Worked examples — Optimization — constrained, unconstrained, real-world problems

4.1.31 · D3 · Maths › Calculus I — Limits & Derivatives › Optimization — constrained, unconstrained, real-world proble

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The scenario matrix

Har optimization problem inhi cells mein se kisi ek mein hoti hai. Hum har row mein kam se kam ek example solve karenge.

Cell	Kya cheez special hai	Answer kahan milta hai	Example
A. Interior max	smooth hilltop, $f^{\prime \prime }<0$	andar ka ek critical point	Ex 1
B. Interior min	smooth valley, $f^{\prime \prime }>0$	andar ka ek critical point	Ex 2
C. Endpoint wins	derivative kabhi zero nahi, ya critical point haar jaata hai	domain ki boundary	Ex 3
D. Degenerate / $f^{\prime \prime }=0$	second-derivative test andha hai	first-derivative sign test chahiye	Ex 4
E. Constrained (substitution)	ek variable ek side relation se khatam kar diya	reduced problem ka interior	Ex 5
F. Constrained (Lagrange)	dono variables rakho, $\nabla f=\lambda \nabla g$ use karo	tangency point	Ex 6
G. Real-world word problem	prose ko symbols mein, phir derivative mein translate karo	jahan bhi recipe le jaaye	Ex 7
H. Exam twist (hidden domain / units)	"obvious" root illegal hai ya galat cheez optimize ho rahi hai	domain / units check ke baad	Ex 8

Koi bhi symbol aane se pehle, ek convention.

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Cell A — Interior maximum

Forecast: aage padhne se pehle sabse bade rectangle ki shape guess karo. (Square? Lamba patla strip?)

1. Naam do aur model banao. Maano sides $x$ aur $y$ hain (metres mein). Area $A=xy$. Yeh step kyun? Jo cheez hum optimize karna chahte hain use symbols mein likhna zaroori hai differentiate karne se pehle.
2. Ek variable mein constrain karo. Fence perimeter hai: $2x+2y=40\Rightarrow y=20-x$. Yeh step kyun? $A=xy$ mein do unknowns hain; 1-D derivative ke liye ek chahiye. Perimeter relation se $y$ eliminate ho jaata hai.
3. Reduce karo. $A(x)=x(20-x)=20x-x^2$, domain $0\le x\le 20$ (ek side negative nahi ho sakti, aur $x=20$ se $y$ ke liye kuch nahi bachta). Yeh step kyun? Ab $A$ ek variable mein plain parabola hai — figure mein red curve dekho, ek ulta bowl.
4. Derivative zero karo. $A^{\prime }(x)=20-2x=0\Rightarrow x=10$. Yeh step kyun? ∩-shaped area curve ke top par slope flat hoti hai, isliye slope ko zero set karte hain.
5. Classify karo. $A^{\prime \prime }(x)=-2<0$ → cap-shaped → second-derivative test se local maximum. Yeh step kyun? Zero slope akele hilltop, dip, ya inflection ho sakta hai; negative curvature hilltop confirm karta hai.
6. Answer. $x=10$, $y=20-10=10$, area $=100{\text{m}}^2$. Optimal pen ek square hai.

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Cell B — Interior minimum

Forecast: kya sabse sasta can tall aur thin hoga, short aur fat, ya kuch beech mein?

1. Model banao. Ek closed cylinder ki surface area: $S=2\pi r^2+2\pi rh$ (do circular lids aur wall). Yeh step kyun? Metal used = total surface area; wahi quantity minimize karni hai.
2. Constrain karo. Volume height fix karta hai: $\pi r^{2}h=500\Rightarrow h=\frac{500}{\pi r^2}$. Yeh step kyun? Do unknowns $r,h$ hain; volume condition $h$ remove kar deta hai.
3. Reduce karo. $S(r)=2\pi r^2+2\pi r\cdot \frac{500}{\pi r^2}=2\pi r^2+\frac{1000}{r}$, domain $r>0$. Yeh step kyun? Ab $S$ sirf $r$ par depend karta hai. Jab $r\to 0^{+}$ toh $\frac{1000}{r}$ term blow up hoti hai; jab $r\to \infty$ toh $2\pi r^2$ term blow up hoti hai — toh minimum beech mein kahi hona chahiye.
4. Derivative zero karo. $S^{\prime }(r)=4\pi r-\frac{1000}{r^2}=0\Rightarrow 4\pi r^3=1000\Rightarrow r^3=\frac{250}{\pi }$. Toh $r={\left(\frac{250}{\pi }\right)}^{1/3}\approx 4.301\text{cm}$. Yeh step kyun? ∪-shaped cost curve ke bottom par slope flat hoti hai.
5. Classify karo. $S^{\prime \prime }(r)=4\pi +\frac{2000}{r^3}>0$ sab $r>0$ ke liye → cup → minimum. Yeh step kyun? Valley confirm karta hai, peak nahi.
6. Answer. $r\approx 4.30\text{cm}$, $h=\frac{500}{\pi r^2}\approx 8.60\text{cm}$ — note karo $h=2r$: optimal can bilkul utna hi tall hai jitna wide hai.

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Cell C — Endpoint jeet jaata hai

Forecast: sabse chhoti value kahan hogi — andar, ya kisi edge par?

1. Differentiate karo. $f^{\prime }(x)=-e^{-x}$. Yeh step kyun? Pehle flat spots dhundhte hain.
2. $f^{\prime }=0$ set karo. $-e^{-x}=0$ ka koi solution nahi hai — $e^{-x}$ hamesha positive hota hai, toh $f^{\prime }(x)<0$ har jagah. Yeh step kyun? Koi interior critical point exist nahi karta; slope hamesha downhill hai (strictly falling red curve dekho).
3. Closed Interval Method par fall back karo. Koi interior candidate nahi hai toh extremes dono endpoints hone chahiye. Evaluate karo: $f(0)=e^0=1$ (maximum), $f(3)=e^{-3}\approx 0.0498$ (minimum). Yeh step kyun? Ek closed interval par continuous function hamesha apne extremes attain karta hai; agar andar nahi toh wall par milenge.
4. Answer. Minimum value $e^{-3}\approx 0.0498$ at $x=3$; maximum $1$ at $x=0$.

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Cell D — Second-derivative test fail ho jaata hai ($f^{\prime \prime }=0$)

Forecast: kya $x=0$ min hai, max hai, ya kuch nahi?

1. Critical points dhundho. $f^{\prime }(x)=4x^3=0\Rightarrow x=0$. Yeh step kyun? Flat-spot hunt.
2. Second-derivative test try karo. $f^{\prime \prime }(x)=12x^2$, toh $f^{\prime \prime }(0)=0$ — inconclusive. Yeh step kyun? Parent ka Taylor argument $f^{\prime \prime }(c)h^2$ use karta tha; agar $f^{\prime \prime }(c)=0$ toh woh term vanish ho jaati hai aur change ka sign nahi bata sakti.
3. Iske badle first-derivative sign test use karo. $f^{\prime }(x)=4x^3$ ka sign har side check karo:
   • $x<0$ ke liye: $4x^3<0$ (downhill).
   • $x>0$ ke liye: $4x^3>0$ (uphill). Yeh step kyun? Ek point ke aas-paas down-then-up bilkul valley ka fingerprint hai — koi curvature number ki zaroorat nahi.
4. Answer. Slope $-$ phir $+$ jaati hai ⇒ $x=0$ par local (aur global) minimum.

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Cell E — Substitution se constrained

Forecast: kya sabse paas wala point vertex $(0,0)$ hai, ya woh side mein kahi shift ho jaata hai?

Picture scene set karti hai: black parabola $y=x^2$, red target dot $(0,\frac{1}{2})$ par, aur red segment jo hum minimize karenge seedha vertex tak neeche jaata hai. Us segment par dhyan rakho — neeche ki algebra prove karegi ki yahi sabse chhota segment hai.

