Question bank — Optimization — constrained, unconstrained, real-world problems

True. Fermat's Theorem ke mutabik, ek differentiable function ek taraf positive slope aur doosri taraf negative slope nahi rakh sakta bina slope zero se guzre, isliye wahan $f^{\prime }=0$ hoga.

False. $f(x)=x^3$ mein $f^{\prime }(0)=0$ hai phir bhi $0$ ek flat inflection hai — graph ussi mein se chadhta rehta hai. Critical points sirf candidates hote hain.

False jaise yeh likha hai — tumhe pehle jaanna chahiye ki $f^{\prime }(c)=0$ aur $f$ $c$ ke paas twice differentiable hai. Second-derivative test sirf critical points ko classify karta hai; $f^{\prime \prime }>0$ akela sirf yeh kehta hai ki graph wahan cup-shaped hai, jo rising slopes par bhi hota hai.

True (Extreme Value Theorem). Closedness + continuity guarantee karti hai ki extremes reach hote hain, chahe critical point par ya endpoint par.

False. $(0,1)$ par, $f(x)=x$ $1$ ki taraf chadhta hai par kabhi reach nahi karta — $1$ supremum (least upper bound) hai phir bhi attained nahi hota. Openness guarantee ko khatam kar deti hai.

False. Tumhe dono endpoints $a$ aur $b$ bhi test karne chahiye; global extreme aksar boundary par baithta hai jahan $f^{\prime }$ ka vanish hona zaroori nahi.

True. Kyunki $\sqrt{\cdot }$ strictly increasing hai $[0,\infty )$ par, jahan bhi $D^2$ sabse chhota hoga, $D$ bhi sabse chhota hoga — lekin $D^2$ messy square-root derivatives se bachata hai.

False. Lagrange condition candidates dhundhti hai — yeh maxima, minima, aur saddles sab ko flag karti hai. Tumhe phir bhi $f$-values compare karne honge (ya geometry check karni hogi) yeh jaanne ke liye ki tumhare paas kya hai.

False. $\lambda$ positive, negative, ya zero ho sakta hai; iska sign tumhe batata hai ki constraint ko loosen karne se optimum kis direction mein change hota hai. Iska magnitude sensitivity hai.

Galat conclusion. $f^{\prime \prime }(0)=0$ ka matlab sirf yeh hai ki test inconclusive hai, yeh nahi ki koi extremum nahi hai. First-derivative sign test dikhata hai $f^{\prime }<0$ $0$ ke left mein aur $f^{\prime }>0$ right mein — isliye $0$ ek genuine minimum hai.

Domain error. Base $12-2x$ ke saath, $x=6$ se $12-2x=0$ milta hai, zero volume deta hai. Yeh $V^{\prime }$ ka ek legitimate root hai lekin boundary par ek minimum hai; asli answer interior root $x=2$ hai (volume $128{\text{cm}}^3$).

Galat. Agar $f^{\prime }$ kabhi vanish nahi karta, to function monotonic hai, isliye maximum simply ek endpoint par baithta hai. Interior critical points ka na hona extrema ko khatam nahi karta.

Ulta. Ek corner ya cusp precisely isliye ek critical point hai kyunki $f^{\prime }(c)$ wahan exist nahi karta. Definition mein dono "slope zero" aur "slope undefined" shamil hain.

Error. Substitution ise ek 1-D problem mein badal deta hai jiske phir bhi endpoints hote hain (variable ki physical limits). Unhe skip karna true extreme miss kar sakta hai, exactly jaise box problem mein.

Quantities ki confusion. Iska matlab hai ki gradient arrows parallel hain (ek doosre ka scalar multiple hai), na ki function values match karti hain. Yeh direction ke baare mein hai, size ya value ke nahi.

Ek critical point ke paas ($f$ wahan twice differentiable hone par), increment deta hai $f(c+h)-f(c)\approx \frac{1}{2}{f}^{\prime \prime }(c)h^2$; kyunki $h^2>0$ kisi bhi small step $h\ne 0$ ke liye, sirf $f^{\prime \prime }(c)$ ka sign change ko upar ya neeche flip karta hai. Taylor Series Expansion aur Fig 1 dekho.

Kyunki critical point par $f^{\prime }(c)=0$ hai, isliye woh poora term vanish ho jaata hai, $h^2$ term ko leading behaviour ke roop mein chhodta hai jo local shape govern karta hai.

Agar $\nabla f$ ka koi component curve ke saath hota, to hum us direction mein slide karke $f$ badha sakte aur feasible rehte — to abhi koi optimum nahi. Sirf jab tangent component zero ho (yaani $\nabla f\parallel \nabla g$) tab hum stuck hain. Lagrange Multipliers (Multivariable) aur Fig 3 dekho.

Ek single-variable function ka ek ordinary derivative hota hai jise hum directly zero set kar sakte hain, same Closed Interval logic use karke; yeh full multivariable machinery ki zaroorat se bachata hai jab constraint solve karna aasaan ho.

Lagrange $x=y$ deta hai (sum ke dono partials equal hain), aur geometrically product $xy$ badhta hai jaise dono factors balance hote hain — yeh AM-GM Inequality hai jo calculus ka disguise pehne hue hai.

Kisi point se ek line tak ka sabse chhota segment hamesha line se right angle par milta hai; agar tumhare answer ka segment perpendicular nahi hai (slopes ka product $-1$ nahi hai), to tumne koi error ki hai. Fig 2 dekho.

Dono ek word problem ko ek symbolic relation mein translate karke aur phir differentiate karke shuru karte hain — calculus machinery identical hai; sirf woh question alag hai jo $f^{\prime }$ se pucha jaata hai.

Us interval ka har point ek critical point hai; function wahan constant hai, isliye woh simultaneously ek (weak) local max aur min dono hai. Koi single "peak" nahi — tum constant value report karte ho.

Endpoint ek boundary maximum hai; Fermat's Theorem apply nahi hota (yeh sirf interior points ke liye hai), exactly isliye Closed Interval Method endpoints ko alag se test karta hai.

Phir $12-2x\to 0$ koi material nahi chhodta; domain collapse hokar single point $x=0$ par aa jaata hai jahan volume $0$ hai. Optimize karne se pehle hamesha confirm karo ki feasible domain non-empty hai.

Equation $\nabla f=\lambda \nabla g$ kisi bhi $\lambda$ se satisfy hoti hai, isliye method koi information nahi deta; tumhe us point ko directly aas-paas ki values inspect karke analyze karna padega.

Agar dono tie karein aur endpoints se zyada hon, to global maximum dono par simultaneously attain hota hai. Global extrema necessarily unique nahi hote.

Careful raho: test smooth function ke local behaviour ko describe karta hai, lekin boundary ek side cut off kar sakti hai. Global picture settle karne ke liye actual values compare karne par fall back karo (Closed Interval Method).