Worked examples — Lagrange multipliers — one and two constraints

4.4.15 · D3 · Maths › Multivariable Calculus › Lagrange multipliers — one and two constraints

Shuru karne se pehle: teen words jo hum baar baar use karenge.

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The scenario matrix

Neeche har cell ek alag tarika hai jisme yeh topic kat sakta hai. Har worked example apne cell(s) ka naam leta hai.

#	Cell class	Kya cheez ise alag banati hai	Example
C1	One constraint, dono signs of $\lambda$	$\pm \lambda$ se max aur min	Ex 1
C2	Constraint level $c\ne 0$ + shadow price	verify karo $\lambda =d{f}^{\ast }/dc$, units	Ex 2
C3	Degenerate: $\nabla g=0$ (singular point)	method chupchap ek extremum miss kar deta hai	Ex 3
C4	Boundary / product-of-vars, $xy=0$ branch	us branch ko check karna zaroori hai jo tune divide karke hata di	Ex 4
C5	Real-world word problem (economics)	words ko $f,g$ mein translate karo; $\lambda$ interpret karo	Ex 5
C6	Two constraints, two real roots (max & min)	$\nabla f$ do normals ke span mein	Ex 6
C7	Two constraints, degenerate: parallel normals	jab method toot jaata hai / koi solution nahi	Ex 7
C8	Limiting behaviour: $\lambda \to 0$	constraint inactive ho jaati hai → unconstrained	Ex 8
C9	Exam twist: unbounded / no extremum exists	Lagrange ek aisa point return karta hai jo curve par saddle hai	Ex 9

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Ex 1 — Dono signs of $\lambda$ (cell C1)

Forecast: andaza lagao ki circle par $f$ sabse bada kahan hoga. (Hint: $f$ mein $y$ ka weight zyada hai.)

Figure s01 (neeche): kali unit circle par char candidate points hain. Red wale $(0,\pm 1)$ maximum hain, kale wale $(\pm 1,0)$ minimum; red arrow upar wale point par $\nabla f$ dikhata hai jo seedha radius ke saath bahar point kar raha hai — $\nabla g$ ke parallel — yahi Lagrange condition hai. Dekho ki $\nabla f$ ka koi sideways (tangential) component nahi wahan.

1. Gradients likho. $\nabla f=(2x,4y)$, $\nabla g=(2x,2y)$. Yeh step kyun? Lagrange dono arrows compare karta hai; hume unhe explicitly chahiye.
2. $\nabla f=\lambda \nabla g$ set karo. $2x=\lambda \cdot 2x$ aur $4y=\lambda \cdot 2y$. Kyun? Extremum par $\nabla f$ ka koi tangential component nahi, isliye woh $\nabla g$ ke parallel hai.
3. Har equation dhyan se padho. Pehle se: $2x(1-\lambda )=0$, isliye $x=0$ ya $\lambda =1$. Doosre se: $2y(2-\lambda )=0$, isliye $y=0$ ya $\lambda =2$. Aisa kyun split karte hain? $x$ ya $y$ se divide karne par $x=0$ / $y=0$ solutions chup chap chhoot jaate — circle ke wahi poles. Kabhi kisi variable ko divide karke mat hataao bina yeh check kiye ki woh zero ho sakta hai.
4. Branches enumerate karo.
   • $\lambda =1\Rightarrow y=0\Rightarrow x=\pm 1$: points $(\pm 1,0)$, $f=1$.
   • $\lambda =2\Rightarrow x=0\Rightarrow y=\pm 1$: points $(0,\pm 1)$, $f=2$.
5. Roles assign karo. $f=1$ ($\lambda =1$ par) minimum hai; $f=2$ ($\lambda =2$ par) maximum hai.

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Ex 2 — Nonzero constraint level & shadow price (cell C2)

Forecast: sabse kareeb point wahan hoga jahan origin se line par perpendicular pada ho. Calculate karne se pehle andaza lagao.

Figure s02 (neeche): kali line constraint $x+2y=10$ hai. Red arrow origin se sabse chhota raasta hai — yeh line ko right angle par red point $(2,4)$ par hit karta hai. Woh perpendicularity yahi hai jo $\nabla f\parallel \nabla g$ yahan force karta hai.

1. Gradients. $\nabla f=(2x,2y)$, $\nabla g=(1,2)$. Kyun? Distance-squared, distance se smoother hai aur uska minimizer same hota hai — differentiate karne ke liye koi square roots nahi.
2. $\nabla f=\lambda \nabla g$: $2x=\lambda$, $2y=2\lambda$. Isliye $x=\lambda /2$, $y=\lambda$. $c$ symbolic kyun rakhein? Hum $f^{\ast }$ ko $c$ ke function ke roop mein chahte hain taaki shadow-price claim test kar sakein.
3. Constraint impose karo $x+2y=c$: $\frac{\lambda }{2}+2\lambda =c\Rightarrow \frac{5\lambda }{2}=c\Rightarrow \lambda =\frac{2c}{5}$. Constraint mein wapas substitute kyun karein? Steps 1–2 sirf answer tak direction fix karte hain (ratio $x:y$); constraint hi woh cheez hai jo line par actual location pin karti hai.
4. Point & value. $\lambda =\frac{2c}{5}$ ko $x=\lambda /2$ aur $y=\lambda$ mein substitute karo: $x=\frac{c}{5},y=\frac{2c}{5}$, isliye $f^{\ast }=\frac{c^2}{25}+\frac{4c^2}{25}=\frac{5c^2}{25}=\frac{c^2}{5}$. $f^{\ast }$ ko $c$ mein formula kyun compute karein? Kyunki is cell ka poora point yahi hai ki $f^{\ast }$ ko $c$ ke respect mein differentiate karein aur $\lambda$ se compare karein — formula se yeh possible hota hai.
5. $c=10$ plug karo: $x=2,y=4$, $f^{\ast }=\frac{100}{5}=20$; distance $=\sqrt{20}=2\sqrt{5}$. Distance bhi kyun report karein? Original question closest point poochh raha tha, isliye $f^{\ast }$ (squared distance) ko actual distance mein translate karte hain.

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Ex 3 — Degenerate: $\nabla g=0$ ek extremum miss kar deta hai (cell C3)

Forecast: yeh curve ek leftward-pointing spike jaisi lagti hai. $x$ sabse chhota kahan hoga? Kya Lagrange ise find kar payega?

Figure s03 (neeche): kali curve cusp $y^2=x^3$ hai; kala arrow "$x$ decrease karo" direction mein point karta hai. Red dot cusp tip $(0,0)$ ko mark karta hai jahan curve ka sharp corner hai aur $\nabla g=0$ — koi normal arrow draw nahi ho sakta, aur yahi wajah hai ki Lagrange ki equation wahan pahunch nahi sakti.

1. Gradients. $\nabla f=(1,0)$, $\nabla g=(-3x^2,2y)$.
2. $\nabla f=\lambda \nabla g$: $1=-3\lambda x^2$ aur $0=2\lambda y$. Doosre se: $\lambda =0$ ya $y=0$. Lekin $\lambda =0$ pehli equation ko $1=0$ bana deta hai — impossible. Isliye $y=0$. Yeh kyun matter karta hai? $y=0$ force karta hai $x^3=0$, yani point $(0,0)$.
3. Point check karo. $(0,0)$ par: $\nabla g=(0,0)=0$. Multiplier equation $1=-3\lambda \cdot 0$ padhi jaaye to $1=0$ — koi finite $\lambda$ kaam nahi karta. Kyun? Cusp par surface ka koi well-defined normal nahi; $\nabla g=0$ exactly woh singular case hai jiske baare mein parent note ne warn kiya tha (Mistake C).
4. Haath se resolve karo. Curve par $x=y^{2/3}\ge 0$, isliye $x$ minimized hai $x=0$ par, cusp $(0,0)$ par. Extremum real hai — Lagrange ki equation bas ise produce nahi kar saki.

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Ex 4 — Variables ka product, $xy=0$ branch (cell C4)

Forecast: $f=xy$ quadrants II/IV mein negative hai, I/III mein positive. Isliye max aur min symmetric $\pm$ values hone chahiye.