1. Distance model banao — lekin square karo. Parabola par kisi general point $(x,x^2)$ se $(0,\frac{1}{2})$ tak Distance${}^2$: $S(x)=x^2+{\left(x^2-\frac{1}{2}\right)}^2.$ Square kyun? ​ increasing hai, toh jo bhi $S$ minimize karta hai woh true distance bhi minimize karta hai — aur algebra polynomial rehti hai (parent ka trick, cell yahan reuse hua).
2. Constraint substitute karo. Humne pehle hi $y=x^2$ use karke $S$ ek variable mein likha — yahi substitution route hai. Domain sab real $x$ hai: parabola dono taraf infinitely stretch karti hai, toh koi endpoints worry nahi. Yeh step kyun? "Point on a curve" problem ko ordinary 1-D optimization mein badalta hai, aur exactly fix karta hai kaun se $x$-values compete kar rahe hain.
3. Expand aur differentiate karo. $S(x)=x^2+x^4-x^2+\frac{1}{4}=x^4+\frac{1}{4}$. $S^{\prime }(x)=4x^3=0\Rightarrow x=0.$ Yeh step kyun? Reduced function par flat-spot hunt. (Notice karo $x^2$ terms ka neat cancellation.)
4. "Infinity par" rule out karo. Kyunki $S(x)=x^4+\frac{1}{4}$ aur $x^4\to +\infty$ jab $x\to \pm \infty$, vertex se door distance badh jaati hai — toh koi better point tails mein nahi chhupa hai. Isliye single interior critical point global minimum hai. Yeh step kyun? Unbounded domain par koi endpoints nahi hote; hume yeh sure karne ke liye limiting behaviour check karna padta hai ki minimum genuinely global hai.
5. Classify karo. $x=0$ ke aas-paas, $S^{\prime }(x)=4x^3$ $-\to +$ jaata hai (same cell-D reasoning, kyunki $S^{\prime \prime }(0)=0$): down phir up, valley. Yeh step kyun? Confirm karta hai ki yeh nearest point hai, farthest nahi.
6. Answer. Nearest point vertex $(0,0)$ hai, distance $\sqrt{S(0)}=\sqrt{\frac{1}{4}}=\frac{1}{2}$ — bilkul figure mein red vertical segment.

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Cell F — Lagrange multipliers se constrained

Forecast: unit circle par kaun si direction "sabse north-east" point karti hai?

1. Gradients set up karo. $g(x,y)=x^2+y^2-1$ ke saath: $\nabla f=(1,1),\nabla g=(2x,2y).$ Yeh step kyun? Ek constrained optimum par, $\nabla f$ constraint curve ke perpendicular hona chahiye — yaani $\nabla g$ ke parallel (Lagrange condition dekho). Red arrow $\nabla f$ north-east point karta hai; optimum wahan hota hai jahan woh outward radius ke saath align ho.
2. $\nabla f=\lambda \nabla g$ likho. Component by component: $1=2\lambda x,1=2\lambda y.$ Yeh step kyun? Parallel vectors scalar multiples hote hain; $\lambda$ woh scalar hai.
3. Solve karo. Dono equations $x=y=\frac{1}{2\lambda }$ deti hain, toh $x=y$. Yeh step kyun? $f$ ke equal partials yahan equal coordinates force karte hain.
4. Constraint apply karo. $x^2+y^2=1$ jab $x=y$ ho toh $2x^2=1\Rightarrow x=\pm \frac{1}{\sqrt{2}}$. Yeh step kyun? Hume circle par land karna hai, sirf tangency satisfy karna kaafi nahi.
5. Evaluate aur answer do.
   • $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$: $f=\sqrt{2}$ → maximum.
   • $\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$: $f=-\sqrt{2}$ → minimum.

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Cell G — Real-world word problem

Forecast: kya woh swimmer ki taraf seedha jaati hai, ya aur aage beach par jaati hai taaki fast sand par zyada waqt bitaaye?

1. Total time model banao. Run distance $=x$ (beach ke saath), swim distance $=\sqrt{(10-x{)}^2+4^2}$. $T(x)=\frac{x}{8}+\frac{\sqrt{(10-x{)}^2+16}}{2},0\le x\le 10.$ Time kyun, distance nahi? Dono legs ki speeds alag hain, toh time $=\text{distance}/\text{speed}$ woh honest quantity hai jo minimize karni hai (yeh exactly Snell's-law / Fermat's-principle logic hai).
2. Differentiate karo. Square root par chain rule use karte hue: $T^{\prime }(x)=\frac{1}{8}+\frac{1}{2}\cdot \frac{-(10-x)}{\sqrt{(10-x{)}^2+16}}.$ Chain rule kyun? Swim distance ek square root ke andar function hai; chain rule "function of a function" handle karta hai.
3. $T^{\prime }=0$ set karo. $d=10-x$ maano: $\frac{1}{8}=\frac{d}{2\sqrt{d^2+16}}\Rightarrow 2\sqrt{d^2+16}=8d\Rightarrow \sqrt{d^2+16}=4d.$ Yeh step kyun? Flat spot sabse fast path mark karta hai; $T^{\prime }=0$ set karna aur fractions clear karna geometry isolate karta hai. Substitution $d=10-x$ sirf algebra tidily karta hai — woh horizontal swim distance ka naam deta hai.
4. Carefully square karo. $\sqrt{d^2+16}=4d$ ko square karna tabhi valid hai jab right side non-negative ho, yaani $d\ge 0$; hum woh restriction rakhte hain aur $d<0$ wala koi bhi root extraneous maan ke discard karenge. Square karne par: $d^2+16=16d^2\Rightarrow 15d^2=16\Rightarrow d=\pm \frac{4}{\sqrt{15}}$. Negative root extraneous hai ($d\ge 0$ fail karta hai), toh $d=\frac{4}{\sqrt{15}}\approx 1.033$. Yeh step kyun? Squaring false solutions invent kar sakta hai; $d\ge 0$ gate aur sign check unhe remove karta hai.
5. Back-substitute karo. $x=10-d\approx 10-1.033=8.967$ m. Yeh step kyun? Humne $d$ ke liye solve kiya; question shore point $x$ poochh raha hai.
6. Classify / endpoints check karo. $T(0)=\frac{\sqrt{116}}{2}\approx 5.385$ s, $T(10)=\frac{10}{8}+\frac{4}{2}=3.25$ s, aur $T(8.967)\approx 3.204$ s — interior point dono edges ko beat karta hai, toh woh $[0,10]$ par minimum hai. Yeh step kyun? Fermat sirf ek candidate deta hai; endpoints compare karna prove karta hai ki woh globally best hai.

