Optimization — constrained, unconstrained, real-world problems: Exercises

4.1.31 · D4 · Maths › Calculus I — Limits & Derivatives › Optimization — constrained, unconstrained, real-world proble

Shuru karne se pehle, teen words jinpe hum zyada rely karenge, plain language mein defined:

Level 1 — Recognition

L1.1 — Critical points pehchaano

$f (x) = x^{3} - 3 x$ ke har critical point batao.

Recall Solution

Kya karein: dhundo jahan slope zero ho. Kyun: critical points hi extrema ke akal mand candidates hote hain. $f^{'} (x) = 3 x^{2} - 3 = 3 (x^{2} - 1) = 3 (x - 1) (x + 1) .$ $f^{'} (x) = 0$ set karo: $x = 1$ ya $x = - 1$ . Slope ek polynomial hai, toh ye kabhi "exist karna band" nahi karta — woh dono hi poori list hain. Answer: critical points at $== x = - 1 ==$ aur $== x = 1 ==$ .

L1.2 — Kaunsa point maximum hai?

Usi $f (x) = x^{3} - 3 x$ ke liye, second-derivative test use karke $x = - 1$ aur $x = 1$ ko classify karo.

Recall Solution

Kya karein: $f^{''}$ compute karo aur har critical point par uska sign dekho. Kyun: $f^{''}$ ka sign batata hai cup (min) hai ya cap (max), seedha Taylor argument se $f (c + h) - f (c) \approx \frac{1}{2} f^{''} (c) h^{2}$ . $f^{''} (x) = 6 x .$

$x = - 1$ par: $f^{''} (- 1) = - 6 < 0$ → cap-shaped → local maximum.
$x = + 1$ par: $f^{''} (1) = + 6 > 0$ → cup-shaped → local minimum. Answer: $x = - 1$ local max hai, $x = 1$ local min hai.

Level 2 — Application

L2.1 — Closed interval par global extrema

$f (x) = x^{3} - 3 x$ ka absolute maximum aur minimum dhundo $[0, 2]$ par.

Recall Solution

Kya karein: Closed Interval Method apply karo — $f$ ko har interior critical point aur dono endpoints par compare karo. Kyun: closed interval par global extreme ek endpoint par bhi chhup sakta hai, chahe wahan slope zero na ho.

$f^{'} (x) = 3 (x - 1) (x + 1)$ ke critical points hain $x = \pm 1$ ; sirf $x = 1$ hi $[0, 2]$ mein hai.
Candidates evaluate karo: $f (0) = 0, f (1) = 1 - 3 = - 2, f (2) = 8 - 6 = 2.$ Answer: absolute maximum $== f (2) = 2 ==$ at $x = 2$ (ek endpoint!); absolute minimum $== f (1) = - 2 ==$ at $x = 1$ .

L2.2 — Ek single-variable word problem

Ek farmer ke paas $40$ m fence hai ek rectangular pen banana ke liye jo seedhi river ke saath lagti ho (river wali side ko fence ki zaroorat nahi). Kaun se dimensions enclosed area maximise karte hain?

Neeche ki figure woh picture setup karti hai jo hume symbols mein translate karni hai: river upar run karti hai (wahan koi fence nahi), $x$ length ki dono red sides river se door jaati hain, aur $y$ length ki ek lavender side uske saath chalti hai. Solution padhne se pehle dekho kaun si sides fenced hain — poora model "sirf teen sides mein fence lagti hai" par hinge karta hai.

Figure — Optimization — constrained, unconstrained, real-world problems

Recall Solution

Draw & name (D, U): figure ki tarah, $x$ = do red sides mein se har ek jo river ke perpendicular hain, $y$ = uske parallel lavender side. Area $A = x y$ . Constrain to one variable (C): sirf teen sides fenced hain (river side, figure mein dashed, free hai), toh $2 x + y = 40 \Rightarrow y = 40 - 2 x$ . $A (x) = x (40 - 2 x) = 40 x - 2 x^{2}, 0 \leq x \leq 20.$ Domain kyun? $x \geq 0$ ek real length hai; $x \leq 20$ rakho taaki $y = 40 - 2 x \geq 0$ rahe. Kill the derivative (K): $A^{'} (x) = 40 - 4 x = 0 \Rightarrow x = 10.$ Discriminate (D): $A^{''} (x) = - 4 < 0$ → cap → maximum. ✓ Check endpoints (C): $A (0) = 0$ , $A (20) = 0$ , $A (10) = 10 \cdot 20 = 200$ . Answer: $x = 10$ m, $y = 40 - 20 = 20$ m, deta hai maximum area $== 200 m^{2} ==$ .