1. Gradients. $\nabla f=(y,x)$, $\nabla g=\left(\frac{x}{2},2y\right)$.
2. $\nabla f=\lambda \nabla g$ set karo: $y=\lambda \frac{x}{2}$, $x=\lambda 2y$. Kyun? Constrained extremum par $\nabla f$ ka ellipse ke saath koi tangential component nahi, isliye woh ellipse ke normal $\nabla g$ ke parallel hona chahiye — wahi parallelism exactly $\nabla f=\lambda \nabla g$ hai.
3. Unhe multiply karo: $y\cdot x=\left(\lambda \frac{x}{2}\right)(2\lambda y)={\lambda }^{2}xy$. Isliye $xy(1-{\lambda }^2)=0$. Divide kyun nahi, factor kyun? $xy$ se divide karne par $xy=0$ branch chhup jaati hai. Factor karne se dono branches bachti hain.
4. Branch A: $xy=0$. Tab $f=0$; points $(\pm 2,0)$ aur $(0,\pm 1)$. Yeh saddle-value candidates hain (na max na min).
5. Branch B: ${\lambda }^2=1$, $\lambda =\pm 1$.
   • $\lambda =1$: $y=x/2$; constraint $\frac{x^2}{4}+\frac{x^2}{4}=1\Rightarrow x^2=2\Rightarrow x=\pm \sqrt{2}$, $y=\pm \frac{\sqrt{2}}{2}$. $f=xy=\sqrt{2}\cdot \frac{\sqrt{2}}{2}=1$.
   • $\lambda =-1$: $y=-x/2$; same magnitude, $f=-1$.

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Ex 5 — Real-world word problem: economics (cell C5)

Forecast: capital 4 guna mehnga hai, to kya firm zyada labour ya capital kharidegi?

1. Translate karo. Constraint $g=x+4y-c=0$ budget level $c=100$ ke saath (poora budget kharcho). Maximize $f=\sqrt{x}\sqrt{y}$. Poora budget kyun? Output dono inputs mein badhta hai, isliye optimum sab kharach kar deta hai. $c$ symbolic kyun rakhein? Hume $f^{\ast }$ chahiye $c$ ke function ke roop mein taaki $\lambda$ interpret kar sakein.
2. Gradients. $\nabla f=\left(\frac{1}{2}{x}^{-0.5}{y}^{0.5},\frac{1}{2}{x}^{0.5}{y}^{-0.5}\right)$, $\nabla g=(1,4)$.
3. Equations. $\frac{1}{2}\sqrt{\frac{y}{x}}=\lambda$ aur $\frac{1}{2}\sqrt{\frac{x}{y}}=4\lambda$.
4. Dono ko divide karo (dono sides positive hain, isliye safe hai): $\frac{\sqrt{y/x}}{\sqrt{x/y}}=\frac{1}{4}\Rightarrow \frac{y}{x}=\frac{1}{4}\Rightarrow x=4y$. Yahan divide kyun? $\lambda$ instantly cancel ho jaata hai, inputs ka ratio milta hai — answer ka economic core yahi hai.
5. Budget $c$ ke terms mein: $x+4y=4y+4y=8y=c\Rightarrow y=\frac{c}{8},x=\frac{c}{2}$. General $c$ kyun solve karein? Taaki optimal-value formula $f^{\ast }(c)$ directly agla step mein nikal aaye.
6. Optimal value $c$ ke function ke roop mein. $f^{\ast }(c)=\sqrt{x}\sqrt{y}=\sqrt{\frac{c}{2}}\sqrt{\frac{c}{8}}=\sqrt{\frac{c^2}{16}}=\frac{c}{4}$. Kyun? Numbers plug karne se pehle yeh general solution hai; ise differentiate karne par shadow price bina kisi guesswork ke milti hai.
7. $c=100$ plug karo. $x=50,y=12.5$, output $f^{\ast }=\frac{100}{4}=25$. Multiplier $\lambda =\frac{1}{2}\sqrt{y/x}=\frac{1}{2}\sqrt{12.5/50}=\frac{1}{2}\cdot \frac{1}{2}=0.25$.

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Ex 6 — Do constraints, do real roots (cell C6)

Forecast: intersection ek tilted circle hai. $z$ iske upar aur neeche ek highest aur lowest point hit karna chahiye.

Figure s04 (neeche): tilted intersection circle (kala) ka side view. Red dot circle ka top hai, $(-1,-1,2)$, jahan $z$ sabse bada hai; kala dot bottom par, $(1,1,-2)$, jahan $z$ sabse chhota hai. Ek closed curve par do extremes — isliye hum do real roots expect karte hain.

1. Gradients. $\nabla f=(0,0,1)$, $\nabla g=(2x,2y,2z)$, $\nabla h=(1,1,1)$.
2. $\nabla f=\lambda \nabla g+\mu \nabla h$: $0=2\lambda x+\mu ,0=2\lambda y+\mu ,1=2\lambda z+\mu$. Combination kyun? Intersection curve par, $\nabla f$ ko do normals ke span (plane) mein lie karna zaroori hai — koi akela normal kaafi nahi.
3. Pehle do subtract karo: $2\lambda (x-y)=0\Rightarrow \lambda =0$ ya $x=y$. Agar $\lambda =0$ to equation 3 deta hai $1=\mu$ lekin equations 1,2 dete hain $\mu =0$ — contradiction. Isliye $x=y$. Do equations subtract kyun? Dono $0$ ke barabar hain, isliye unka difference ek step mein unknown $\mu$ eliminate kar deta hai aur $x$ aur $y$ ke beech ek clean relation haath mein aa jaata hai — teen variables ko ek ki taraf collapse karne ka sabse tez raasta.
4. Points find karne ke liye constraints use karo. $h$: $2x+z=0\Rightarrow z=-2x$. $g$: $2x^2+z^2=6\Rightarrow 2x^2+4x^2=6\Rightarrow 6x^2=6\Rightarrow x=\pm 1$. Ab $g,h$ kyun laao? Step 3 ne curve par ek direction fix ki; do constraint equations actual points pin karti hain jahan woh direction circle se milti hai.
   • $x=1$: $y=1,z=-2$: point $(1,1,-2)$, $z=-2$ (minimum).
   • $x=-1$: $y=-1,z=2$: point $(-1,-1,2)$, $z=2$ (maximum).
5. Multipliers recover karo. $x=-1,y=-1,z=2$ lo. Equation 1 se: $\mu =-2\lambda x=2\lambda$. Equation 3 mein substitute karo: $1=2\lambda z+\mu =2\lambda (2)+2\lambda =6\lambda \Rightarrow \lambda =\frac{1}{6}$, phir $\mu =2\lambda =\frac{1}{3}$. $\lambda ,\mu$ ki parwah kyun? Yeh confirm karte hain ki $\nabla f$ sach mein do normals ke span mein baithta hai (ek valid solution exist karta hai), aur $\lambda ,\mu$ do shadow prices hain — $z^{\ast }$ ki sensitivity sphere aur plane ko respectively relax karne ke respect mein.
6. *Do roots kyun? Tilted circle ka $z$-direction mein ek upar aur ek neeche hota hai; dono signs $x=\pm 1$ exactly wahi do points hain — ek max, ek min.

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Ex 7 — Do constraints, degenerate parallel normals (cell C7)

Forecast: calculate karne se pehle do planes ko dhyan se dekho.

1. Constraints padho. $h=2x+2y+2z-5=2(x+y+z)-5$. Lekin $g=0$ kehta hai $x+y+z=1$, isliye $h=2(1)-5=-3\ne 0$. Pehle check kyun? Dono planes parallel hain ($\nabla h=2\nabla g$) aur coincide nahi karte — isliye feasible set empty hai.
2. Gradients se confirm karo. $\nabla g=(1,1,1)$, $\nabla h=(2,2,2)=2\nabla g$. Yeh linearly dependent hain (ek doosre ka stretch hai). Fatal kyun? Lagrange ke two-constraint formula ko $\nabla g,\nabla h$ independent chahiye taaki unka span ek genuine plane ho. Agar woh parallel hain, to "plane" ek single line mein collapse ho jaati hai aur method ki assumption fail ho jaati hai (parent Mistake C, two-constraint version).
3. Conclusion. Koi feasible point exist nahi karta ⇒ koi extremum nahi. Lagrange sahi taur par koi solution produce karne se mana kar deta hai.

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Ex 8 — Limiting behaviour $\lambda \to 0$: constraint inactive ho jaati hai (cell C8)

Forecast: $f$ ka unconstrained minimum $(3,4)$ hai, jahan $x+y=7$. $c=7$ par constraint kya karti hai?

1. Gradients. $\nabla f=(2(x-3),2(y-4))$, $\nabla g=(1,1)$.
2. Equations. $2(x-3)=\lambda$, $2(y-4)=\lambda$. Isliye $x-3=y-4\Rightarrow y=x+1$.
3. Constraint. $x+(x+1)=c\Rightarrow x=\frac{c-1}{2},y=\frac{c+1}{2}$.
4. Multiplier. $\lambda =2(x-3)=2\left(\frac{c-1}{2}-3\right)=c-7$. Yeh kyun dekhen? $\lambda$ measure karta hai ki constraint kitni "hard push" kar rahi hai. Jab $c\to 7$, $\lambda \to 0$.
5. Interpret karo. $c=7$ par constrained minimizer $(3,4)$ hai — unconstrained minimizer, jo pehle se hi line par lie karta hai. Constraint kuch cost nahi karti, isliye uski shadow price zero hai. Yeh $\lambda =0$ limit Unconstrained optimization — critical points se link karti hai.

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Ex 9 — Exam twist: Lagrange point jo maximum nahi hai (cell C9)

Forecast: hyperbola sketch karo. Kya $x$ ka is par koi maximum hai hi?

Figure s05 (neeche): kala hyperbola do branches mein hai. Red dot $(1,0)$ par hai jahan Lagrange land karta hai; kala annotation dikhata hai ki $x$ bina bound ke badhta rehta hai jab tum right branch par upar slide karte ho — isliye woh red dot branch par $x$ ka minimum hai, maximum nahi.

1. Gradients. $\nabla f=(1,0)$, $\nabla g=(2x,-2y)$. Kyun? Hum dono arrows "twist" problem mein bhi banaate hain — trap result interpret karne mein hai, setup mein nahi.
2. Equations. $1=2\lambda x$ aur $0=-2\lambda y$. Doosra deta hai $\lambda =0$ (reject, kyunki yeh equation 1 khatam kar deta hai) ya $y=0$. Isliye $y=0$. $\lambda =0$ reject kyun? Agar $\lambda =0$, equation 1 padhi jaaye to $1=0$, impossible — isliye sirf live branch $y=0$ hai.
3. Constraint impose karo. $y=0$ ke saath: $x^2-1=0\Rightarrow x=\pm 1$. Candidates $(\pm 1,0)$, $\lambda =\frac{1}{2x}=\pm \frac{1}{2}$. Yahan $\lambda$ kyun compute karein? Yeh confirm karta hai ki dono candidates legitimate stationary points hain (finite non-zero $\lambda$); problem yeh nahi ki Lagrange fail hua, balki yeh ki points ka kya matlab hai.
4. Honestly classify karo. Right branch ko parametrize karo $x=\sqrt{1+y^2}$ ke roop mein. Jab $|y|\to \infty$, $x\to \infty$ — $x$ bina bound ke badhta hai. Isliye us branch par $x$ ka koi maximum nahi; point $(1,0)$ wahan hai jahan $x$ sabse chhota hai (vertex), yani yeh ek minimum hai, maximum nahi. Parametrize kyun? Yeh implicit curve ko ek variable ke explicit function mein convert karta hai taaki hum $x$ ka behaviour directly dekh sakein aur max-vs-min guessing ki jagah settle kar sakein.
5. Conclusion. Jaise problem rakha gaya hai usmein hyperbola par $x$ ka koi maximum nahi ($\sup x=+\infty$). Lagrange ne sach mein $(1,0)$ ek stationary point ke roop mein return kiya, lekin yeh branch minimum hai; $(-1,0)$ left branch par analogous vertex hai. Seekh: Lagrange point ko label karne se pehle hamesha pucho ki kya extremum exist bhi kar sakta hai (kya objective feasible set par bounded hai?).

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Recall Self-test

Unit circle par, $f=x^2+2y^2$ do alag $\lambda$ values kyun deta hai? ::: Kyunki max ($(0,\pm 1)$ par, $\lambda =2$) aur min ($(\pm 1,0)$ par, $\lambda =1$) alag constrained extrema hain, har ek ka apna parallel-scaling factor hota hai. Kaunsa ek check Ex 7 ko instantly kill kar deta? ::: Test karo ki $\nabla g$ aur $\nabla h$ linearly independent hain ya nahi — yahan $\nabla h=2\nabla g$, isliye feasible set empty ho sakta hai aur method applicable nahi hai.

Connections

• Lagrange multipliers — one and two constraints — parent theory jise yeh examples exercise karte hain.
• Gradient and directional derivative — upar har $\nabla f$ arrow.
• Tangent planes and normal vectors — $\nabla g,\nabla h$ as normals (Ex 6, 7).
• Unconstrained optimization — critical points — $\lambda \to 0$ limit (Ex 8).
• KKT conditions — kya karna hai jab constraints inequalities hon / boundaries appear karein (Ex 9).
• Level sets and contours — Ex 1, 4 ke peeche tangency picture.
• Dual problem and shadow prices — Ex 2, 5 mein $\lambda$ interpretation.