Optimization - 8-masala yechimi | MathModel.uz

4.8-bo'lim : Optimallashtirish

8. Bizda 50 sm dan 20 sm gacha bo'lgan karton bo'lagi bor va biz burchaklarni kesib, qutichani hosil qilish uchun yon tomonlarini katlamoqchimiz. Maksimal hajmni beradigan qutining balandligini aniqlang.

Two sketches side by side with an arrow pointing from the left sketch to the right one. On the left is the flat piece of cardboard, 50 cm wide and 20 cm tall, with a small square cut out of each of the four corners and dashed fold lines marked. On the right is the open-topped box that results when the four sides are folded up.

Yechim

Birinchi qadam muammoning tezkor eskizini qilishdir.

$Two sketches side by side with an arrow pointing from the left sketch to the right one. On the left is the flat piece of cardboard with a square of side $h$ cut out of each of the four corners, leaving a width of $50 - 2h$ across the bottom and a height of $20 - 2h$ up the side. On the right is the open topped box that results when the sides are folded up, with dimensions $50 - 2h$ by $20 - 2h$ by $h$.$

2-qadam

Eslatmalardagi bu kabi muammoda bo'lgani kabi, cheklov haqiqatan ham qutining o'lchamidir va bu rasmda hisobga olingan, shuning uchun biz qilishimiz kerak bo'lgan yagona narsa biz maksimal darajaga keltirmoqchi bo'lgan hajm tenglamasini o'rnatishdir.

\[V\left( h \right) = h\left( {50 - 2h} \right)\left( {20 - 2h} \right) = 4{h^3} - 140{h^2} + 1000h\]

3-qadam

Buning uchun muhim nuqta(lar)ni topish hozir unchalik qiyin bo'lmasligi kerak, shuning uchun bu ish,

\[V'\left( h \right) = 12{h^2} - 280h + 1000\hspace{0.75in}h = \frac{35 \pm 5\sqrt {19}}{3} = 4.4018,\,\,18.9315\]

Yuqoridagi rasmdan biz $h$ chegaralari $h = 0$ va $h = 10$ bo'lishi kerakligini ko'rishimiz mumkin (eng katta $h$ ½ kichikroq tomon bo'lishi mumkin). E'tibor bering, ularning hech biri haqiqatan ham jismoniy ma'noga ega emas, lekin ular $h$ uchun limitlarni ta'minlaydi.

Shunday qilib, bizda $0 \le h \le 10$ bo'lishi kerak va bu ikkinchi muhim nuqtani yo'q qiladi va shuning uchun biz tashvishlanishimiz kerak bo'lgan yagona muhim nuqta - $h = 4.4018$

4-qadam

Bizda $h$-da limitlar mavjud bo'lganligi sababli, ovoz balandligi funksiyasini ulab, bizda maksimal darajaga ega ekanligini tezda tekshirishimiz mumkin.

\[V\left( 0 \right) = 0\hspace{0.75in}V\left( {4.4018} \right) = 2030.34\hspace{0.75in}V\left( {10} \right) = 0\]

Shunday qilib, biz maksimal hajmni olish uchun qutining balandligi $\require{bbox} \bbox[2pt,border:1px solid black]{{h = 4.4018}}$ bo'lishi kerakligini ko'rishimiz mumkin.