Level 3 — Analysis

L3.1 — Curve par closest point ( $D^{2}$ chuno)

Parabola $y = x^{2}$ par woh point dhundo jo point $(0, 3)$ ke sabse paas ho.

Neeche ki figure trap aur answer dono ek saath dikhati hai: butter-yellow point $(0, 0)$ seedha coral target ke neeche hai aur lagta hai sabse paas, lekin parabola ke flanks par dono mint points genuinely closer hain. Dashed distance lines compare karo — slanted ones seedhe neeche wale se chote hain. Neeche ki algebra batayegi kyun.

Recall Solution

Kya karein aur kyun: hum distance $D$ ki jagah squared distance $S$ minimise karte hain. Kyunki $\cdot$ kabhi sirf badhta hai, jo bhi $x$ $S$ ko smallest banata hai wahi $D$ ko bhi smallest banata hai — aur squaring ugly square root khatam kar deta hai. $S (x) = (x - 0)^{2} + (y - 3)^{2} = x^{2} + (x^{2} - 3)^{2} .$ Expand karo: $S (x) = x^{2} + x^{4} - 6 x^{2} + 9 = x^{4} - 5 x^{2} + 9$ . Kill the derivative: $S^{'} (x) = 4 x^{3} - 10 x = 2 x (2 x^{2} - 5) = 0.$ Roots: $x = 0$ ya $x^{2} = \frac{5}{2} \Rightarrow x = \pm 5/2$ . Discriminate: $S^{''} (x) = 12 x^{2} - 10$ .

$x = 0$ par: $S^{''} (0) = - 10 < 0$ → squared distance ka local max (figure mein butter point). Reject karo.
$x = \pm 5/2$ par: $S^{''} > 0$ → minima (mint points). Unbounded domain par global check: yahan $x$ poore $R$ par hai, jiske koi endpoints nahi hain — toh hume rule out karna hoga ki door kahin smaller values na hon. Jab $∣ x ∣ \to \infty$ , leading $x^{4}$ term dominate karta hai, toh $S (x) \to + \infty$ . Isliye squared distance dono directions mein bina bound ke badhti hai aur interior minima se beat nahi kar sakti. Kyunki do symmetric minima equal, finite values dete hain aur baki sab bada hai, woh global minima hain. Minimum distance-squared compute karo: $S (\pm 5/2) = \frac{25}{4} - 5 \cdot \frac{5}{2} + 9 = \frac{25}{4} - \frac{50}{4} + \frac{36}{4} = \frac{11}{4}$ . $x^{2} = \frac{5}{2}$ ke saath, $y = x^{2} = \frac{5}{2}$ . Answer: do closest points $(\pm 5/2, \frac{5}{2})$ , har ek distance $== 11/4 = \frac{11}{2} \approx 1.658 ==$ par.

Level 4 — Synthesis

L4.1 — Minimum material ka cylinder (do variables, ek constraint)

Ek closed cylindrical can ko fixed volume $V = 1000 cm^{3}$ hold karna hai. Woh radius $r$ aur height $h$ dhundo jo total surface area (top + bottom + side) minimise kare.

