Critical points — finding, classifying: Worked examples

4.4.12 · D3 · Maths › Multivariable Calculus › Critical points — finding, classifying

Scenario matrix

Har critical-point problem inhi cells mein se kisi ek mein aata hai. Last column us example ka naam deta hai jo use nail karta hai.

#	Case class	Tricky kyun hai	Example
A	$D > 0, f_{xx} > 0$	saaf minimum	Ex 1
B	$D > 0, f_{xx} < 0$	saaf maximum	Ex 2
C	$D < 0$	saddle ( $f_{xx}$ ka sign irrelevant)	Ex 3
D	Mixed term $f_{x y} \neq = 0$	min jaisa lagta hai, saddle hai	Ex 4
E	Kai critical points	har ek ko alag classify karo	Ex 5
F	$D = 0$ degenerate	Hessian chup hai — directly inspect karo	Ex 6
G	Critical points ki poori line/curve	infinitely many; $\nabla f$ ek set par vanish hota hai	Ex 7
H	Constrained extremum (koi free critical point nahi)	boundary / Lagrange, Hessian nahi	Ex 8
I	Word problem with units	translate → optimise → units check karo	Ex 9

Jin prerequisites pe hum lean karte hain: Gradient and Directional Derivatives, Hessian Matrix, Quadratic Forms and Definiteness, Lagrange Multipliers, Taylor Series in Several Variables.

$f (x, y) = x^{2} + x y + y^{2}$ (cell A) Pehle Forecast. Dono variables squared hain aur $+$ signs ke saath. Guess: origin par minimum? Ruko — $x y$ term use bigaad sakta hai. (Forecast kyun? Compute karne se pehle signs padhna ek prediction set karta hai test karne ke liye — agar $D$ baad mein negative nikle toh hum jaanenge ki $x y$ term jeet gaya.)

Step 1 — gradient. $f_{x} = 2 x + y, f_{y} = x + 2 y .$ Yeh step kyun? Critical points wahan rehte hain jahan ground har direction mein flat ho, yaani jahan $\nabla f = 0$ ho (parent definition).

Step 2 — $\nabla f = 0$ solve karo. $2 x + y = 0$ se $y = - 2 x$ milta hai. $x + 2 y = 0$ mein substitute karo: $x + 2 (- 2 x) = - 3 x = 0 \Rightarrow x = 0$ , phir $y = 0$ . Ek hi critical point $(0, 0)$ . Yeh step kyun? Dono partials ko ek saath vanish karna chahiye; ek linear system cleanly solve ho jaata hai.

Step 3 — Hessian & discriminant. $f_{xx} = 2, f_{y y} = 2, f_{x y} = 1$ , toh $D = f_{xx} f_{y y} - f_{x y}^{2} = (2) (2) - 1^{2} = 3 > 0.$ Yeh step kyun? $D$ decide karta hai ki yeh extremum hai ya nahi (parent mnemonic: "D decides, then $f_{xx}$ describes").

Step 4 — type padho. $D > 0$ aur $f_{xx} = 2 > 0 \Rightarrow$ local minimum. Yeh step kyun? $D > 0$ ne humein pehle hi bata diya ki point genuinely extremum hai (quadratic form definite hai, har direction mein same sign); ab $f_{xx}$ describe karta hai ki kis type ka — positive matlab surface bowl ki tarah upar cup karti hai, toh yeh minimum hai.

Verify. $x y$ term itna strong nahi tha. Completing the square se sanity-check karo: $x^{2} + x y + y^{2} = (x + \frac{y}{2})^{2} + \frac{3}{4} y^{2} \geq 0,$ jo $0$ ke barabar sirf $(0, 0)$ par hai. Positive coefficients ke saath squares ka sum ⇒ genuine minimum. ✓

Case B — ek saaf maximum

Worked example Ex 2 ·

$f (x, y) = - x^{2} - y^{2} + 4 x + 6 y$ (cell B) Forecast. Negative squares har direction mein neeche kheenchte hain, linear terms sirf peak ko origin se shift karte hain. Guess: pehle quadrant mein kahin ek single maximum. (Forecast kyun? Expected type aur rough location ko abhi naam dena matlab algebra ek leap of faith ki jagah ek check ban jaata hai.)

