Worked examples — Absolute extrema on closed bounded regions

This page is the workout room for the three-step recipe. The parent note gave you two clean examples. Real problems are messier: the region might have corners, or be a full disk, or the boundary might touch a place where $f$ misbehaves. Below we first draw a map of every kind of case the topic can throw at you, then work enough examples that every square on the map gets stamped.

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The scenario matrix

Every absolute-extrema problem is some combination of region shape and twist. Here is the full grid — each cell is a distinct thing that can trip you up.

Cell	Region / feature	The twist it introduces	Covered in
C1	Rectangle (polygon)	Four straight edges + four corners as mandatory candidates	Ex 1
C2	Triangle (slanted edge)	A boundary edge that is not horizontal/vertical → substitute a line	Ex 2
C3	Disk (smooth curve)	Boundary is a circle → parametrize $x=\cos t,y=\sin t$, all $t$	Ex 3
C4	Interior critical point outside $D$	Step 1 gives a point you must discard	Ex 4
C5	Degenerate: no interior critical point at all	$\nabla f=0$ has no solution → everything lives on the fence	Ex 5
C6	Point of non-differentiability inside $D$	A sharp cone/valley where $\nabla f$ doesn't exist but is still a candidate	Ex 6
C7	Tie / multiple winners	Max (or min) attained at several points at once	Ex 7
C8	Real-world word problem	Translate "temperature on a plate" into the recipe, carry units	Ex 8
C9	Exam twist: constraint hits quadrant signs	On the circle the winner sits in a specific quadrant; sign of $\cos t,\sin t$ matters	Ex 9

We work Examples 1–9 below; the "Covered in" column is your checklist.

Recall The recipe, one line (from the parent)

Interior ($\nabla f=0$ inside, plus non-differentiable points) → Boundary (reduce to 1 variable, include endpoints/corners) → Compare all values.

Before we start, three symbols the parent used but let us pin down from zero:

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Example 1 — Rectangle with four corners (Cell C1)

Forecast (guess first!): The $-2x$ term tilts the bowl. Where do you think the lowest point sits — inside, on an edge, or at a corner? Write your guess before reading on.

Step 1 — interior critical points. $f_x=2x-2=0\Rightarrow x=1,f_y=2y=0\Rightarrow y=0.$ Why this step? Interior extrema need the uphill arrow to vanish; solving $\nabla f=0$ finds the flat spots. Point $(1,0)$ is inside $D$ ($-1<1<2$, $-1<0<2$) ✓. Value $f(1,0)=1+0-2=-1$.

Step 2 — walk the four edges (each a 1-variable job). Why this step? The bowl's tilt might push the highest value onto the fence; we must check all four sides.

• Bottom $y=-1,x\in [-1,2]$: $g(x)=x^2-2x+1=(x-1{)}^2$. Critical $x=1$: $g=0$. Ends: $g(-1)=4,g(2)=1$.
• Top $y=2,x\in [-1,2]$: $g(x)=x^2-2x+4=(x-1{)}^2+3$. Critical $x=1$: $3$. Ends: $g(-1)=7,g(2)=4$.
• Left $x=-1,y\in [-1,2]$: $g(y)=1+y^2+2=y^2+3$. Min at $y=0$: $3$. Ends: $g(-1)=4,g(2)=7$.
• Right $x=2,y\in [-1,2]$: $g(y)=4+y^2-4=y^2$. Min at $y=0$: $0$. Ends: $g(-1)=1,g(2)=4$.

The four corners appear above as edge endpoints: $(-1,-1)\to 4$, $(2,-1)\to 1$, $(-1,2)\to 7$, $(2,2)\to 4$.

Step 3 — compare. Candidate values: $\{-1,0,3,4,7,1\}$. $\boxed{\text{abs max}=7\text{ at }(-1,2);\text{abs min}=-1\text{ at }(1,0).}$

Verify: The min $-1$ is the interior critical value; the max is the corner farthest from the low point $(1,0)$. Distance${}^2$ from $(1,0)$ to $(-1,2)$ is $4+4=8$, the largest corner distance — consistent with $f$ being (distance to $(1,0)$)${}^2-1$. Indeed $f=(x-1{)}^2+y^2-1$, so min $=-1$ and max $=8-1=7$. ✓

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Example 2 — Triangle with a slanted edge (Cell C2)

Forecast: $f=x+y$ has no flat spot anywhere (its arrow never vanishes). So everything must be on the fence — where on this triangle is $x+y$ biggest?

