Keling, ushbu muammoning muhim nuqtalarini topaylik. Bu quyidagi tizimni hal qilishni anglatadi.

\[\begin{align*}{f_x} & = 0 : \,\,\, - 2x{y^2} + 8x = 2x\left( {4 - {y^2}} \right) = 0\hspace{0.25in} \to \hspace{0.25in}y = \pm 2\,\,\,{\mbox{or}}\,\,\,x = 0\\ {f_y} & = 0 : \,\,\,\,9{y^2} - 2{x^2}y + 16y - 20 = 0\end{align*}\]

Yuqorida ko'rsatilganidek, bizda birinchi tenglamadan uchta mumkin bo'lgan variant mavjud. