Absolute extrema on closed bounded regions

4.4.14 · Maths › Multivariable Calculus

YEH KAAM KARTA HI KYU HAI

"Closed" AUR "Bounded" dono kyun chahiye?

Bounded hata do: $D = R^{2}$ par, $f = x$ ka koi max nahi — yeh $+ \infty$ tak bhaag jaata hai.
Closed hata do: open disk $x^{2} + y^{2} < 1$ par, $f = x$ kabhi apna sup $1$ tak nahi pahunchta (boundary point exclude hai).

Dono conditions "escape routes" band karti hain — value infinity tak nahi bhaag sakti, aur kisi missing edge ki taraf chhup nahi sakti.

Yahan second-derivative test kyun nahi? Kyunki hamare paas candidate points ki ek finite list hai aur hum bas actual values compare karte hain. Saddle vs. extremum classify karne ki zaroorat hi nahi — direct comparison jeet jaata hai.

Figure — Absolute extrema on closed bounded regions

Worked Example 1 — ek rectangle par

$f (x, y) = x^{2} - 2 x y + 2 y$ ke absolute extrema $D = {0 \leq x \leq 3, 0 \leq y \leq 2}$ par dhundho.

Step 1 — interior. $f_{x} = 2 x - 2 y = 0, f_{y} = - 2 x + 2 = 0.$ Yeh step kyun? Interior extrema ko ek flat tangent plane chahiye, toh dono partials ko $0$ set karo. $f_{y}$ se: $x = 1$ . $f_{x}$ se: $y = x = 1$ . Point $(1, 1)$ $D$ ke andar hai. ✓ $f (1, 1) = 1 - 2 + 2 = 1$ .

Step 2 — boundary (4 edges). Kyun? Fence extremum hold kar sakti hai chahe interior flat-free ho.

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

Step 3 — compare. Candidate values: ${1, 0, 9, 0, 4, 0, 4, 9, 1}$ . $abs max = 9 at (3, 0); abs min = 0 at (0, 0) and (2, 2) .$ Yeh step kyun? Hum finite candidate list mein se bas sabse bada aur sabse chhota pick karte hain.

Worked Example 2 — ek disk par (Lagrange / parametrize use karo)

$f (x, y) = x^{2} + y^{2} - x - y + 1$ ke absolute extrema disk $x^{2} + y^{2} \leq 1$ par dhundho.

Step 1 — interior. $f_{x} = 2 x - 1 = 0 \Rightarrow x = \frac{1}{2}, f_{y} = 2 y - 1 = 0 \Rightarrow y = \frac{1}{2} .$ Kya $(\frac{1}{2}, \frac{1}{2})$ andar hai? $\frac{1}{4} + \frac{1}{4} = \frac{1}{2} < 1$ ✓. $f (\frac{1}{2}, \frac{1}{2}) = \frac{1}{4} + \frac{1}{4} - \frac{1}{2} - \frac{1}{2} + 1 = \frac{1}{2} .$

Step 2 — boundary $x^{2} + y^{2} = 1$ . $x^{2} + y^{2} = 1$ substitute karo: $f = 1 - x - y + 1 = 2 - (x + y) .$ Yeh step kyun? Constraint quadratic terms ko khatam kar deta hai, ek clean linear cheez bachti hai. Parametrize karo $x = cos t, y = sin t$ : $f = 2 - (cos t + sin t) = 2 - 2 sin (t + \frac{π}{4})$ . Kyunki $sin$ $[- 1, 1]$ mein range karta hai: $f_{m a x}^{b d r y} = 2 + 2 (at x = y = - \frac{1}{2}), f_{m i n}^{b d r y} = 2 - 2 .$

Step 3 — compare. Values: interior $0.5$ , boundary $2 - 2 \approx 0.586$ , $2 + 2 \approx 3.414$ . $abs max = 2 + 2; abs min = \frac{1}{2} at (\frac{1}{2}, \frac{1}{2}) .$

Common Mistakes

Recall Feynman: ek 12-saal ke bacche ko samjhao

Socho ek ऊबड़-खाबड़ trampoline ek fenced backyard ke upar khinchi gayi hai. Tum trampoline ke sabse oopar aur sabse neeche ke points chahte ho. Sirf do tarah ke spots ho sakte hain jo sabse oopar/neeche ho sakte hain: ya toh yard ke andar koi flat dimple, ya phir kahin fence ke saath. Toh tum saare flat dimples mark karo, phir poori fence ke saath chalo aur wahan ke bumps mark karo, aur finally ek kadam peeche ho ke jo tumne mark kiya uska sabse upar aur sabse neeche point dhundho. Ho gaya!

Flashcards

Kaunsa theorem guarantee karta hai ki absolute max & min exist karta hai?

Extreme Value Theorem —

f

ke liye jo closed aur bounded set

D

par continuous ho.

D

mein guaranteed extrema ke liye kaunsi do properties chahiye?

Closed (boundary contain karta ho) aur bounded (ek disk mein fit ho).

Interior extremum

\nabla f = 0

kyun force karta hai?

Agar

\nabla f \neq = 0

ho toh tum

+ \nabla f

ke along move karke

f

badha sakte ho, toh woh max tha hi nahi (similarly min ke liye).

Absolute extrema ke liye 3-step recipe kya hai?

(1) Interior critical points

\nabla f = 0

D

ke andar; (2) boundary par

f

ko 1-var mein reduce karo, crit pts + endpoints/corners dhundho; (3) saari values compare karo.

Kya hum absolute extrema ke liye Hessian second-derivative test use karte hain?

Nahi — hum sirf finite candidate list par actual function values compare karte hain.

Disk

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

ki boundary par

f

ko simplify karne ki trick kya hai?

Constraint substitute karo aur/ya

x = cos t, y = sin t

parametrize karo taaki yeh 1-variable ho jaaye.

Corners/endpoints ko candidates mein kyun include karna zaroori hai?

Closed interval par ek 1-var function apne extrema endpoints par attain kar sakta hai, sirf jahan derivative zero ho wahan nahi.

Example 1 mein absolute max kahan tha aur uski value kya thi?

(3, 0)

par

9

, boundary par (interior critical point par nahi).

Connections

Critical points and gradient — interior step $\nabla f = 0$ solve karta hai.
Second derivative test (Hessian) — local classification ke liye use hota hai, yahan nahi.
Lagrange multipliers — boundary constraint handle karne ka alternate tarika.
Extreme Value Theorem (1D) — is method ka single-variable ancestor.
Closed and bounded sets (topology) — guarantee ke liye dono kyun chahiye.
Parametrization of curves — boundary ko 1-var problem mein badalna.

YEH KAAM KARTA HI KYU HAI

KAISE KAREIN: algorithm