Question bank — Isoperimetric problems — constraints (Lagrange multipliers in variational sense)

False. Plain E–L saari admissible functions par ghoomta hai, including wo jo $K=\ell$ ko tod deti hain; tumhe augmented $F=f-\lambda g$ ko optimise karna hoga taaki variations constraint surface par bani rahein.

True. Ek integral constraint ek number ki worth ki restriction impose karta hai, isliye ek constant $\lambda$ kaafi hai — compare karo pointwise constraints $g(x,y)=0$ se jinhe $\lambda (x)$ chahiye hoti hai.

False. $\lambda$ constraint ke zariye fix hota hai; $\ell$ badlo aur $\lambda$ bhi generally badlega — actually $\lambda =d{J}^{\ast }/d\ell$ hai jo marginal rate of change hai (dekho Shadow Price / Envelope Theorem).

False. Ek knob dono kaam nahi kar sakta — $K$ ko fixed rakhna aur $J$ ki stationarity probe karna; tumhe two-parameter family $\bar{y}+{ϵ}_1{\eta }_1+{ϵ}_2{\eta }_2$ chahiye taaki multiplier emerge ho sake.

True. Beltrami first integral ko integrate karne par $(x-x_0{)}^2+(y-C{)}^2={\lambda }^2$ milta hai, toh $\lambda$ literally circle ka radius hai.

False. Fixed length par centre of mass ki height minimise karne se $y=\lambda +C\cosh \left(\frac{x-x_0}{C}\right)$ milta hai — ek hyperbolic cosine (Catenary Curve), jo sirf bottom ke paas parabolic dikhta hai.

True. Teen unknowns hain $C_1,C_2,\lambda$, jinhe teen conditions match karti hain: do boundary values $y(a),y(b)$ aur constraint $\int gdx=\ell$.

$F$ ki E–L equation sign se form mein nahi badlti; sign sirf $\lambda$ ki numerical value ko flip karta hai. Lekin agar tum chahte ho ki $\lambda$ apna shadow-price meaning $d{J}^{\ast }/d\ell$ rakhe, consistent raho — sign "galat" nahi hai, bas ek bookkeeping convention hai jise mid-problem mix nahi karna chahiye.

Error: ${\eta }_1$ $\delta K=0$ enforce karne mein kharach ho jaata hai, isliye wo free nahi hai. Sirf ${\eta }_2$ essentially arbitrary rehta hai, aur usi degree of freedom par fundamental lemma act karta hai taaki bracket zero ho jaaye.

Error: kyunki boundary values fixed hain, har bump ${\eta }_i(a)={\eta }_i(b)=0$ satisfy karta hai, toh wo boundary terms exactly zero hain. Wo contribute nahi karte, vanish ho jaate hain.

Error: $\lambda$ constraint ${\int }_a^{b}gdx=\ell$ se determine hota hai, $J$ ko re-optimise karne se nahi. Solved $y(x)$ (jo abhi bhi $\lambda$ contain karta hai) ko $K$ mein plug karo aur use $\ell$ ke barabar set karo.

Error: Beltrami Identity ek first integral hai sirf tab jab $F=f-\lambda g$ mein koi explicit $x$ na ho. Agar $f$ ya $g$ directly $x$ par depend karta hai, toh tumhe poora E–L equation solve karna hoga.

Error: length yahan constraint hai, objective nahi. Dido fixed length par area maximise karta hai; straight line sabse choti curve hai, jo zero enclosed area deti hai — bilkul opposite extremal.

Use delete karna problem badal deta hai; $\lambda \times$(constraint) add karne se surface $K=\ell$ par optimum unchanged rehta hai (kyunki wahan added term ek fixed constant hai) jabki ek hard constrained search ko free search mein convert karta hai.

Kyunki two-parameter family functional problem ko $({ϵ}_1,{ϵ}_2)$-space mein ek ordinary constrained extremum mein compress kar deti hai, jahan finite-dimensional rule directly apply hota hai; variational statement usi ki limit hai.

Envelope theorem ke zariye, $\lambda =d{J}^{\ast }/d\ell$: yeh batata hai ki budget relaxed karne per unit par optimal objective kitna improve hota hai — constraint dhila karne ki marginal value.

Ek integral constraint ek global scalar condition impose karta hai, jo ek scalar $\lambda$ se match hoti hai; ek pointwise condition har $x$ par hold karti hai, isliye use har jagah enforce karne ke liye puri function $\lambda (x)$ chahiye.

Augmented Beltrami integral closed form mein solvable hai aur $(x-x_0{)}^2+(y-C{)}^2={\lambda }^2$ deta hai — sirf constant curvature wali curve — jo Isoperimetric Inequality ka equality case hai.

Beltrami ek first integral hai: usne explicit $x$ ki absence use karke E–L equation ko ek baar already integrate kar liya hai, differentiation ka ek order ek integration constant $C$ ke badle mein de diya.

Tab $\lambda =0$ aur constraint inactive hai — isoperimetric solution unconstrained se coincide karti hai, bilkul waise jaise ek slack Lagrange constraint zero multiplier deta hai.

Unhe join karne wali sirf wahi curve us length ki hai jo straight segment hai, toh "search" degenerate hai — ek unique admissible function hai aur optimise karne ki koi jagah nahi; extremal forced hai.

Koi admissible curve exist nahi karti — tum do points ko straight line se choti arc se connect nahi kar sakte, isliye constrained problem empty hai aur koi solution nahi hai.

Curve straight chord ki taraf tight hoti jaati hai; $\cosh$ solution mein $C\to \infty$, hyperbolic cosine ko nearly straight line mein flatten kar deta hai (iska parabola-jaisa bottom poori visible curve ban jaata hai).

Constant $\lambda$ ki jagah multiplier function $\lambda (x)$ lo aur $F=f-\lambda (x)g$ banao; constraint tab har $x$ par hold karni chahiye, sirf average mein nahi.

Koi interior extremal return nahi hota; optimum tab admissible set ki boundary par hota hai (jaise ek endpoint configuration), jo E–L-plus-multiplier machinery nahi dekh sakti — yeh sirf interior stationary curves find karta hai.