Absolute extrema on closed bounded regions: Question bank

4.4.14 · D5 · Maths › Multivariable Calculus › Absolute extrema on closed bounded regions

Neeche har reveal ek poora reasoning sentence hai — kabhi bare "true" ya "false" nahi. Point hai kyun.

Setup jo traps se pehle chahiye

Questions samajhne se pehle, teen words ko precisely pin down karna zaroori hai.

Figure — Absolute extrema on closed bounded regions

True ya false — justify karo

Kisi bhi bounded set par ek continuous function absolute max attain karta hai.

False — bounded enough nahi hai; set closed bhi hona chahiye. Open disk

x^{2} + y^{2} < 1

par,

f = x

approaches

1

but kabhi reach nahi karta kyunki boundary point missing hai. Dekho Closed and bounded sets (topology).

Kisi bhi closed set par ek continuous function absolute max attain karta hai.

False — closed enough nahi hai; ye bounded bhi hona chahiye. Closed set

R^{2}

par,

f = x

+ \infty

tak bhaag jaata hai, koi maximum nahi.

Agar

D

ke interior mein har jagah

\nabla f \neq = 0

hai, toh

f

D

par koi absolute extrema nahi hain.

False — iska sirf matlab hai ki extrema interior mein nahi hain; woh phir bhi exist karte hain aur boundary par hona chahiye, Extreme Value Theorem ke guarantee se (jo closed bounded sets ke liye har dimension mein hold karta hai).

D

par

f

ka absolute maximum hamesha ek interior critical point hota hai.

False — ye boundary par baith sakta hai. Upar restate kiye gaye Worked Example 1 mein, max value

9

corner

(3, 0)

par aayi, na ki interior critical point

(1, 1)

par.

Jab ek single interior critical point mil jaaye, tum ruk sakte ho.

False — tumhe poori boundary walk karni hogi aur compare karna hoga, kyunki ek boundary point interior candidate ko beat kar sakta hai; interior critical point sirf candidate list mein ek entry hai.

Agar

f

poore

D

par differentiable hai, toh saare candidate points

\nabla f = 0

solve karne se aate hain.

False — differentiability interior mein "corner/kink" candidates ko rule out karti hai, lekin boundary phir bhi candidates provide karti hai (uske apne critical points, plus endpoints aur corners).

Yeh confirm karne ke liye ki kaun sa candidate absolute max hai, Hessian second-derivative test zaroori hai.

False — absolute extrema ke liye tum simply candidates ki finite list mein

f

ke numerical values compare karte ho; Second derivative test (Hessian) sirf local interior behaviour classify karta hai aur boundary winner ko galat bhi label kar sakta hai.

Har woh point jahan

D

ke andar

\nabla f = 0

ho,

f

D

par absolute extremum hai.

False — ek critical point saddle ya sirf ek local feature ho sakta hai; ye sirf ek candidate hai. Iska status comparison se decide hota hai, na is baat se ki ye critical hai.

Disk boundary par,

x^{2} + y^{2} = 1

f

mein substitute karna boundary analysis khatam kar deta hai.

False — substitution

f

ko simplify karta hai lekin ek remaining freedom (angle) phir bhi vary karta hai; tumhe uss ek variable par optimise karna hoga, e.g.

x = cos t, y = sin t

via Parametrization of curves.

Ek function do alag points par apna absolute minimum attain kar sakta hai.

True — ties forbidden nahi hain. Restate kiye gaye Worked Example 1 mein minimum value

0

dono

(0, 0)

aur

(2, 2)

par attain hua tha.

Agar

D

mein ek hole hai (ek non-simply-connected region), tum sirf outer fence walk karo.

False — holed region ki boundary ke do parts hote hain, outer fence aur hole ke around inner rim. Dono closed

D

ki boundary ka hissa hain, isliye extrema inner rim par chhup sakte hain; tumhe har boundary component walk karni hogi.

Error dhundho

f = x^{2} - 2 x y + 2 y

ka interior critical point

(1, 1)

hai jahan

f = 1

hai, isliye absolute minimum

1

hai." Galti kahan hai?

Interior value

1

ko kabhi boundary values se compare nahi kiya gaya; true minimum

0

hai (boundary par), isliye

1

ko minimum kehna Step 3, comparison ko poora skip kar deta hai.

"Edge

y = 2

par,

f = (x - 2)^{2}

ka critical point

x = 2

hai jahan

f = 0

hai, isliye edge ki extreme values sirf

0

hain." Kya galat hai?

Sirf edge ka interior critical point check kiya gaya; endpoints

x = 0

aur

x = 3

(

f = 4

aur

f = 1

) drop ho gaye, aur unme se ek edge ka maximum hai.

f = x^{2} + y^{2}

x^{2} + y^{2} \leq 4

: boundary par

f = x^{2} + y^{2} = 4

constant hai, isliye boundary kuch contribute nahi karta." Kya ye reasoning safe hai?

Conclusion (boundary value constant

4

hai) yahan correct hai, lekin phrase "contributes nothing" dangerous hai — woh constant

4

phir bhi ek candidate hai aur absolute max nikalta hai; tumhe use comparison mein include karna hoga.

"Gradient saddle point par zero hota hai, isliye saddle kabhi absolute extremum nahi ho sakta." Sahi hai?

Galat reasoning — saddle local extremum nahi hota, lekin ek restricted region par ye phir bhi sabse badi ya sabse chhoti seen value tie kar sakta hai; chahe kuch bhi ho, hum classify nahi karte, hum values compare karte hain.

"Maine boundary ko

x = cos t, y = sin t

for

t \in [0, π]

se parametrise kiya aur extremes dhundh liye." Kya miss kiya?

Aadha circle — poori boundary ke liye

t \in [0, 2 π]

chahiye;

π

par rokna silently lower semicircle discard kar deta hai jahan ek extremum chhup sakta hai.

"Square

D

par maine har edge ke extremes dhundhe lekin chaar corners ko kisi bhi edge se belong nahi maana, isliye maine unhe ignore kiya." Fix?

Har corner do edges ka shared endpoint hai aur mandatory candidate hai; corners ignore karna answer miss kar sakta hai, jaise Worked Example 1 mein corner

(3, 0)

par max dikhata hai.

"Mera region ek annulus hai (ek disk jisme ek chhoti disk removed hai). Maine interior par aur outer circle par optimise kiya." Kya missing hai?

Removed hole ko bound karne wala inner circle bhi boundary ka hissa hai; use skip karna ek aisa extremum miss kar sakta hai jo hole ke rim par ho.

Why questions

Interior extremum

\nabla f = 0

kyun force karta hai?

Agar

\nabla f \neq = 0

ho toh tum direction

+ \nabla f

mein step le sakte ho

f

increase karne ke liye (ya

- \nabla f

mein decrease karne ke liye), jo contradict karta hai ki woh point max (ya min) tha; dekho Critical points and gradient.

Hum boundary ko single-variable problem mein kyun reduce karte hain instead of

\nabla f = 0

use karne ke?

Boundary par

f

ek curve par constrained hai, isliye uska unconstrained gradient generally wahan zero nahi hota; parametrising use ek free variable mein collapse kar deta hai jahan ordinary 1-variable calculus apply hoti hai, ya tum Lagrange multipliers use karo.

Hum har candidate ko classify karne ke bajaye actual values compare kyun karte hain?

Hamare paas ek finite candidate list hai, aur absolute max/min simply un actual numbers mein se sabse bada/chhota hai — classification (saddle vs. extremum) ek alag, local question ka jawab hai.

D

closed aur bounded hona kyun zaroori hai, sirf ek nahi?

Bounded value ko

\pm \infty

tak bhaagne se rokta hai; closed use ek missing boundary point ki taraf sneaking karne se rokta hai — guarantee hold karne ke liye dono escape routes band hone chahiye.

Extreme Value Theorem ke liye continuity essential kyun hai?

Ek jump ya spike

f

ko ek aisi value approach karne deta hai jo kabhi attain nahi hoti (e.g. peak par ek removed point); continuity aisi gaps forbid karti hai isliye supremum actually reach hota hai.

Recipe ko Hessian kyun nahi chahiye chahe woh local-extrema chapters mein appear kare?

Hessian ka jawab hai "kya ye critical point local max/min/saddle hai?"; absolute problem instead poochhta hai "kaun se candidate ki sabse badi value hai?", aur woh direct comparison se settle hota hai.

Holed region boundary step ko longer kyun banata hai?

Kyunki "boundary" ka matlab hai

D

ke saare edges ek saath — ek hole poora ek extra inner curve add kar deta hai, isliye tumhe har separate boundary component par parametrise aur optimise karna hoga, phir har result ko comparison mein include karna hoga.

Edge cases

Agar

f

poore

D

par constant hai, toh uske absolute extrema kya hain?

Har point simultaneously absolute max aur absolute min hai; extreme value woh single constant hai, jo har jagah attain hoti hai.

Agar interior critical point

D

ke bahar ho toh?

Tum use discard karte ho — sirf woh interior critical points $D$ ke andar candidates hain;

D

ke baahir

\nabla f = 0

ka solution is problem ke liye irrelevant hai.

Agar interior critical point exactly

D

ki boundary par ho toh?

Ye automatically boundary analysis mein capture ho jaata hai, isliye tum use wahan handle karo; ye phir bhi ek valid candidate hai, bas ek boundary point count hota hai.

Kya ek single point (degenerate region

D = {P}

) ke absolute extrema ho sakte hain?

Haan — ek-point set closed aur bounded hai, aur

f (P)

trivially dono max aur min hai; theorem phir bhi apply hota hai.

Agar

f

D

ke andar ek corner (non-differentiable point) ho toh?

Tab

\nabla f

wahan exist nahi kar sakta, isliye woh point candidate list mein ek non-differentiable interior candidate ki tarah add hota hai, un points ke saath jahan

\nabla f = 0

hai.

Agar do adjacent boundary edges ek corner par milti hain jahan

f

smooth hai, toh bhi kya corner list karna zaroori hai?

Haan —

f

ki smoothness is fact ko nahi hataati ki har edge ek closed interval hai jiska endpoint (corner) extreme value hold kar sakta hai.

Ek annulus

1 \leq x^{2} + y^{2} \leq 4

ke liye, kitni boundary curves walk karni hain?

Do — outer circle

x^{2} + y^{2} = 4

aur inner circle

x^{2} + y^{2} = 1

; dono region ko bound karti hain, isliye har ek ko parametrise aur search karna hoga.

Recall Ek-line survival checklist

$D$ ke andar interior critical points ➜ har boundary component (outer fence aur koi bhi inner rims) ko 1-var mein reduce karo saare endpoints/corners ke saath ➜ har value compare karo; kabhi classify mat karo, kabhi fence skip mat karo, kabhi corner drop mat karo.

Connections

Absolute extrema on closed bounded regions — woh parent method jise ye traps stress-test karti hain.
Critical points and gradient — interior candidates ka source.
Second derivative test (Hessian) — woh tool jiske misuse ke against ye traps warn karti hain.
Lagrange multipliers — alternative boundary handling.
Extreme Value Theorem (1D) — existence guarantee; har dimension mein identical hypotheses.
Closed and bounded sets (topology) — kyun dono hypotheses matter karte hain, aur holed regions ke liye "boundary" ka kya matlab hai.
Parametrization of curves — har fence ko 1-var problem mein convert karna.