 2's flat-spot hunt.
• Extreme Value Theorem (1D) — the 1-D max/min used on every edge.
• Parametrization of curves — freezing one variable per edge.
• Second derivative test (Hessian) — deliberately not used here.
• Lagrange multipliers — an alternative for curved boundaries.
• Closed and bounded sets (topology) — why the whole hunt is guaranteed to succeed.