Exercises — Lagrange multipliers — one and two constraints

KYA karte hain: sirf words ko standard template mein translate karo. (a) Sab kuch ek side le jao: $g(x,y)=x^2+y^2-25$, aur constraint hai $g=0$. (b) $\nabla f=(3,4)$ (constant — ek plane ke slopes kabhi nahi badalte) aur $\nabla g=(2x,2y)$. (c) Lagrange ki condition: $(3,4)=\lambda (2x,2y),x^2+y^2=25.$ Yeh 3 equations hain ($3=2\lambda x$, $4=2\lambda y$, aur constraint) 3 unknowns $(x,y,\lambda )$ mein. Is count ko pehchaanna hi L1 ka poora point hai.

(a) False. Ek constraint par, $\nabla f$ usually nonzero hota hai; use surface se bahar point karna hota hai, $\nabla g$ ke parallel. $\nabla f=0$ set karna unconstrained rule hai (Unconstrained optimization — critical points). (b) False. $\nabla g$ surface ke perpendicular (normal) point karta hai, uske saath nahi. (c) True. $\lambda =d{f}^{\ast }/dc$ shadow price hai (Dual problem and shadow prices).

$3=2\lambda x$ aur $4=2\lambda y$ se hum $\lambda$ ke terms mein coordinates solve karte hain: $x=\frac{3}{2\lambda },y=\frac{4}{2\lambda }=\frac{2}{\lambda }.$ Divide kyun karte hain: dono component equations already $x,y$ de dete hain; hume bas ek aur relation chahiye ($\lambda$ ka), jo constraint supply karta hai. $x^2+y^2=25$ mein substitute karo: $\frac{9}{4{\lambda }^2}+\frac{4}{{\lambda }^2}=25\Rightarrow \frac{9}{4{\lambda }^2}+\frac{16}{4{\lambda }^2}=\frac{25}{4{\lambda }^2}=25\Rightarrow {\lambda }^2=\frac{1}{4},\lambda =\pm \frac{1}{2}.$ $\lambda =+\frac{1}{2}$ lo: $x=3,y=4$. Tab $f=3(3)+4(4)=9+16=25$. Doosra sign $(-3,-4)$ deta hai, $f=-25$ (min). Max $=25$ at $(3,4)$. Sanity check: $(3,4)$ khud $(3,4)$ ki direction hai, radius-5 circle par scale ki gayi — plane exactly $\nabla f$ ke saath sabse tezi se chadh'ti hai, toh sabse ऊंचा point wahi hai jahan radius same direction mein point kare. ✔

$f=x^2+y^2+z^2$ minimize karo (squared distance — square root se aasaan, same minimizer) subject to $g=2x+y+2z-9=0$. $\nabla f=(2x,2y,2z)=\lambda (2,1,2)=\nabla g\cdot \lambda .$ Componentwise: $2x=2\lambda \Rightarrow x=\lambda$; $2y=\lambda \Rightarrow y=\lambda /2$; $2z=2\lambda \Rightarrow z=\lambda$. Plane mein substitute karo: $2\lambda +\frac{\lambda }{2}+2\lambda =9\Rightarrow \frac{9\lambda }{2}=9\Rightarrow \lambda =2$. Toh $(x,y,z)=(2,1,2)$, aur $f^{\ast }=4+1+4=9$. Minimum squared distance $=9$ (distance $3$). Yeh min kyun hai max nahi: ek flat plane par, origin se distance ka koi upper bound nahi hota lekin ek single lowest point hota hai — perpendicular ka foot.

$\nabla f=(2x,4y)$, $\nabla g=(2x,2y)$. Condition $\nabla f=\lambda \nabla g$: $2x=\lambda 2x,4y=\lambda 2y.$ Case A — $x\ne 0$: pehli equation $\lambda =1$ deti hai. Tab doosri banti hai $4y=2y\Rightarrow 2y=0\Rightarrow y=0$. Constraint: $x^2=1$, toh $x=\pm 1$. Points $(\pm 1,0)$, $f=1$. Case B — $y\ne 0$: doosri $\lambda =2$ deti hai. Tab pehli banti hai $2x=4x\Rightarrow x=0$. Constraint: $y^2=1$, $y=\pm 1$. Points $(0,\pm 1)$, $f=2$. Case C — $x=0$ aur $y=0$: impossible, woh point circle par nahi hai. Toh koi degenerate solution miss nahi hoti. Classify: minimum $f=1$ at $(\pm 1,0)$ (figure mein red dots), maximum $f=2$ at $(0,\pm 1)$ (mint dots). Yeh exactly wahan baithe hain jahan $f$ ke ellipse-shaped level curves circle ko kiss karti hain (Level sets and contours) — tangency $\nabla f\parallel \nabla g$ ka geometric chehra hai. $x,y$ ke sign par split karna kyun zaroori hai: agar tumne blindly $2x=2\lambda x$ se $x$ cancel kiya hota toh $\lambda =2$ branch (jiske liye $x=0$ chahiye) khote ja lete. Har factor ke zero hone ko handle karna L3 ka analysis skill hai.

Parent se, optimum par $x=y=c/2$ aur $\lambda =2x=c$. Optimal value differentiate karo: $\frac{d{f}^{\ast }}{dc}=\frac{d}{dc}\left(\frac{c^2}{2}\right)=c$. Indeed $\lambda =c$. ✔ Interpretation: jaise-jaise $c$ badhta hai line origin se door jaati hai, toh minimum squared distance badhta hai — ek positive $\lambda$ matlab "constraint ko baahir ki taraf relax karne se $f^{\ast }$ baDa ho jaata hai." BaDa $c$ → steep sensitivity. Yeh shadow-price statement concretely bana ke dikhaya gaya hai.