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Cell H — Exam twist (illegal root / wrong objective)

Forecast: guess karo ki field river ke saath zyada wide hoga ya perpendicular.

1. Model aur constrain karo. Fence length $L=2x+y$ (do ends length $x$ ke, ek far side length $y$ ki). Area: $xy=200\Rightarrow y=\frac{200}{x}$. Yeh step kyun? Sirf teen sides fenced hain — geometry sahi padhna exam ki aadhi ladaai hai.
2. Reduce karo. $L(x)=2x+\frac{200}{x}$, domain $x>0$. Yeh step kyun? Ek variable, differentiate karne ke liye ready.
3. Derivative zero karo. $L^{\prime }(x)=2-\frac{200}{x^2}=0\Rightarrow x^2=100\Rightarrow x=\pm 10$. Sign factor kyun nikalo? Do algebraic roots appear hote hain: $x=10$ aur $x=-10$.
4. Twist — illegal root reject karo. Fence length negative nahi ho sakti, toh $x=-10$ domain $x>0$ ke bahar hai; discard karo. Sirf $x=10$ bachta hai. Yeh step kyun? Yahi classic exam trap hai: $L^{\prime }$ ka ek genuine root jo physically impossible hai.
5. Classify aur answer do. $L^{\prime \prime }(x)=\frac{400}{x^3}>0$ at $x=10$ → minimum. Phir $y=\frac{200}{10}=20$, aur $L=2(10)+20=40$ m. Yeh step kyun? Positive curvature valley confirm karta hai; actual fence length units ke saath report karo.

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Recall Main kaun si cell mein hoon? (self-quiz)

Function $[a,b]$ par strictly increasing hai — max kahan hai? ::: Right endpoint $b$ par (Cell C). $f^{\prime }(c)=0$ aur $f^{\prime \prime }(c)=0$ — agli tool? ::: First-derivative sign test (Cell D). Do variables ek equation se tied hain, ek eliminate karna hai — kaun sa method? ::: Substitution (Cell E). Do variables, constraint explicitly solve karna mushkil — kaun sa method? ::: Lagrange multipliers, $\nabla f=\lambda \nabla g$ (Cell F). Ek length ke liye $x=-10$ mila — kya karo? ::: Reject karo; woh physical domain ke bahar hai (Cell H). Do speeds mein time optimize karna — distance minimize karo ya time? ::: Time, kyunki speeds alag hain (Cell G).

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Connections

• Optimization — constrained, unconstrained, real-world problems — parent recipe jo yeh examples drill karte hain.
• Critical Points & Fermat's Theorem — kyun pehle $f^{\prime }=0$ dhundhte hain.
• Second Derivative & Concavity — Cells A, B, H mein ∪/∩ classification.
• Closed Interval Method (Global Extrema) — Cells C aur G decide karta hai.
• Taylor Series Expansion — explain karta hai $f^{\prime \prime }=0$ kyun khamosh hai (Cell D).
• Lagrange Multipliers (Multivariable) — Cell F ka tangency method.
• AM-GM Inequality — Cell A ki "equal is best" morality.
• Related Rates — prose ko derivative equations mein badhalne ka sibling skill.

Case Map