Absolute extrema on closed bounded regions: Worked examples

4.4.14 · D3 · Maths › Multivariable Calculus › Absolute extrema on closed bounded regions

Yeh page workout room hai teen-step recipe ke liye. Parent note ne aapko do clean examples diye the. Real problems zyada messy hote hain: region ke corners ho sakte hain, ya poora disk ho sakta hai, ya boundary kisi aisi jagah touch kar sakti hai jahan $f$ misbehave karta ho. Neeche pehle hum har tarah ke case ka ek map draw karte hain jo yeh topic aap par throw kar sakta hai, phir itne examples karte hain ki map ka har square stamped ho jaye.

The scenario matrix

Har absolute-extrema problem region shape aur twist ka koi na koi combination hota hai. Yeh raha poora grid — har cell ek alag cheez hai jo aapko trip kar sakti hai.

Cell	Region / feature	Twist jo introduce hota hai	Covered in
C1	Rectangle (polygon)	Chaar seedhi edges + chaar corners mandatory candidates ke roop mein	Ex 1
C2	Triangle (slanted edge)	Ek boundary edge jo horizontal/vertical nahin hai → ek line substitute karo	Ex 2
C3	Disk (smooth curve)	Boundary ek circle hai → parametrize karo $x = cos t, y = sin t$ , sab $t$ ke liye	Ex 3
C4	Interior critical point outside $D$	Step 1 ek aisa point deta hai jise aapko discard karna hai	Ex 4
C5	Degenerate: koi bhi interior critical point nahin	$\nabla f = 0$ ka koi solution nahin → sab kuch fence par hi hota hai	Ex 5
C6	$D$ ke andar non-differentiability ka point	Ek sharp cone/valley jahan $\nabla f$ exist nahin karta lekin phir bhi candidate hai	Ex 6
C7	Tie / multiple winners	Max (ya min) kai points par ek saath achieve hota hai	Ex 7
C8	Real-world word problem	"Temperature on a plate" ko recipe mein translate karo, units carry karo	Ex 8
C9	Exam twist: constraint quadrant signs ko hit karta hai	Circle par winner ek specific quadrant mein baithta hai; $cos t, sin t$ ka sign matter karta hai	Ex 9

Hum Examples 1–9 neeche work karte hain; "Covered in" column aapki checklist hai.

Recall Recipe, ek line mein (parent se)

Interior ( $\nabla f = 0$ andar, plus non-differentiable points) → Boundary (1 variable mein reduce karo, endpoints/corners include karo) → Compare sari values.

Shuru karne se pehle, teen symbols jo parent ne use kiye — inhe zero se pin down karte hain:

Example 1 — Rectangle with four corners (Cell C1)

Forecast (pehle andaza lagao!): $- 2 x$ term bowl ko tilt karta hai. Aapke hisaab se lowest point kahan hoga — andar, kisi edge par, ya corner par? Aage padhne se pehle apna guess likho.

Step 1 — interior critical points. $f_{x} = 2 x - 2 = 0 \Rightarrow x = 1, f_{y} = 2 y = 0 \Rightarrow y = 0.$ Yeh step kyun? Interior extrema ke liye uphill arrow ka vanish hona zaroori hai; $\nabla f = 0$ solve karne se flat spots milte hain. Point $(1, 0)$ $D$ ke andar hai ( $- 1 < 1 < 2$ , $- 1 < 0 < 2$ ) ✓. Value $f (1, 0) = 1 + 0 - 2 = - 1$ .

Step 2 — charon edges par chalo (har ek 1-variable job hai). Yeh step kyun? Bowl ka tilt highest value ko fence par push kar sakta hai; hame charon sides check karni hongi.

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

Charon corners upar edge endpoints ke roop mein aate hain: $(- 1, - 1) \to 4$ , $(2, - 1) \to 1$ , $(- 1, 2) \to 7$ , $(2, 2) \to 4$ .

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

Verify: Min $- 1$ interior critical value hai; max woh corner hai jo low point $(1, 0)$ se sabse door hai. $(1, 0)$ se $(- 1, 2)$ tak distance $^{2}$ hai $4 + 4 = 8$ , sabse bada corner distance — $f$ ke (distance to $(1, 0)$ ) $^{2} - 1$ hone ke saath consistent hai. Sach mein $f = (x - 1)^{2} + y^{2} - 1$ , toh min $= - 1$ aur max $= 8 - 1 = 7$ . ✓

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

Forecast: $f = x + y$ ka kahin koi flat spot nahin hai (iska arrow kabhi vanish nahin hota). Toh sab kuch fence par hona chahiye — is triangle par $x + y$ sabse bada kahan hai?

Figure — Absolute extrema on closed bounded regions

Figure s01 — kya dikhata hai aur kaise padhein. Orange-tinted triangle hamara region $D$ hai. Origin ke paas orange arrow gradient $\nabla f = (1, 1)$ hai: yeh "upar aur daayein" point karta hai, har jagah same kyunki $f$ ek flat tilted plane hai. Teal hypotenuse highlighted hai kyunki $f = 2$ uski poori length par hai — winners ki ek poori edge. Plum dot $(0, 0)$ par sabse nichla point hai. Picture padhte hue, koi bhi algebra se pehle solution obvious hai: plane ka uphill arrow maximum ko door wali edge tak kheench leta hai aur minimum ko nazdiki corner tak.

Step 1 — interior. $f_{x} = 1 \neq = 0, f_{y} = 1 \neq = 0.$ Yeh step kyun? Critical point ke liye dono partials ko simultaneously vanish karna hoga, $\nabla f = (f_{x}, f_{y}) = (0, 0)$ . Yahan $f_{x} = 1$ aur $f_{y} = 1$ dono constantly $1$ hain — na to koi kabhi zero hai, toh pair kabhi dono ek saath vanish nahin ho sakti. Koi interior critical point exist nahin karta. (Cell C5 ki jhalak.) Boundary par jaao.

Step 2 — teen edges. Yeh step kyun? Poora action fence par hai.

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) $(2, 0)$ aur $(0, 2)$ se guzarne wali line hai $x + y = 2$ . Line substitute kyun? Is edge par $f = x + y = 2$ identically hai — bilkul flat! Toh hypotenuse par har point $f = 2$ deta hai.

Step 3 — compare. Values: ${0, 2}$ , $2$ poori ek edge par hit hota hai. $abs max = 2 (whole hypotenuse); abs min = 0 at (0, 0) .$

Verify: $f = x + y$ "diagonal direction mein kitna door" measure karta hai. Corner $(0, 0)$ origin-corner se sabse nazdik hai ( $0$ ); hypotenuse hi line $x + y = 2$ hai, toh naturally $f = 2$ wahan. Slanted edge ko uski line equation substitute karke handle kiya. ✓

Example 3 — Disk, circle ko parametrize karo (Cell C3)

Forecast: $f = x y$ positive hai jahan $x, y$ same sign share karte hain, negative jahan differ karte hain. Guess: max aur min $4 5^{\circ}$ diagonals par boundary par hote hain. Confirm karte hain.

Figure s02 — kya dikhata hai aur kaise padhein. Black circle boundary $x^{2} + y^{2} = 2$ hai. Dotted contour lines $f = x y$ ke level sets hain (teal jahan $f < 0$ , orange jahan $f > 0$ ). Dekho kaise contours axes ke paas chipke hain aur diagonals ke saath phoolte hain — hame bata rahe hain ki $f$ $(\pm 1, \pm 1)$ corners ki taraf sabse tezi se badhta hai. Orange dots "same-sign" diagonal par do maxima hain; teal dots "opposite-sign" diagonal par do minima hain; plum dot center par woh saddle hai jise hum sirf record karte hain. Figure chaar boundary winners ek symmetric cross mein predict karta hai — Cell C7 tie naturally aa jaata hai.

Step 1 — interior. $f_{x} = y = 0, f_{y} = x = 0 \Rightarrow (0, 0) .$ Yeh step kyun? Andar flat spot. $(0, 0)$ disk ke andar hai. $f (0, 0) = 0$ . (Yeh saddle hai, lekin hum classify nahin karte — bas value record karte hain.)

Step 2 — boundary $x^{2} + y^{2} = 2$ . Parametrize kyun? Circle smooth hai, koi corners nahin; ek dial $t$ poore circle ko sweep karta hai. Set karo $x = 2 cos t, y = 2 sin t$ $t \in [0, 2 π]$ ke saath taaki poori fence cover ho: $f = x y = 2 cos t sin t = sin (2 t) .$ Identity $2 cos t sin t = sin 2 t$ kyun? Kyunki $sin$ ka range $[- 1, 1]$ obvious hai — yeh "product ka max find karo" ko "sine wave ki top padho" mein badal deta hai. Jab $t$ $[0, 2 π]$ par chalta hai, argument $2 t$ $[0, 4 π]$ par chalta hai, toh $sin (2 t)$ do full waves complete karta hai aur certainly dono extremes hit karta hai: $sin (2 t) \in [- 1, 1] : max = 1 at t = \frac{π}{4}, \frac{5 π}{4}, min = - 1 at t = \frac{3 π}{4}, \frac{7 π}{4} .$

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

Verify: $t = \frac{π}{4}$ par: $x = y = 2 \cdot \frac{1}{2} = 1$ , $f = 1$ ✓. Charon winners diagonals par hain — forecast se match karta hai. Do maxima, do minima (Cell C7 tie circle par naturally aata hai).

Example 4 — Interior critical point $D$ ke BAHAR girta hai (Cell C4)

Forecast: $- 6 x$ bowl ka bottom bahut daayein kheenchta hai — shayad chote se unit disk ke bahar. Kisi critical point ka kya hota hai jo aap use nahin kar sakte?

Step 1 — interior. $f_{x} = 2 x - 6 = 0 \Rightarrow x = 3, f_{y} = 2 y = 0 \Rightarrow y = 0.$ Candidate $(3, 0)$ : kya yeh andar hai? $3^{2} + 0 = 9 > 1$ ✗. Discard kyun? Step 1 sirf woh solutions rakhta hai jo $D$ ke andar hain; $(3, 0)$ bahut bahar hai, toh yeh candidate nahin hai. Hame zero usable interior points milte hain.

Step 2 — boundary $x^{2} + y^{2} = 1$ . Substitute kyun? Circle par $x^{2} + y^{2} = 1$ hai, toh $f = 1 - 6 x$ . Parametrize karo $x = cos t, y = sin t$ $t \in [0, 2 π]$ ke saath: $f = 1 - 6 cos t .$ $cos t \in [- 1, 1]$ (ek period mein poora range), aur $f$ decrease karta hai jab $cos t$ badhta hai: $max : cos t = - 1 (t = π) \Rightarrow f = 7 at (- 1, 0), min : cos t = 1 (t = 0) \Rightarrow f = - 5 at (1, 0) .$

Step 3 — compare. Sirf boundary candidates survive karte hain. $abs max = 7 at (- 1, 0); abs min = - 5 at (1, 0) .$

Verify: $f = (x - 3)^{2} + y^{2} - 9$ = (distance to $(3, 0)$ ) $^{2} - 9$ . Disk ka $(3, 0)$ se closest point hai $(1, 0)$ (distance $2$ ): $f = 4 - 9 = - 5$ ✓. Farthest hai $(- 1, 0)$ (distance $4$ ): $f = 16 - 9 = 7$ ✓. Discarded critical point $f$ ka global min tha poore $R^{2}$ par — irrelevant kyunki woh field se bahar hai.

Example 5 — Degenerate: koi interior critical point nahin (Cell C5)

Forecast: Ek disk par tilted plane. Flat plane kabhi andar flat spot nahin rakhti — toh extremes do opposite edge points par hone chahiye. Kaun se?

Step 1 — interior. $f_{x} = 3 \neq = 0, f_{y} = - 4 \neq = 0.$ Yeh step kyun? Confirm karo ki koi interior critical point nahin hai: $\nabla f = (3, - 4)$ ek constant nonzero arrow hai, toh na partial kabhi vanish hota hai — pair kabhi $(0, 0)$ nahin ho sakti. Uphill direction har jagah same hai.

Step 2 — boundary $x^{2} + y^{2} = 25$ . Parametrize karo $x = 5 cos t, y = 5 sin t$ $t \in [0, 2 π]$ ke saath: $f = 15 cos t - 20 sin t .$ Ek sinusoid mein combine kyun? Sum $a cos t + b sin t$ equals $R cos (t - ϕ)$ with amplitude $R = a^{2} + b^{2}$ — aur puri period $t \in [0, 2 π]$ par woh cosine apna poora range $[- R, R]$ sweep karta hai. Yahan $R = 1 5^{2} + 2 0^{2} = 625 = 25$ . $f \in [- 25, 25] .$

Step 3 — compare. $abs max = 25; abs min = - 25.$ Max jahan point $\nabla f = (3, - 4)$ ke saath align karta hai: $(x, y) = 5 \cdot \frac{( 3 , - 4 )}{5} = (3, - 4)$ . Min opposite par: $(- 3, 4)$ .

Verify: $f (3, - 4) = 9 + 16 = 25$ ✓, $f (- 3, 4) = - 9 - 16 = - 25$ ✓. Aur $3^{2} + (- 4)^{2} = 25$ toh $(3, - 4)$ circle par hai ✓. Poora answer fence par tha — empty-interior case cleanly handle ho gaya.

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

Forecast: $f$ bas origin se distance hai. Iska lowest value obviously center par hai — lekin kya center ek " $\nabla f = 0$ " point hai? Dhyan se.

Step 1 — interior. Origin se door partials compute karo: $f_{x} = \frac{x}{x ^{2} + y ^{2}}, f_{y} = \frac{y}{x ^{2} + y ^{2}} .$ Dono ko $= 0$ set karne ke liye $x = 0, y = 0$ chahiye — lekin wahan formula zero se divide karta hai. Yeh kyun matter karta hai: $(0, 0)$ par graph ek sharp cone tip hai; $f$ wahan differentiable nahin hai, toh $\nabla f$ exist nahin karta. Recipe explicitly kehti hai: jahan $f$ differentiable nahin, woh points bhi candidates hain. $(0, 0)$ record karo: $f = 0$ .

Step 2 — boundary (chaar edges). Kyun? Distance baahir ki taraf badhti hai, toh max fence par hai. Symmetry se har edge aisi hai jaise $x = 1, y \in [- 1, 1]$ : $f = 1 + y^{2}$ , $y = 0$ par minimum ( $= 1$ ), $y = \pm 1$ par maximum ( $= 2$ ). Corners $(\pm 1, \pm 1)$ $2$ dete hain.

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

Verify: $2 \approx 1.414$ ; corner $(1, 1)$ ki distance $1 + 1 = 2$ ✓. Min ek non-differentiable point par tha jise ek andha " $\nabla f = 0$ " search miss kar deta — poora Cell C6 ka point yahi hai. ✓

Example 7 — Chaar-taraf ka tie (Cell C7)

Forecast: Origin se distance-squared. Lowest center par, highest... kitne corners tie karte hain?

Step 1 — interior. $f_{x} = 2 x = 0, f_{y} = 2 y = 0 \Rightarrow (0, 0)$ , andar, $f = 0$ . Yahan smooth hai (koi cone nahin — square, square root ke square ke andar hai, toh $f$ ek genuine paraboloid hai, har jagah differentiable).

Step 2 — boundary. Har edge $x = 1$ : $f = 1 + y^{2}$ , max $2$ at $y = \pm 1$ , min $1$ at $y = 0$ . Charon corners $(\pm 1, \pm 1)$ $2$ dete hain.

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

Verify: Max chaar distinct points par ek saath achieve hota hai — ek genuine tie. $f (1, 1) = 1 + 1 = 2$ ✓. Square ki symmetry ties guarantee karti hai; sirf ek corner report karna incomplete answer hoga.

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

Forecast: Heating $x$ ke saath badhti hai phir $- x^{2}$ ise wapas kheenchta hai; $2 x y$ term bade $x$ aur bade $y$ ko ek saath reward karta hai. Compute karne se pehle hot corner guess karo.

Step 1 — interior. $T_{x} = 4 - 2 x + 2 y = 0, T_{y} = 2 x = 0.$ $T_{y}$ se: $x = 0$ . Phir $T_{x}$ se: $4 + 2 y = 0 \Rightarrow y = - 2$ . Point $(0, - 2)$ — plate ke bahar ( $y \geq 0$ ). Discard kyun? $D$ mein nahin; koi usable interior critical point nahin (Cell C4 ki echo).

Step 2 — chaar edges (positions cm mein, temperatures $^{\circ}$ C mein). Yeh step kyun? Koi interior candidate nahin hone par, sabse hot aur cold spots plate ke rim par hone chahiye.

Bottom $y = 0$ cm, $x \in [0, 3]$ cm: $T = 20 + 4 x - x^{2}$ , $T^{'} = 4 - 2 x = 0 \Rightarrow x = 2$ 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: $T = 20 + 4 x - x^{2} + 4 x = 20 + 8 x - x^{2}$ , $T^{'} = 8 - 2 x = 0 \Rightarrow x = 4$ cm ( $[0, 3]$ ke bahar, 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 — $x$ -terms sab vanish ho jaate hain).
Right $x = 3$ cm, $y \in [0, 2]$ cm: $T = 20 + 12 - 9 + 6 y = 23 + 6 y$ , $y$ mein increasing. Ends: $T (3, 0) = 23^{\circ}$ C, $T (3, 2) = 35^{\circ}$ C.