Recall Solution

Name & model (U): ek closed cylinder ki surface area hai $A = do circular ends 2 π r^{2} + curved side 2 π r h .$ Constrain to one variable (C): volume constraint $π r^{2} h = 1000$ deta hai $h = \frac{1000}{π r ^{2}}$ . Substitute karo: $A (r) = 2 π r^{2} + 2 π r \cdot \frac{1000}{π r ^{2}} = 2 π r^{2} + \frac{2000}{r}, r > 0.$ Kill the derivative (K): $A^{'} (r) = 4 π r - \frac{2000}{r ^{2}} = 0 \Rightarrow 4 π r^{3} = 2000 \Rightarrow r^{3} = \frac{500}{π} .$ $r = (\frac{500}{π})^{1/3} \approx 5.419 cm .$ Discriminate (D): $A^{''} (r) = 4 π + \frac{4000}{r ^{3}} > 0$ sab $r > 0$ ke liye → cup → minimum. ✓ (Koi endpoints worry karne wale nahi kyunki $r \to 0^{+}$ aur $r \to \infty$ dono $A \to \infty$ bhejte hain.) Height: $h = \frac{1000}{π r ^{2}}$ . $r^{3} = 500/ π$ use karke milta hai $h = \frac{1000}{π r ^{2}} = \frac{1000 r}{π r ^{3}} = \frac{1000 r}{500} = 2 r \approx 10.839 cm$ . Answer: $r \approx== 5.419 == cm$ , $h = 2 r \approx== 10.839 == cm$ . Khoobsoorat baat, optimal can mein height diameter ke barabar hoti hai ( $h = 2 r$ ).

Level 5 — Mastery

L5.1 — Lagrange + AM–GM cross-check

$f (x, y, z) = x y z$ ko maximise karo subject to $x + y + z = 12$ jahan $x, y, z > 0$ ho. Phir answer ko AM–GM inequality se confirm karo.

Recall Solution

Route 1 — Lagrange multipliers. Hum chahte hain largest $f$ jabki plane $g (x, y, z) = x + y + z - 12 = 0$ par stuck hain. Geometrically optimum wahan hai jahan $\nabla f$ constraint surface ke perpendicular point kare, yaani $\nabla g$ ke parallel ho: $\nabla f = λ \nabla g .$ Yahan $\nabla f = (y z, x z, x y)$ aur $\nabla g = (1, 1, 1)$ , toh $y z = λ, x z = λ, x y = λ .$ $y z = x z$ se (aur $z > 0$ ke saath) milta hai $y = x$ . $x z = x y$ se (aur $x > 0$ ke saath) milta hai $z = y$ . Toh $x = y = z$ . Constraint force karta hai $3 x = 12 \Rightarrow x = y = z = 4$ , deta hai $f_{m a x} = 4 \cdot 4 \cdot 4 === 64 == .$

Route 2 — AM–GM check. Arithmetic Mean–Geometric Mean inequality kehti hai, positive numbers ke liye, $\frac{x + y + z}{3} \geq 3 x y z, equality iff x = y = z .$ Yahan left side fixed hai: $\frac{12}{3} = 4$ . Toh $3 x y z \leq 4 \Rightarrow x y z \leq 64$ , equality exactly jab $x = y = z = 4$ ho. Same answer, bina calculus ke. ✓

Kyun dono routes agree karte hain: Lagrange ne ek smooth function ka critical point dhunda; AM–GM ne prove kiya ki woh global maximum hai. Saath mein woh dikhate hain ki $64$ sirf ek candidate nahi — woh true maximum hai.

Recall Feynman recap: har level actually kya test kar raha tha?

L1 — kya tum flat spots spot kar sakte ho aur hill ko dip se alag kar sakte ho?
L2 — kya tum ek story ko single-variable area (ya cost) function mein convert karke recipe crank kar sakte ho (endpoints dekho)?
L3 — kya tum ek smarter target chun sakte ho ( $D^{2}$ ) aur obvious point par trust karne se inkaar kar sakte ho?
L4 — kya tum do variables ko ek mein collapse karne ke liye constraint use kar sakte ho?
L5 — kya tum same cheez do tareekon se solve karke prove kar sakte ho ki woh really extreme hai?

Connections

Critical Points & Fermat's Theorem — L1 yahan hai: kyun flat spots hi akel candidates hain.
Second Derivative & Concavity — cup/cap sign test jo L1–L4 mein use hua.
Closed Interval Method (Global Extrema) — L2 mein endpoint comparison.
Taylor Series Expansion — $\frac{1}{2} f^{''} (c) h^{2}$ idea jo second-derivative test ke peeche hai.
Lagrange Multipliers (Multivariable) — L4 aur L5.
AM-GM Inequality — L5 mein non-calculus proof.
Related Rates — words ko derivative equations mein translate karne ki sibling skill.

Concept Map

Exercises ek staircase banate hain — har level pichle skill ko reuse karta hai aur ek naya demand add karta hai, sab master recipe ke guide mein: