Visual walkthrough — Absolute extrema on closed bounded regions

4.4.14 · D2 · Maths › Multivariable Calculus › Absolute extrema on closed bounded regions

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

Har symbol ka naam pehle bata deta hoon, phir use karte hain.

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Step 1 — Terrain aur fence dekho

KYA. Koi bhi algebra se pehle, dekho hum kis cheez par khade hain. Rectangle $D$ ek flat floor plan hai; uske upar ek curved surface hai jiska height $f$ hai.

KYUN. Extreme Value Theorem ka 2D wala version promise karta hai ki ek continuous height ek closed, bounded field par zaroor kisi jagah ek genuine highest aur lowest value pe pahunchegi. Hamara $f$ ek polynomial hai, isliye yeh har jagah smooth hai — koi gaps nahi, koi cliffs nahi — toh promise hold karta hai. Hum hunting jaane ke liye allowed hain; guarantee hai ki koi khazana milega zaroor.

PICTURE. Floor rectangle hai; colored surface height hai. Notice karo yeh dips aur rises karta hai — humara kaam hai extreme dips aur rises dhundhna bina infinitely many points check kiye.

Poori strategy: top/bottom sirf do tarah ki jagah par chhup sakta hai — fence ke andar ek flat dimple, ya fence par kahi. Toh hum sirf in dono ko search karte hain, aur kahi nahi.

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Step 2 — Andar ke flat spots dhundho (the gradient)

KYA. Field ke andar, ek highest ya lowest point par ground locally flat honi chahiye — surface wahan ek level tangent plane rakha hoga. "Dono directions mein flat" matlab dono partial slopes zero hain.

Gradient kyun? Agar gradient arrow kisi jagah zero nahi hota, tum us direction mein ek tiny step le sakte ho aur higher climb kar sakte ho — toh woh jagah summit nahi ho sakta tha. Flat interior extrema isliye force karte hain $\nabla f=0$. (Dekho Critical points and gradient.)

Hamare $f=x^2-2xy+2y$ ke liye:

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

Solve karo. Doosri equation deti hai $-2x+2=0\Rightarrow x=1$. Pehli mein daalo: $2(1)-2y=0\Rightarrow y=1$. Ek flat spot: $(1,1)$.

Kya yeh fence ke andar hai? $0<1<3$ aur $0<1<2$ — haan, strictly andar. ✓ Uski height: $f(1,1)=1^2-2(1)(1)+2(1)=1-2+2=1.$

PICTURE. Gradient arrows sab flat point se dur flow karte hain; bilkul $(1,1)$ par arrow zero ho jaata hai.

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Step 3 — Bottom edge $y=0$ par chalo

KYA. Ab hum fence par chalte hain. Bottom edge par, $y$ $0$ par stuck hai, aur sirf $x$ $0$ se $3$ tak move kar sakta hai. Ek variable freeze karna surface ko ek plain 1-variable curve mein badal deta hai (dekho Parametrization of curves).

KYUN. Boundary points ka ek poora curve hai jahan interior ka "flat" rule apply nahi hota — fence ek extreme pin kar sakta hai jab wahan ground sloped ho. Isliye hum $f$ ko har edge par reduce karke ek single-variable function mein karna zaroori hai aur ordinary 1-D max/min use karna hoga (from Extreme Value Theorem (1D)).

$y=0$ substitute karo:

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

$g(x)=x^2$ $[0,3]$ par sirf increase karta hai (uska slope $2x$ positive hai). Toh is edge par koi interior critical point nahi — extremes dono endpoints par hain: $g(0)=f(0,0)=0,g(3)=f(3,0)=9.$

PICTURE. Bottom edge ek rising parabola mein lift hui; max right corner par hai.

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Step 4 — Top edge $y=2$ par chalo

KYA. $y=2$ freeze karo; $x$ ko $0$ se $3$ tak slide karo.

KYUN. Same reason — 1-D reduce karo. Yeh edge zyada interesting hai kyunki reduced curve ka actually interval ke andar ek low point hai.

$y=2$ substitute karo:

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

Tidy form $(x-2{)}^2$ keh raha hai: ek bowl jo $x=2$ par bottom karta hai, height $0$. Check karo ki yeh edge ka interior critical point hai: derivative $2(x-2)=0\Rightarrow x=2$, aur $0<2<3$ ✓. Phir endpoints: $h(2)=f(2,2)=0,h(0)=f(0,2)=4,h(3)=f(3,2)=1.$

PICTURE. Top edge ek bowl mein lift hui; uski lowest touch $(2,2)$ par hai, height $0$.

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Step 5 — Left aur Right edges par chalo

KYA. Do vertical edges bache hain. Left edge $x=0$ par, sirf $y$ move karta hai; right edge $x=3$ par, sirf $y$ move karta hai.

KYUN. Same 1-D reduction; ab doosra variable frozen hai.

Left, $x=0$:

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

Ek seedha climbing line — extremes ends par: $f(0,0)=0$, $f(0,2)=4$.

Right, $x=3$:

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

Ek seedha descending line (slope $-4<0$) — extremes ends par: $f(3,0)=9$, $f(3,2)=1$.

PICTURE. Left edge upar jaati hai, right edge neeche — dono seedhi ramps hain, isliye sirf unke endpoints matter karte hain.

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Step 6 — Har candidate ko compare karo

KYA. Saari recorded heights ikatthi karo aur simply sabse badi aur choti padho.

KYUN. Ab hamare paas finite list of suspects hai. Koi classifying nahi, koi Hessian nahi — direct comparison sab decide karta hai. (Comparing, classifying se better hai: winner ek boundary corner ho sakta hai jise Hessian score hi nahi kar sakta.)

Kahan	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	$(3,2)$	$1$

Sabse badi height $9$ hai; sabse choti $0$ hai (do baar achieve hoti hai). $\boxed{\text{abs max}=9\text{ at }(3,0);\text{abs min}=0\text{ at }(0,0)\text{ and }(2,2).}$

PICTURE. Floor plan jisme har candidate dotted hai; winner (max) aur dono lowest (min) circled hain.

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Step 7 — Degenerate cases (taaki tum kabhi hairan na ho)

KYA. Agar field ya function unusual ho toh kya? Teen quick pictures edge behaviours ki.

KYUN. Contract yeh hai: har scenario dikhaya jaana chahiye, taaki tum kisi aisi cheez se na takrao jiske baare mein bataaya nahi gaya.

• Flat spot fence ke bahar land kare. Agar $\nabla f=0$ solve karne par, maan lo $(5,5)$ aata, tum use discard kar doge — yeh $D$ mein nahi hai. Tab extrema poori tarah boundary par hote hain.
• Koi interior flat spot hi nahi. Kuch $f$ mein $\nabla f\ne 0$ har jagah hota hai (e.g. $f=x$). Tab Step 2 kuch nahi deta aur saare candidates fence se aate hain.
• Ties. Hamara min $0$ do points par hota hai, $(0,0)$ aur $(2,2)$. Yeh allowed hai — "the" minimum ek value hai; yeh kai jagah attain ho sakti hai.

PICTURE. Left: ek bahar wala critical point crossed out. Middle: koi flat spot nahi wala tilted plane. Right: same lowest height par do alag spots.

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Ek-picture summary

Upar sab kuch ek single frame mein: field, ek interior flat spot, char edges har ek 1-D curve mein reduced, winner par star ($9$) aur tied minima par dots ($0$).

Recall Feynman: poora walk simple shabdon mein

Hamare paas ek bumpy sheet thi jo ek rectangular fenced yard par stretch thi, aur hum uske highest aur lowest points chahte the. Pehle humne find kiya jahan sheet yard ke andar flat hai — "uphill kis taraf hai?" poochh ke — the gradient — aur woh ek jagah dhundh ke jahan uphill kahi point nahi karta: $(1,1)$, height $1$. Phir hum fence ki charon sides par chale. Har side par ek coordinate frozen hai, isliye sheet ek simple 1-D curve ban jaati hai: bottom ek rising parabola thi (corner $(3,0)$ par peak, height $9$), top ek bowl tha ($(2,2)$ par $0$ touch karta hai), left ek rising ramp tha, right ek falling ramp — ramps sirf apne ends par matter karte hain. Humne har flat spot, har corner aur har edge-dip par height likh li. Aakhir mein humne sirf list dekhi aur point kiya: highest $9$ hai $(3,0)$ par; lowest $0$ hai, aur yeh do jagah hota hai, $(0,0)$ aur $(2,2)$. Koi fancy tests nahi — bas andar search karo, fence par chalo, aur compare karo.

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Connections

• Critical points and gradient — Step 2 ka flat-spot hunt.
• Extreme Value Theorem (1D) — har edge par use hone wala 1-D max/min.
• Parametrization of curves — har edge par ek variable freeze karna.
• Second derivative test (Hessian) — yahan deliberately use nahi kiya gaya.
• Lagrange multipliers — curved boundaries ke liye ek alternative.
• Closed and bounded sets (topology) — isliye poora hunt succeed hone ki guarantee hai.