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

Verify: Units check — $x, y$ cm mein enter hote hain aur constant $20$ plus har term $^{\circ}$ C mein calibrated hai, toh output temperature hai. $T (3, 2) = 20 + 12 - 9 + 12 = 35^{\circ}$ C ✓. $T (0, y) = 20^{\circ}$ C sab $y$ ke liye ✓ (cold "spot" poori left edge hai — ek Cell C7 tie ek word problem ke andar chhupa hua). Hottest corner ne forecast se match kiya: bada $x$ , bada $y$ . ✓

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

Forecast: Phir ek tilted plane → sirf boundary. Gradient $(1, 2)$ first quadrant mein point karta hai (dono components positive), toh max upar-aur-daayein hona chahiye. Exact point nail karte hain aur iska quadrant sign-by-sign confirm karte hain.

Figure s03 — kya dikhata hai aur kaise padhein. Black circle boundary $x^{2} + y^{2} = 5$ hai. Orange arrow $\nabla f = (1, 2)$ hai, origin se draw kiya gaya; iska tip boundary point $(1, 2)$ par exactly land karta hai (orange dot) — dono coordinates positive → Quadrant I, maximum. Teal dot antipode $(- 1, - 2)$ par hai jiske dono coordinates negative → Quadrant III, minimum. Plum dotted lines level sets $x + 2 y = c$ hain; yeh gradient ke perpendicular chalti hain, toh circle par chalne se $f$ sabse tezi se tab badhta hai jab aap arrow ke saath jaate ho. Picture sign argument ko visual banati hai: winner wahan hai jahan position vector $\nabla f$ ke same direction mein point karta hai.

Step 1 — interior. $f_{x} = 1 \neq = 0, f_{y} = 2 \neq = 0.$ Yeh step kyun? Dono partials constant aur nonzero hain, toh $\nabla f = (1, 2)$ kabhi $(0, 0)$ nahin banta: koi interior critical point nahin (Cell C5 flavour). Sare candidates fence par hain.

Step 2 — boundary $x^{2} + y^{2} = 5$ . Parametrize karo $x = 5 cos t, y = 5 sin t$ $t \in [0, 2 π]$ ke saath: $f = 5 cos t + 25 sin t = 5 (cos t + 2 sin t) .$ $cos t + 2 sin t$ ka amplitude $R = 1^{2} + 2^{2} = 5$ hai, aur puri period par yeh $\pm R$ hit karta hai, toh $f \in [- 5 \cdot 5, 5 \cdot 5] = [- 5, 5] .$ Max tab hota hai jab $(x, y)$ $\nabla f = (1, 2)$ ke saath point karta hai: $(x, y) = 5 \cdot \frac{( 1 , 2 )}{5} = (1, 2) .$

Sign-by-sign analysis (C9 point). Maximum par, $x = 1 > 0$ aur $y = 2 > 0$ — ek $(+, +)$ pair, toh yeh Quadrant I mein hai. Minimum bilkul opposite point hai $(- 1, - 2)$ : yahan $x < 0$ aur $y < 0$ , ek $(-, -)$ pair → Quadrant III. Iska general rule: circle par linear $f = a x + b y$ ke liye, maximum us quadrant mein baithta hai jiska sign pattern $(a, b) = \nabla f$ se match karta hai, aur minimum us quadrant mein jisme opposite signs hain.

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

Verify: $(1, 2)$ circle par: $1 + 4 = 5$ ✓. $f (1, 2) = 1 + 4 = 5$ ✓, $f (- 1, - 2) = - 1 - 4 = - 5$ ✓. Lagrange multipliers se compare karo: condition $\nabla f = λ \nabla g$ kehti hai $(1, 2) = λ (2 x, 2 y)$ , yaani $(x, y) ∥ (1, 2)$ — same answer, different route, same sign conclusion.

Checklist: kya humne har cell stamp kiya?

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

Connections

Absolute extrema on closed bounded regions — parent recipe jo har example follow karta hai.
Critical points and gradient — Step 1 $\nabla f = 0$ solve karta hai (Ex 1, 3, 4, 8).
Second derivative test (Hessian) — deliberately unused: hum values compare karte hain, classify nahin (note: Ex 3 ka saddle humne kabhi label nahin kiya).
Lagrange multipliers — alternate boundary handling, Ex 9 mein dikhaya.
Extreme Value Theorem (1D) — guarantee karta hai ki 1-var edge sub-problems ke apne max/min hain.
Closed and bounded sets (topology) — kyun guarantee upar ke har $D$ par hold karti hai.
Parametrization of curves — circles ko single-dial $t$ problems mein banana (Ex 3, 4, 5, 9).