MoreOptimization - 6-masala yechimi | MathModel.uz

4.9-bo'lim : Ko'proq optimallashtirish

6. Quvur bo'lagi 18 fut kenglikdagi koridor bo'ylab olib ketilmoqda. Yo'lakning oxirida to'g'ri burchakli burilish mavjud va koridor 12 metr kengligida torayadi. Koridorda burilish bo'ylab ko'tarilishi mumkin bo'lgan eng uzun quvur (har doim gorizontal holatda) nima?

A sketch of the hallway described in the problem, seen from above. The wide hallway is 18 feet across and runs vertically, the narrow hallway is 12 feet across and runs horizontally, and they meet at a right-angled turn. The pipe is drawn as a straight line lying diagonally across the inside corner of the turn, touching the outer wall of each hallway and resting against the inner corner.

Yechim

Keling, barcha muhim miqdorlar berilgan quvur va koridorlarning tezkor eskizini boshlaylik.

$A sketch of the pipe and the hallways seen from above with all the important quantities added in. The wide hallway is 18 feet across and runs vertically, the narrow hallway is 12 feet across and runs horizontally, and they meet at a right-angled turn. The pipe runs as a straight line from the point $A$ on the outer wall of the wide hallway, through the inner corner $B$, out to the point $C$ on the outer wall of the narrow hallway. The piece from $A$ to $B$ is labeled ${L_2}$, the piece from $B$ to $C$ is labeled ${L_1}$, and the angle $\theta $ that the pipe makes with the walls is marked at both ends.$

2-qadam

Keyinchalik, bizdan optimallashtirish so'raladigan cheklov va tenglamani o'rnatishimiz kerak.

Ushbu bo'limning eslatmalarida shunga o'xshash muammoni muhokama qilganimizdek, biz quvurning umumiy uzunligini minimallashtirishimiz kerak. Biz minimallashtirishimiz kerak bo'lgan tenglama, u holda,

\[L = {L_1} + {L_2}\]

Shuningdek, biz eslatmalarda muhokama qilganimizdek, aslida ikkita limitlar mavjud: ikkita koridorning kengligi. Biz ularni $\theta $ burchagi nuqtai nazaridan osongina hal qilishimiz mumkin.

\[{L_1} = 12\sec \theta \hspace{0.5in}{L_2} = 18\csc \theta \]

Eslatmalar muammosida muhokama qilinganidek, bizda $0 < \theta < {\frac{\pi}{2}}$ bo'lishi kerakligini ham bilamiz.

3-qadam

Bu erda qilishimiz kerak bo'lgan yagona narsa, biz minimallashtirishimiz mumkin bo'lgan $\theta $ nuqtai nazaridan funktsiyani olish uchun uzunlik funksiyasiga ikkita cheklovimizni ulashdir.

\[L\left( \theta \right) = 12\sec \theta + 18\csc \theta \]

4-qadam

Uzunlik funksiyasining hosilasi,

\[L'\left( \theta \right) = 12\sec \theta \tan \theta - 18\csc \theta \cot \theta \]

Keyinchalik, biz buni nolga tenglashtirishimiz va $0 < \theta < {\frac{\pi}{2}}$ diapazonidagi kritik nuqtani olish uchun $\theta $ uchun buni hal qilishimiz kerak.

\[\begin{align*}12\sec \theta \tan \theta & = 18\csc \theta \cot \theta \\ \frac{{\sec \theta \tan \theta }}{{\csc \theta \cot \theta }} & = \frac{{18}}{{12}}\\ {\tan ^3}\theta & = \frac{3}{2}\end{align*}\]

Bizga kerak bo'lgan muhim nuqta: $\theta = {\tan ^{ - 1}}\left( {\sqrt[3]{{{\frac{3}{2}}}}} \right) = 0.8528$.

5-qadam

Bu minimumni beradigan qiymat ekanligini tekshirish boshqa muammolarga qaraganda biroz qiyinroq.

Ushbu bo'lim uchun eslatmalarda ta'kidlanganidek, biz $\theta \to 0$-ni ko'chirsak, bizda $L \to \infty $ bor va $\theta \to {\frac{\pi}{2}}$-ni harakatlantirganda bizda $L \to \infty $ mavjud. Shuning uchun, $\theta = 0.8528$ radianining har ikki tomonida biz diapazon oxirigacha harakat qilganimizda quvur uzunligi cheksizgacha oshadi.

Shuning uchun, bu burchak bizga quvurning minimal uzunligini berishi kerak va biz burchakka mos keladigan eng katta quvurdir.

6-qadam

Biz burchakka o'rnatishimiz mumkin bo'lgan eng katta quvur,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{L\left( {0.8528} \right) = 42.1409{\mbox{ feet}}}}\]