Step 1 — gradient. $f_{x} = - 2 x + 4, f_{y} = - 2 y + 6.$ Kyun? Flat ground condition.

Step 2 — solve. $- 2 x + 4 = 0 \Rightarrow x = 2$ ; $- 2 y + 6 = 0 \Rightarrow y = 3$ . Critical point $(2, 3)$ . Kyun? Har partial linear hai, toh ek-ek solution.

Step 3 — Hessian. $f_{xx} = - 2, f_{y y} = - 2, f_{x y} = 0$ , toh $D = (- 2) (- 2) - 0 = 4 > 0$ . Yeh step kyun? Yahan first derivatives zero hain, toh woh shape ke baare mein kuch nahi bataate; second derivatives (Hessian) curvature carry karte hain, aur $D$ unhe ek single number mein package karta hai jo extremum-vs-saddle decide karta hai.

Step 4 — type. $D > 0$ aur $f_{xx} = - 2 < 0 \Rightarrow$ local maximum. Yeh step kyun? Jab $D > 0$ ho toh hum jaante hain yeh extremum hai; $f_{xx}$ ka sign phir type describe karta hai — negative matlab surface upar se cap ho gayi (ek dome), yaani maximum.

Verify. Completing the square: $f = - (x - 2)^{2} - (y - 3)^{2} + 13$ . $f$ zyada se zyada $13$ ho sakta hai (dono squares zero), jo $(2, 3)$ par achieve hota hai. Peak height $f (2, 3) = 13$ . ✓

Figure — Critical points — finding, classifying

Upar ki figure dekho — teen "clean" landscapes contour maps ke roop mein draw ki hain. Left (Ex 1): closed rings magenta dot par tighten ho rahe hain, toh har direction wahan se dur jaane par chadhti hai — ek bowl / minimum. Middle (Ex 2): same rings lekin colours inverted, har direction orange dot se dur girती hai — ek dome / maximum. Right (Ex 3, aage bana): contours ek cross banate hain, closed loops nahi — woh hyperbolic pattern ek saddle ka fingerprint hai, jahan do directions aapas mein disagree karte hain.

Case C — ek saaf saddle

Worked example Ex 3 ·

$f (x, y) = x^{2} - 4 y^{2}$ (cell C) Forecast. $x$ ke along upar, $y$ ke along neeche. Yeh mismatch saddle ki definition hai — lekin $D$ se confirm karwa lete hain. (Forecast kyun? Compute karne se pehle dono squares par opposite signs spot karna humein $D < 0$ anticipate karne deta hai aur discriminant ko confirmation ki tarah treat karne deta hai.)

Step 1 — gradient. $f_{x} = 2 x, f_{y} = - 8 y \Rightarrow$ critical point $(0, 0)$ . Yeh step kyun? Hume pehle flat spot locate karna hai: $\nabla f = 0$ solve karo, aur $2 x = 0, - 8 y = 0$ $(0, 0)$ dete hain.

Step 2 — Hessian. $f_{xx} = 2, f_{y y} = - 8, f_{x y} = 0$ , toh $D = (2) (- 8) - 0 = - 16 < 0$ . Yeh step kyun? Gradient yahan zero hai, toh woh shape ke baare mein kuch nahi bata sakta; second derivatives (Hessian) curvature collect karte hain aur $D = det H$ woh sign-test hai jo hum padhte hain.

Step 3 — type. $D < 0 \Rightarrow$ saddle, aur yahan $f_{xx}$ irrelevant hai — parent warn karta hai ki $D < 0$ hone ke baad use kabhi mat dekhna. Yeh step kyun? $D < 0$ matlab quadratic form indefinite hai — kuch directions mein positive, kuch mein negative — jo precisely ek saddle hai; aur koi check zaroori nahi.

Verify. Origin se do paths lo:

$x$ ke along ( $y = 0$ set karo): $f = x^{2}$ ⇒ upar curves.
$y$ ke along ( $x = 0$ set karo): $f = - 4 y^{2}$ ⇒ neeche curves. Ek path upar, ek neeche ⇒ saddle. ✓ Yeh ek indefinite quadratic form ka geometric meaning hai (Quadratic Forms and Definiteness).

Case D — mixed-term disguise

Worked example Ex 4 ·

$f (x, y) = x^{2} + y^{2} - 3 x y$ (cell D — parent ka "steel-manned mistake") Forecast. Dono pure second derivatives positive hain ( $f_{xx} = f_{y y} = 2$ ). Naive aankhein chillati hain "minimum!". Guess karo ki cross-term $- 3 x y$ itna strong hai ki use overturn kar sake. (Forecast kyun? Pehle tempting "minimum" guess commit karna aane wale reversal ko memorable banata hai — tum trap ko spring hote feel karte ho rather than use padhte hue.)

Step 1 — gradient. $f_{x} = 2 x - 3 y, f_{y} = 2 y - 3 x .$ Yeh step kyun? Kuch bhi classify karne se pehle humein dhundna hai jahan ground flat ho, yaani $\nabla f = 0$ solve karo; yeh dono partials compute karne se shuru hota hai.

Step 2 — solve. $2 x - 3 y = 0$ aur $2 y - 3 x = 0$ . Pehle se, $x = \frac{3}{2} y$ ; plug in: $2 y - \frac{9}{2} y = - \frac{5}{2} y = 0 \Rightarrow y = 0$ , toh $x = 0$ . Sirf ek hi critical point $(0, 0)$ . Yeh step kyun? Dono partials ek saath vanish hone chahiye; pair ko simultaneously solve karna single candidate point pin down karta hai.

Step 3 — discriminant. $f_{xx} = 2, f_{y y} = 2, f_{x y} = - 3$ : $D = (2) (2) - (- 3)^{2} = 4 - 9 = - 5 < 0.$ Yeh step yahan sabse zyada kyun matter karta hai? Yahi woh jagah hai jahan "dono positive ⇒ min" fail hota hai. Mixed partial $f_{x y}$ cross-curvature measure karta hai; jab $f_{x y}^{2} > f_{xx} f_{y y}$ ho toh woh dominate karta hai aur $D$ negative ho jaata hai.

Step 4 — type. $D < 0 \Rightarrow$ saddle. Yeh step kyun? $D < 0$ quadratic form ko indefinite banata hai: completed-square coefficients ke opposite signs hote hain, toh $f$ kuch directions mein badhti hai aur kuch mein girती hai. Yahi saddle ki definition hai — aur crucially $f_{xx}$ ka sign yahan koi role nahi khaelta, jo exactly woh reason hai jisse "dono positive ⇒ min" reasoning collapse hoti hai.

Verify. Diagonal direction $y = x$ lo: $f = x^{2} + x^{2} - 3 x^{2} = - x^{2}$ ⇒ neeche. Anti-diagonal $y = - x$ lo: $f = x^{2} + x^{2} + 3 x^{2} = 5 x^{2}$ ⇒ upar. Ek taraf neeche, ek taraf upar ⇒ saddle, despite $f_{xx} > 0$ . ✓

Upar ki figure $f$ ko origin se do diagonal paths ke along plot karti hai, 2-D surface ko do familiar 1-D curves mein badal kar. Magenta curve hai $f$ along $y = x$ : yeh neeche khulaati hai ( $f = - x^{2}$ ). Violet curve hai $f$ along $y = - x$ : yeh upar khulaati hai ( $f = 5 x^{2}$ ). Orange dot par ek doosre ko cross karte dono curves — ek frown, ek smile — yeh visual proof hai ki $f_{xx} > 0$ wala point phir bhi saddle ho sakta hai: cross-term hi ek direction ko flip karta hai.

Case E — kai critical points, har ek ko classify karo

Worked example Ex 5 ·

$f (x, y) = x^{3} + y^{3} - 3 x y$ (cell E) Forecast. Har variable mein cubic, $- 3 x y$ se coupled. Cubics ek se zyada flat spot de sakte hain. Guess: do critical points, ek saddle aur ek extremum. (Forecast kyun? Predict karna ki kitne critical points expect hain, guard karta hai ki hum zyada jaldi na ruk jaayein — agar algebra sirf ek root de toh hum jaante hain re-check karna hai.)

Step 1 — gradient. $f_{x} = 3 x^{2} - 3 y, f_{y} = 3 y^{2} - 3 x .$ Yeh step kyun? Critical points exactly wahan hain jahan $\nabla f = 0$ ho, toh pehla kaam do partial derivatives compute karna hai — $x$ mein differentiate karo ( $y$ fixed rakh ke) aur $y$ mein ( $x$ fixed rakh ke) — unhe zero set karne aur flat spots hunt karne se pehle.

Step 2 — system solve karo. Dono ko zero set karo: $x^{2} = y, y^{2} = x .$ $y = x^{2}$ ko $y^{2} = x$ mein substitute karo: $x^{4} = x \Rightarrow x (x^{3} - 1) = 0 \Rightarrow x = 0$ ya $x = 1$ .

$x = 0 \Rightarrow y = 0$ : point $(0, 0)$ .
$x = 1 \Rightarrow y = 1$ : point $(1, 1)$ . Substitute kyun? Yeh do equations ko ek polynomial mein collapse karta hai, saare roots expose karte hue.

Step 3 — Hessian (point ke saath badalta hai). $f_{xx} = 6 x, f_{y y} = 6 y, f_{x y} = - 3$ , toh $D = (6 x) (6 y) - (- 3)^{2} = 36 x y - 9.$ Yeh step kyun? Kyunki $f_{xx} = 6 x$ aur $f_{y y} = 6 y$ position ke saath change karte hain, discriminant ab ek fixed number ki jagah $(x, y)$ mein ek formula hai — hum ise symbolically compute karte hain taaki har critical point par alag se plug in kar sakein.

Step 4 — har cell alag classify karo.

$(0, 0)$ par: $D = 36 (0) - 9 = - 9 < 0 \Rightarrow$ saddle.
$(1, 1)$ par: $D = 36 (1) - 9 = 27 > 0$ , aur $f_{xx} = 6 (1) = 6 > 0 \Rightarrow$ local minimum. Ek-ek kyun? $D$ ab position ka function hai; same formula alag points par alag verdicts deta hai.

Verify. Heights: $f (0, 0) = 0$ aur $f (1, 1) = 1 + 1 - 3 = - 1$ . Minimum $(1, 1)$ saddle se neeche baitha hai — consistent (ek valley ka bottom nearby pass se neeche). ✓

Case F — $D = 0$ degenerate trap

Worked example Ex 6 ·

$f (x, y) = x^{2} y^{2}$ (cell F) Forecast. Do squares ka product $\geq 0$ hai. Guess: minimum jahan bhi zero hai? Lekin dekhna $D$ kaise collapse hota hai. (Forecast kyun? $f \geq 0$ anticipate karna humein direct inspection trust karne ke liye prime karta hai jab Hessian test — jaisa humein suspect tha — kuch nahi bata pata.)

Step 1 — gradient. $f_{x} = 2 x y^{2}, f_{y} = 2 x^{2} y .$ Kyun? Har variable mein alag product rule.

Step 2 — solve. $2 x y^{2} = 0$ aur $2 x^{2} y = 0$ . Dono hold karte hain jab bhi $x = 0$ ya $y = 0$ ho. Toh dono coordinate axes critical hain! Origin $(0, 0)$ par focus karo (interesting isolated-looking wala; axes Case G territory hain). Yeh step kyun? Hum har point dhundh rahe hain jahan $\nabla f = 0$ ho; har partial ko factor karna dikhata hai ki solution set ek akela point nahi balki do poori lines hain, jo khud mein warning hai ki yeh problem degenerate hai.

Step 3 — origin par Hessian. $f_{xx} = 2 y^{2}, f_{y y} = 2 x^{2}, f_{x y} = 4 x y$ . $(0, 0)$ par teeno $0$ hain: $D = (0) (0) - 0^{2} = 0.$ Test yahan kyun marta hai. Taylor expansion ka quadratic part bilkul zero hai — second-order term koi information carry nahi karta. Humein higher-order terms dekhne chahiye (Taylor Series in Several Variables).

Step 4 — directly inspect karo. $f = x^{2} y^{2} = (x y)^{2} \geq 0$ , aur $f = 0$ dono axes par. Toh origin ek (non-strict) minimum hai — lekin ek flat, degenerate wala, Case A ke tidy bowl jaisa nahi. Yeh step kyun? $D = 0$ ke saath Hessian chup hai, toh hum minimum ki definition par wapas jaate hain — kya $f$ nearby kabhi choti hoti hai? Kyunki $(x y)^{2}$ kabhi $0$ se neeche nahi ja sakta, origin genuinely ek minimum hai.

Verify. $y = x$ ke along: $f = x^{4} \geq 0$ ⇒ upar. Kisi bhi axis ke along: $f \equiv 0$ ⇒ flat. Kabhi negative nahi ⇒ minimum, inspection se confirm kiya, $D$ se nahi. ✓

Case G — critical points ki poori line

Worked example Ex 7 ·

$f (x, y) = (x - y)^{2}$ (cell G) Forecast. Yeh ek "trough" hai jo ek line ke along run karta hai. Guess: ek critical point nahi balki poora set. (Forecast kyun? Ek set predict karna, ek point ki jagah, humein galat tarike se ek single isolated minimum declare karne se rokta hai jab algebra ek poori line return kare.)

Step 1 — gradient. $f_{x} = 2 (x - y), f_{y} = - 2 (x - y) .$ Yeh step kyun? Critical points wahan hain jahan $\nabla f = 0$ ho; hum pehle differentiate karte hain taaki dekh sakein kaun se points dono partials ko vanish karte hain — aur yahan chain rule common factor $(x - y)$ nikaal leta hai.

Step 2 — solve. Dono vanish hote hain jab bhi $x - y = 0$ ho, yaani poori line $y = x$ ke along. Infinitely many critical points. Line kyun? Function sirf combination $x - y$ "dekhta" hai; $y = x$ ke along move karo aur kuch nahi badlta.

Step 3 — Hessian. $f_{xx} = 2, f_{y y} = 2, f_{x y} = - 2$ , toh $D = (2) (2) - (- 2)^{2} = 4 - 4 = 0$ har jagah. $D = 0$ phir kyun? Ek degenerate (rank-deficient) Hessian exactly wahi dikhta hai jo ek flat trough dikhta hai — $H$ ki ek eigenvalue zero hai (Hessian Matrix).

Step 4 — inspect karo. $f = (x - y)^{2} \geq 0$ , line $y = x$ par zero. Line ke perpendicular ground upar curves karti hai; line ke along flat hai. $y = x$ ka har point ek (non-strict) minimum hai. Yeh step kyun? Kyunki $D = 0$ Hessian test ko mute kar deta hai, hum definition par wapas jaate hain: $(x - y)^{2}$ kabhi negative nahi hota aur line par $0$ equals karta hai, toh us line ka har point minimum hai.

Verify. Point $(1, 1)$ lo (line par): $f = 0$ . $(1, 0)$ par perpendicularly step off karo: $f = (1 - 0)^{2} = 1 > 0$ . Line ke along $(2, 2)$ par step karo: $f = 0$ , unchanged. Ek valley floor. ✓

Upar ki figure trough ka contour map hai. Magenta diagonal woh line $y = x$ hai jahan har point critical hai — notice karo ki contour colour us par constant hai, dikhata hai ki ground us direction mein bilkul flat hai. Violet arrow line ke across point karta hai, jahan contours bunch together hote hain aur $f$ $(x - y)^{2}$ ki tarah chadh jaata hai. Flat floor aur rising walls wala ek trough: infinite line of minima aisi dikhti hai.

Case H — constrained extremum (koi free critical point nahi)

Worked example Ex 8 ·

$f (x, y) = x y$ ko circle $x^{2} + y^{2} = 8$ par maximize karo (cell H) Forecast. Freely, $f = x y$ ka sirf origin par saddle hai (check: $f_{x} = y, f_{y} = x \Rightarrow (0, 0)$ ; $D = (0) (0) - 1^{2} = - 1 < 0$ ). Toh koi bhi max constraint curve par rehni chahiye. Guess: maximum jahan $∣ x ∣ = ∣ y ∣$ ho. (Forecast kyun? Pehle se realise karna ki free critical point saddle hai humein bata deta hai ki answer boundary par hona chahiye — yeh Hessian ki jagah Lagrange ki taraf le jaata hai.)

Step 1 — Lagrange set up karo. Constraint $g = x^{2} + y^{2} - 8 = 0$ ke saath, Lagrange Multipliers ki method demand karti hai $\nabla f = λ \nabla g$ : $y = λ (2 x), x = λ (2 y) .$ Lagrange kyun, Hessian kyun nahi? Hessian test interior flat spots dhundta hai; yahan extremum ek boundary curve par forced hai, toh free $\nabla f = 0$ machinery apply nahi hoti.

Step 2 — solve karo. Do equations multiply karo: $x y = 4 λ^{2} x y$ . Agar $x y \neq = 0$ toh $4 λ^{2} = 1 \Rightarrow λ = \pm \frac{1}{2}$ . $λ = \frac{1}{2}$ lo: phir $y = x$ ; constraint mein daalo $x^{2} + x^{2} = 8 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$ . Points $(2, 2)$ aur $(- 2, - 2)$ . Yeh step kyun? Lagrange condition plus constraint teen equations ka system hai $x, y, λ$ mein; pehle $λ$ eliminate karne ke liye multiply karo, phir size fix karne ke liye constraint use karo, yeh saare candidate points tak pahunchne ka saaf raasta hai.

Step 3 — evaluate & compare karo. $f (2, 2) = 4$ , $f (- 2, - 2) = 4$ . ( $λ = - \frac{1}{2}$ branch $y = - x$ deta hai, points $(2, - 2), (- 2, 2)$ with $f = - 4$ , yeh minima hain.) Compare kyun? Ek closed curve par extreme values $f$ ko saare candidate points par evaluate karte hue milte hain.

Step 4 — conclusion. Maximum value $4$ , $(2, 2)$ aur $(- 2, - 2)$ par; minimum $- 4$ doosre pair par. Yeh step kyun? Optimisation actual extreme value aur us jagah ke baare mein poochhta hai jahan woh occur ho; saare candidates list karne ke baad, hum simply largest ( $4$ ) aur smallest ( $- 4$ ) function values pick karte hain answer state karne ke liye.

Verify. Circle parametrise karo: $x = 8 cos θ, y = 8 sin θ$ , toh $f = 8 cos θ sin θ = 4 sin (2 θ)$ , jiska max $4$ hai. ✓ Matches.

Case I — units ke saath word problem

Worked example Ex 9 · Sabse sasta open-top box

(cell I) Ek rectangular open-top box mein $V = 32 m^{3}$ hold karna zaroori hai. Material costs uniform hain. Surface area $S$ minimize karo (base + char sides).

Forecast. Guess karo ki box cube nahi hoga (koi top nahi!) — yeh height se zyada chauda hona chahiye. (Forecast kyun? Ek physical prediction — squat, cubic nahi — ek sanity target deta hai: agar algebra ek tall thin box return kare toh hum jaanenge koi sign ya constraint galat tha.)

Step 1 — do free variables se model banao. Base $x \times y$ (metres) aur height $h$ lo. Volume $h = \frac{32}{x y}$ fix karta hai. Open-top area: $S (x, y) = x y + 2 x h + 2 y h = x y + \frac{64}{y} + \frac{64}{x} [m^{2}] .$ $h$ substitute kyun? Yeh teen variables ko do mein reduce karta hai, ek constrained problem ko $(x, y)$ mein free critical-point problem mein badal kar.

Step 2 — gradient. $S_{x} = y - \frac{64}{x ^{2}}, S_{y} = x - \frac{64}{y ^{2}} .$ Yeh step kyun? Sabse sasta box $S$ ka minimum hai, aur minima critical points par baithte hain; toh hum $\nabla S$ compute karte hain $\nabla S = 0$ solve karne ke liye.

Step 3 — solve karo. Dono ko zero set karo: $y = \frac{64}{x ^{2}}$ aur $x = \frac{64}{y ^{2}}$ . Symmetry se try karo $x = y$ : phir $x = \frac{64}{x ^{2}} \Rightarrow x^{3} = 64 \Rightarrow x = 4$ . Toh $x = y = 4$ m, aur $h = \frac{32}{16} = 2$ m. Yeh step kyun, aur symmetry guess kyun? Dono partials ek saath vanish hone chahiye; do equations $x \leftrightarrow y$ swap karne ke under symmetric hain, toh extremum symmetric line $x = y$ par expected hai, jo system ko ek solvable cubic mein collapse karta hai.

Step 4 — confirm karo ki minimum hai (Hessian). $S_{xx} = \frac{128}{x ^{3}}, S_{y y} = \frac{128}{y ^{3}}, S_{x y} = 1$ . $(4, 4)$ par: $S_{xx} = S_{y y} = \frac{128}{64} = 2$ , toh $D = (2) (2) - 1^{2} = 3 > 0, S_{xx} = 2 > 0 \Rightarrow minimum. ✓$ Check kyun? Word problem mein sahi type ka critical point chahiye — genuine minimum, saddle nahi.

Verify (value & units). $S (4, 4) = 16 + \frac{64}{4} + \frac{64}{4} = 16 + 16 + 16 = 48 m^{2}$ . Volume check: $4 \cdot 4 \cdot 2 = 32 m^{3}$ ✓. Units: area $m^{2}$ mein ✓. Base $4 \times 4$ height $2$ se zyada chauda hai — forecast se match karta hai (koi top nahi ⇒ chota aur squat). ✓

Active Recall

Recall Ex 4 mein

$f_{xx}$ aur $f_{y y}$ dono $+ 2$ the. Saddle kyun? Kyunki $f_{x y} = - 3$ ne $D = 4 - 9 = - 5 < 0$ banaya. Mixed term ki cross-curvature dominate kar gayi; $D < 0$ $f_{xx}$ ke sign ko override karta hai.

Recall Second derivative test kab chup ho jaata hai, aur phir kya karte hain?

Jab $D = 0$ ho (Ex 6, Ex 7) — quadratic Taylor term vanish ho jaata hai ya rank-deficient hota hai. Function ko directly inspect karo ya higher-order terms use karo.

Recall Ex 8 mein Hessian ki jagah Lagrange multipliers kyun chahiye the?

Extremum ek constraint curve par forced tha; free $\nabla f = 0$ condition sirf interior flat points dhundti hai, aur yahan sirf interior critical point saddle tha.

Recall Box problem (Ex 9) mein optimal box cube kyun nahi hai?

Koi top nahi hone ki wajah se, sirf base aur char sides material cost karte hain, toh base ( $4 \times 4$ ) spread out karna aur height low ( $2$ ) rakhna faaydemand hai, rather than saare edges equalize karna.

Connections

Critical points — finding, classifying — parent topic jinhe yeh examples exhaust karte hain.
Hessian Matrix — degenerate ( $D = 0$ ) cases rank-deficient Hessians hain (Ex 6, 7).
Quadratic Forms and Definiteness — min/max/saddle ↔ definite/indefinite forms (Ex 3, 4).
Lagrange Multipliers — constrained extrema (Ex 8).
Taylor Series in Several Variables — $D = 0$ kyun higher-order inspection force karta hai.
Gradient and Directional Derivatives — $\nabla f = 0$ har example shuru karta hai.

Worked examples — Critical points — finding, classifying

Scenario matrix

Case A — ek saaf minimum

Case B — ek saaf maximum

Case C — ek saaf saddle

Case D — mixed-term disguise

Case E — kai critical points, har ek ko classify karo

Case F — $D = 0$ degenerate trap

Case G — critical points ki poori line

Case H — constrained extremum (koi free critical point nahi)

Case I — units ke saath word problem

Active Recall

Connections