Figure s01 — what it shows and how to read it. The orange-tinted triangle is our region $D$. The orange arrow near the origin is the gradient $\nabla f=(1,1)$: it points "up and to the right," the same everywhere because $f$ is a flat tilted plane. The teal hypotenuse is highlighted because $f=2$ along its entire length — a whole edge of winners. The plum dot at $(0,0)$ is the single lowest point. Reading the picture, the solution is obvious before any algebra: the plane's uphill arrow drags the maximum to the far edge and the minimum to the near corner.

Step 1 — interior. $f_x=1\ne 0,f_y=1\ne 0.$ Why this step? A critical point needs both partials to vanish simultaneously, $\nabla f=(f_x,f_y)=(0,0)$. Here $f_x=1$ and $f_y=1$ are each constantly $1$ — neither is ever zero, so the pair can never both vanish. No interior critical point exists. (Foreshadows Cell C5.) Move to the boundary.

Step 2 — three edges. Why this step? The whole action is on the fence.

• Bottom $y=0,x\in [0,2]$: $f=x$, increasing → ends $f(0,0)=0,f(2,0)=2$.
• Left $x=0,y\in [0,2]$: $f=y$, increasing → ends $f(0,0)=0,f(0,2)=2$.
• Slanted (hypotenuse) the line through $(2,0)$ and $(0,2)$ is $x+y=2$. Why substitute the line? On this edge $f=x+y=2$ identically — flat! So every point on the hypotenuse gives $f=2$.

Step 3 — compare. Values: $\{0,2\}$, with $2$ hit along an entire edge. $\boxed{\text{abs max}=2\text{ (whole hypotenuse)};\text{abs min}=0\text{ at }(0,0).}$

Verify: $f=x+y$ measures "how far along the diagonal direction." The corner $(0,0)$ is nearest the origin-corner ($0$); the hypotenuse is the line $x+y=2$, so of course $f=2$ there. A slanted edge handled by substituting its line equation. ✓

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Example 3 — Disk, parametrize the circle (Cell C3)

Forecast: $f=xy$ is positive where $x,y$ share a sign, negative where they differ. Guess: max and min live at $45^{\circ }$ diagonals on the boundary. Let's confirm.

Figure s02 — what it shows and how to read it. The black circle is the boundary $x^2+y^2=2$. The dotted contour lines are level sets of $f=xy$ (teal where $f<0$, orange where $f>0$). Notice how the contours hug the axes and swell along the diagonals — telling us $f$ grows fastest toward the corners $(\pm 1,\pm 1)$. The orange dots on the "same-sign" diagonal are the two maxima; the teal dots on the "opposite-sign" diagonal are the two minima; the plum dot at the center is the saddle we merely record. The figure predicts four boundary winners in a symmetric cross — a Cell C7 tie appearing for free.

Step 1 — interior. $f_x=y=0,f_y=x=0\Rightarrow (0,0).$ Why this step? Flat spot inside. $(0,0)$ is inside the disk. $f(0,0)=0$. (It's a saddle, but we don't classify — we just record the value.)

Step 2 — boundary $x^2+y^2=2$. Why parametrize? The circle is smooth with no corners; one dial $t$ sweeps it all. Set $x=\sqrt{2}\cos t,y=\sqrt{2}\sin t$ with $t\in [0,2\pi ]$ so the whole fence is covered: $f=xy=2\cos t\sin t=\sin (2t).$ Why the identity $2\cos t\sin t=\sin 2t$? Because $\sin$ has an obvious range $[-1,1]$ — it turns "find the max of a product" into "read off the top of a sine wave." As $t$ runs over $[0,2\pi ]$, the argument $2t$ runs over $[0,4\pi ]$, so $\sin (2t)$ completes two full waves and certainly hits both extremes: $\sin (2t)\in [-1,1]:\max =1\text{ at }t=\frac{\pi }{4},\frac{5\pi }{4},\min =-1\text{ at }t=\frac{3\pi }{4},\frac{7\pi }{4}.$

Step 3 — compare. Interior $0$; boundary $\pm 1$. $\boxed{\text{abs max}=1\text{ at }(1,1)\&(-1,-1);\text{abs min}=-1\text{ at }(1,-1)\&(-1,1).}$

Verify: At $t=\frac{\pi }{4}$: $x=y=\sqrt{2}\cdot \frac{1}{\sqrt{2}}=1$, $f=1$ ✓. All four winners sit on the diagonals — matching the forecast. Two maxima, two minima (a Cell C7 tie appears naturally on the circle).

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Example 4 — Interior critical point falls OUTSIDE $D$ (Cell C4)

Forecast: The $-6x$ tugs the bowl's bottom far to the right — probably outside the little unit disk. What happens to a critical point you can't use?

Step 1 — interior. $f_x=2x-6=0\Rightarrow x=3,f_y=2y=0\Rightarrow y=0.$ Candidate $(3,0)$: is it inside? $3^2+0=9>1$ ✗. Why discard? Step 1 keeps only solutions inside $D$; $(3,0)$ is far outside, so it is not a candidate. We get zero usable interior points.

Step 2 — boundary $x^2+y^2=1$. Why substitute? On the circle $x^2+y^2=1$, so $f=1-6x$. Parametrize $x=\cos t,y=\sin t$ with $t\in [0,2\pi ]$: $f=1-6\cos t.$ $\cos t\in [-1,1]$ (its full range over one period), and $f$ decreases as $\cos t$ grows: $\max :\cos t=-1(t=\pi )\Rightarrow f=7\text{at}(-1,0),\min :\cos t=1(t=0)\Rightarrow f=-5\text{at}(1,0).$

Step 3 — compare. Only boundary candidates survive. $\boxed{\text{abs max}=7\text{ at }(-1,0);\text{abs min}=-5\text{ at }(1,0).}$

Verify: $f=(x-3{)}^2+y^2-9$ = (distance to $(3,0)$)${}^2-9$. Closest point of the disk to $(3,0)$ is $(1,0)$ (distance $2$): $f=4-9=-5$ ✓. Farthest is $(-1,0)$ (distance $4$): $f=16-9=7$ ✓. The discarded critical point was the global min of $f$ on all of ${\mathbb{R}}^2$ — irrelevant because it's off the field.

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Example 5 — Degenerate: no interior critical point (Cell C5)

Forecast: A plane tilted over a disk. A flat plane never has a flat spot inside — so the extremes must be at two opposite edge points. Which?

Step 1 — interior. $f_x=3\ne 0,f_y=-4\ne 0.$ Why this step? Confirm there is no interior critical point: $\nabla f=(3,-4)$ is a constant nonzero arrow, so neither partial ever vanishes — the pair can never be $(0,0)$. The uphill direction is the same everywhere.

Step 2 — boundary $x^2+y^2=25$. Parametrize $x=5\cos t,y=5\sin t$ with $t\in [0,2\pi ]$: $f=15\cos t-20\sin t.$ Why combine into one sinusoid? A sum $a\cos t+b\sin t$ equals $R\cos (t-\varphi )$ with amplitude $R=\sqrt{a^2+b^2}$ — and over a full period $t\in [0,2\pi ]$ that cosine sweeps its whole range $[-R,R]$. Here $R=\sqrt{15^2+20^2}=\sqrt{625}=25$. $f\in [-25,25].$

Step 3 — compare. $\boxed{\text{abs max}=25;\text{abs min}=-25.}$ Max where the point aligns with $\nabla f=(3,-4)$: $(x,y)=5\cdot \frac{(3,-4)}{5}=(3,-4)$. Min at the opposite: $(-3,4)$.

Verify: $f(3,-4)=9+16=25$ ✓, $f(-3,4)=-9-16=-25$ ✓. And $3^2+(-4{)}^2=25$ so $(3,-4)$ is on the circle ✓. The whole answer lived on the fence — the empty-interior case handled cleanly.

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Example 6 — Non-differentiable point inside (Cell C6)

Forecast: $f$ is just the distance from the origin. Its lowest value is obviously at the center — but is the center a "$\nabla f=0$" point? Careful.

Step 1 — interior. Compute partials away from the origin: $f_x=\frac{x}{\sqrt{x^2+y^2}},f_y=\frac{y}{\sqrt{x^2+y^2}}.$ Setting both $=0$ needs $x=0,y=0$ — but there the formula divides by zero. Why this matters: at $(0,0)$ the graph is a sharp cone tip; $f$ is not differentiable there, so $\nabla f$ doesn't exist. The recipe explicitly says: points where $f$ isn't differentiable are also candidates. Record $(0,0)$: $f=0$.