$\nabla f=(0,0,1)$, $\nabla g=(2x,2y,-1)$, $\nabla h=(1,1,1)$. Condition $\nabla f=\lambda \nabla g+\mu \nabla h$: $0=2\lambda x+\mu ,0=2\lambda y+\mu ,1=-\lambda +\mu .$ Pehli do subtract karo: $2\lambda (x-y)=0$. Case $\lambda \ne 0$: $x=y$. (Case $\lambda =0$ eqn 1 se $\mu =0$ force karta hai lekin eqn 3 se $\mu =1$ — contradiction, toh discard karo.) $x=y$ ke saath constraints: $z=2x^2$ aur $2x+z=3$, hence $2x+2x^2=3$, yaani $2x^2+2x-3=0\Rightarrow x=\frac{-1\pm \sqrt{7}}{2}.$ Numerically $x\approx 0.823$ ya $x\approx -1.823$. Phir $z=3-2x$:

• $x\approx 0.823\Rightarrow z\approx 1.354$ — $z$ ka minimum.
• $x\approx -1.823\Rightarrow z\approx 6.646$ — $z$ ka maximum. Do answers kyun: intersection ek closed loop hai jo neeche dip karta hai phir ऊंचा utha hai (figure dekho); $z$ ka ek lowest aur ek highest point hai us par. Dono Lagrange satisfy karte hain kyunki dono woh jagahein hain jahan $\nabla f$, $\mathrm{span}\{\nabla g,\nabla h\}$ mein baithta hai. Min par multipliers ($x=y\approx 0.823$): $0=2\lambda x+\mu$ aur $1=-\lambda +\mu$ se, $\mu =-2\lambda x,1=-\lambda -2\lambda x=-\lambda (1+2x)\Rightarrow \lambda =\frac{-1}{1+2x}.$ $2x=-1+\sqrt{7}$ ke saath, $1+2x=\sqrt{7}$, toh $\lambda =-1/\sqrt{7}\approx -0.378$ aur $\mu =-2\lambda x=2x/\sqrt{7}\cdot ?$ — compute karo: $\mu =-2\lambda x=\frac{2x}{1+2x}=\frac{-1+\sqrt{7}}{\sqrt{7}}\approx 0.622$.

$g=x+y+z-3$. $\nabla f=(yz,xz,xy)=\lambda (1,1,1)$. Toh $yz=xz=xy=\lambda$. $yz=xz$ se: $z(y-x)=0$; kyunki $z>0$, $y=x$. $xz=xy$ se: $x(z-y)=0$; kyunki $x>0$, $z=y$. Hence $x=y=z$. Constraint: $3x=3\Rightarrow x=1$. Toh $(1,1,1)$, $f=1\cdot 1\cdot 1=1$. Interpretation: fixed sum wale positive numbers ka product tab sabse baDa hota hai jab woh equal hon — yeh AM–GM inequality hai jo Lagrange se nikal rahi hai. Maximum $=1$.

Maano $A=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$ toh $f={\mathbf{x}}^{\top }A\mathbf{x}$ jahan $\mathbf{x}=(x,y)$. Tab $\nabla f=2A\mathbf{x}$ aur, $g={\mathbf{x}}^{\top }\mathbf{x}-1$ ke saath, $\nabla g=2\mathbf{x}$. Lagrange demand karta hai $2A\mathbf{x}=\lambda 2\mathbf{x}$, yaani $\boxed{A\mathbf{x}=\lambda \mathbf{x}.}$ Yeh exactly eigenvalue equation hai: constrained critical directions $A$ ke eigenvectors hain, aur $\lambda$ eigenvalue hai. Aise point par value: $f={\mathbf{x}}^{\top }A\mathbf{x}={\mathbf{x}}^{\top }(\lambda \mathbf{x})=\lambda {\mathbf{x}}^{\top }\mathbf{x}=\lambda \cdot 1=\lambda$. Toh unit circle par $f$ eigenvalue ke barabar hoti hai. $f$ ka maximum isliye largest eigenvalue ${\lambda }_{\max }$ hai, uske eigenvector par achieve hota hai; minimum ${\lambda }_{\min }$ hai. ∎ Yeh deep kyun hai: shadow-price meaning ($\lambda =$ $f$ ki value) literal ho jaati hai, aur Lagrange ne abhi eigenvalues ki variational characterization derive kar di hai.

$\nabla f=(1,0)$, $\nabla g=(-3x^2,2y)$. $(0,0)$ par: $\nabla g=(0,0)=0$. Lagrange $\nabla f=\lambda \nabla g$ maangta hai, yaani $(1,0)=\lambda (0,0)=(0,0)$ — kisi bhi $\lambda$ ke liye impossible. Toh algebra cusp par koi solution produce nahi karta, jabki curve genuinely wahan apna leftmost point (minimum of $x$) rakhta hai $x=0$ par. Conclusion: Lagrange yeh extremum miss kar jaata hai kyunki uska core assumption — $\nabla g\ne 0$, toh surface ka well-defined normal ho — ek singular point par fail ho jaata hai. Practice mein Fix: hamesha alag se woh points inspect karo jahan $\nabla g=0$ ho aur koi bhi boundaries hon, jaisa KKT conditions aur parent Mistake C warn karte hain.