MoreOptimization - 3-masala yechimi | MathModel.uz

4.9-bo'lim : Ko'proq optimallashtirish

3. $x = 3 - 2{y^2}$ da $\left( { - 4,0} \right)$ ga eng yaqin nuqta(lar)ni toping.

Yechim

Keling, ushbu vaziyatning qisqacha chizmasi bilan boshlaylik. Quyida funksiya grafigining eskizi hamda $\left( { - 4,0} \right)$ nuqtasi keltirilgan. Ko'rib turganimizdek, javob sifatida ikkita ball olishni kutishimiz mumkin, farq faqat $y$ koordinatasidagi belgidir.

$A sketch of the parabola $x = 3 - 2{y^2}$, which opens to the left with vertex at $\left( {3,0} \right)$, together with the point $\left( { - 4,0} \right)$ marked on the x-axis. Two points on the parabola, labeled $\left( {x,y} \right)$ above the x-axis and $\left( {x, - y} \right)$ below it, are marked and dashed lines are drawn from each of them back to $\left( { - 4,0} \right)$, showing that there will be two answers that differ only by the sign of the y-coordinate.$

2-qadam

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

Bunday holda, cheklov bizga berilgan tenglamadir. Nuqta grafikda yotishi kerak va shuning uchun ham tenglamani qondirishi kerak.

\[x = 3 - 2{y^2}\]

Bizdan grafikdagi nuqta (yoki nuqta) va $\left( { - 4,0} \right)$ nuqtasi orasidagi masofani minimallashtirish talab qilinmoqda. Buni $\left( { - 4,0} \right)$ va $\left( {x,y} \right)$ orasidagi masofaga qarab amalga oshirishimiz mumkin. Bu ikki nuqta orasidagi masofa,

\[d = \sqrt {{{\left( {x + 4} \right)}^2} + {y^2}} \]

Ushbu bo'lim uchun eslatmalarda muhokama qilganimizdek, masofa kvadratini minimallashtiradigan nuqta masofaning o'zini ham minimallashtiradi va shuning uchun ildiz bilan ishlamaslik uchun biz masofaning kvadratini minimallashtiramiz yoki,

\[{d^2} = {\left( {x + 4} \right)^2} + {y^2}\]

3-qadam

Endi bizda bu nuqtadan qanday davom etish bo'yicha ikkita tanlov bor. Birinchi variant - biz berilgan tenglamani $x$ ga kvadrat masofada qo'shish va 4 ni olishdir.^th Biz ishlashimiz kerak bo'lgan $y$ uchun darajali ko'phad. Ikkinchisi, ${y^2}$ uchun tenglamani yechish va uni masofaning kvadratiga kiritish va 2 ni olishdir.^nd Biz ishlashimiz kerak bo'lgan $x$ uchun darajali ko'phad. Ikkinchi variant ishlash uchun "yaxshiroq" ko'phad beradi, shuning uchun biz buni qilamiz.

\[{y^2} = {\frac{1}{2}}\left( {3 - x} \right) = {\frac{3}{2}} - {\frac{1}{2}}x\]

Buni masofaning kvadratiga ulash, beradi,

\[f\left( x \right) = {d^2} = {\left( {x + 4} \right)^2} + {\frac{3}{2}} - {\frac{1}{2}}x = {x^2} + {\frac{15}{2}}x + {\frac{35}{2}}\]

4-qadam

Buning uchun muhim nuqta(lar)ni topish hozircha unchalik qiyin bo'lmasligi kerak. Mana lotin.

\[f'\left( x \right) = 2x + {\frac{15}{2}}\]

Bundan biz bitta muhim nuqtaga o'xshaymiz: $x = - {\frac{15}{4}} = - 3.75$.

5-qadam

Masofa kvadrati funksiyasining ikkinchi hosilasi:

\[f''\left( x \right) = 2\]

Bundan ikkinchi hosila har doim ijobiy ekanligini ko'rishimiz mumkin. Shunday qilib, kvadratik masofa har doim konkav bo'ladi va shuning uchun biz 4-bosqichda olingan yagona kritik nuqta nisbiy minimal bo'lishi kerak va shuning uchun $x$ qiymati bo'lishi kerak, bu $\left( { - 4,0} \right)$ ga eng yaqin nuqtalarni beradi.

6-qadam

Va nihoyat, biz faqat haqiqiy nuqtalarni beradigan $y$ qiymatlarini aniqlashimiz kerak.

\[{y^2} = {\frac{3}{2}} - {\frac{1}{2}}\left( { - {\frac{15}{4}}} \right) = {\frac{27}{8}}\hspace{0.5in} \Rightarrow \hspace{0.5in}y = \pm \sqrt {{\frac{27}{8}}} = \pm 1.8371\]

Shunday qilib, grafikdagi $\left( { - 4,0} \right)$ ga eng yaqin bo'lgan ikkita nuqta:

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\left( { - {\frac{15}{4}},\sqrt {{\frac{27}{8}}} } \right)\hspace{0.5in} \& \hspace{0.5in}\left( { - {\frac{15}{4}}, - \sqrt {{\frac{27}{8}}} } \right)}}\]