Step 2 — boundary (four edges). Why? Distance grows outward, so the max is on the fence. By symmetry each edge is like $x=1,y\in [-1,1]$: $f=\sqrt{1+y^2}$, minimized at $y=0$ ($=1$), maximized at $y=\pm 1$ ($=\sqrt{2}$). Corners $(\pm 1,\pm 1)$ give $\sqrt{2}$.

Step 3 — compare. Values $\{0,1,\sqrt{2}\}$. $\boxed{\text{abs max}=\sqrt{2}\text{ at the four corners};\text{abs min}=0\text{ at }(0,0).}$

Verify: $\sqrt{2}\approx 1.414$; corner $(1,1)$ has distance $\sqrt{1+1}=\sqrt{2}$ ✓. The min sat at a non-differentiable point that a blind "$\nabla f=0$" search would have skipped — the whole point of Cell C6. ✓

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Example 7 — A four-way tie (Cell C7)

Forecast: Distance-squared from origin. Lowest at center, highest at... how many corners tie?

Step 1 — interior. $f_x=2x=0,f_y=2y=0\Rightarrow (0,0)$, inside, $f=0$. Smooth here (no cone — the square is inside the square root's square, so $f$ is a genuine paraboloid, differentiable everywhere).

Step 2 — boundary. Each edge $x=1$: $f=1+y^2$, max $2$ at $y=\pm 1$, min $1$ at $y=0$. All four corners $(\pm 1,\pm 1)$ give $2$.

Step 3 — compare. $\{0,1,2\}$. $\boxed{\text{abs max}=2\text{ at all four corners};\text{abs min}=0\text{ at }(0,0).}$

Verify: The max is attained at four distinct points simultaneously — a genuine tie. $f(1,1)=1+1=2$ ✓. Symmetry of the square guarantees ties; reporting only one corner would be an incomplete answer.

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Example 8 — Word problem, carry units (Cell C8)

Forecast: Heating grows with $x$ then $-x^2$ pulls it back; the $2xy$ term rewards big $x$ and big $y$ together. Guess the hot corner before computing.

Step 1 — interior. $T_x=4-2x+2y=0,T_y=2x=0.$ From $T_y$: $x=0$. Then $T_x$: $4+2y=0\Rightarrow y=-2$. Point $(0,-2)$ — outside the plate ($y\ge 0$). Why discard? Not in $D$; no usable interior critical point (echoes Cell C4).

Step 2 — four edges (positions in cm, temperatures in ${}^{\circ }$C). Why this step? With no interior candidate, the hottest and coldest spots must sit on the plate's rim.

• Bottom $y=0$ cm, $x\in [0,3]$ cm: $T=20+4x-x^2$, $T^{\prime }=4-2x=0\Rightarrow x=2$ cm: $T=20+8-4=24{}^{\circ }$C. Ends: $T(0,0)=20{}^{\circ }$C, $T(3,0)=20+12-9=23{}^{\circ }$C.
• Top $y=2$ cm, $x\in [0,3]$ cm: $T=20+4x-x^2+4x=20+8x-x^2$, $T^{\prime }=8-2x=0\Rightarrow x=4$ cm (outside $[0,3]$, discard). Ends: $T(0,2)=20{}^{\circ }$C, $T(3,2)=20+24-9=35{}^{\circ }$C.
• Left $x=0$ cm, $y\in [0,2]$ cm: $T=20{}^{\circ }$C (constant — the $x$-terms all vanish).
• Right $x=3$ cm, $y\in [0,2]$ cm: $T=20+12-9+6y=23+6y$, increasing in $y$. Ends: $T(3,0)=23{}^{\circ }$C, $T(3,2)=35{}^{\circ }$C.

Step 3 — compare the candidate temperatures $\{24,20,23,35\}{}^{\circ }$C. $\boxed{\text{hottest}=35{}^{\circ }\text{C}\text{ at }(3,2)\text{ cm};\text{coldest}=20{}^{\circ }\text{C}\text{ along the edge }x=0.}$

Verify: Units check — $x,y$ enter in cm and the constant $20$ plus every term is calibrated in ${}^{\circ }$C, so the output is a temperature. $T(3,2)=20+12-9+12=35{}^{\circ }$C ✓. $T(0,y)=20{}^{\circ }$C for all $y$ ✓ (the cold "spot" is the whole left edge — a Cell C7 tie hiding inside a word problem). Hottest corner matched the forecast: big $x$, big $y$. ✓

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Example 9 — Exam twist: quadrant signs on the circle (Cell C9)

Forecast: A tilted plane again → boundary only. The gradient $(1,2)$ points into the first quadrant (both components positive), so the max should sit up-and-to-the-right. Let's nail the exact point and confirm its quadrant sign-by-sign.

Figure s03 — what it shows and how to read it. The black circle is the boundary $x^2+y^2=5$. The orange arrow is $\nabla f=(1,2)$, drawn from the origin; its tip lands exactly on the boundary point $(1,2)$ (orange dot) — both coordinates positive → Quadrant I, the maximum. The teal dot at the antipode $(-1,-2)$ has both coordinates negative → Quadrant III, the minimum. The plum dotted lines are level sets $x+2y=c$; they run perpendicular to the gradient, so sliding along the circle raises $f$ fastest exactly when you head along the arrow. The picture makes the sign argument visual: the winner is where the position vector points the same way as $\nabla f$.

Step 1 — interior. $f_x=1\ne 0,f_y=2\ne 0.$ Why this step? Both partials are constant and nonzero, so $\nabla f=(1,2)$ never becomes $(0,0)$: no interior critical point (Cell C5 flavour). All candidates are on the fence.

Step 2 — boundary $x^2+y^2=5$. Parametrize $x=\sqrt{5}\cos t,y=\sqrt{5}\sin t$ with $t\in [0,2\pi ]$: $f=\sqrt{5}\cos t+2\sqrt{5}\sin t=\sqrt{5}(\cos t+2\sin t).$ Amplitude of $\cos t+2\sin t$ is $R=\sqrt{1^2+2^2}=\sqrt{5}$, and over the full period it hits $\pm R$, so $f\in [-\sqrt{5}\cdot \sqrt{5},\sqrt{5}\cdot \sqrt{5}]=[-5,5].$ The max occurs when $(x,y)$ points along $\nabla f=(1,2)$: $(x,y)=\sqrt{5}\cdot \frac{(1,2)}{\sqrt{5}}=(1,2).$

Sign-by-sign analysis (the C9 point). At the maximum, $x=1>0$ and $y=2>0$ — a $(+,+)$ pair, so it lives in Quadrant I. The minimum is the exact opposite point $(-1,-2)$: here $x<0$ and $y<0$, a $(-,-)$ pair → Quadrant III. The general rule this reveals: for a linear $f=ax+by$ on a disk, the maximum sits in the quadrant whose sign pattern matches $(a,b)=\nabla f$, and the minimum in the quadrant with the opposite signs.

Step 3 — compare. $\boxed{\text{abs max}=5\text{ at }(1,2)(\text{Q I});\text{abs min}=-5\text{ at }(-1,-2)(\text{Q III}).}$

Verify: $(1,2)$ on circle: $1+4=5$ ✓. $f(1,2)=1+4=5$ ✓, $f(-1,-2)=-1-4=-5$ ✓. Compare with Lagrange multipliers: the condition $\nabla f=\lambda \nabla g$ says $(1,2)=\lambda (2x,2y)$, i.e. $(x,y)\parallel (1,2)$ — same answer, different route, same sign conclusion.

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Checklist: did we stamp every cell?

Recall Matrix coverage

C1 rectangle+corners ::: Example 1 C2 slanted edge ::: Example 2 C3 disk parametrize ::: Example 3 C4 interior point outside $D$ ::: Example 4 C5 no interior critical point ::: Example 5 C6 non-differentiable point ::: Example 6 C7 tie / multiple winners ::: Examples 3, 7, 8 C8 word problem with units ::: Example 8 C9 quadrant-sign twist ::: Example 9

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Connections

• Absolute extrema on closed bounded regions — the parent recipe every example follows.
• Critical points and gradient — Step 1 solves $\nabla f=0$ (Ex 1, 3, 4, 8).
• Second derivative test (Hessian) — deliberately unused: we compare values, not classify (note Ex 3's saddle we never labelled).
• Lagrange multipliers — alternate boundary handling, shown in Ex 9.
• Extreme Value Theorem (1D) — guarantees the 1-var edge sub-problems have their own max/min.
• Closed and bounded sets (topology) — why the guarantee holds on every $D$ above.
• Parametrization of curves — turning circles into single-dial $t$ problems (Ex 3, 4, 5, 9).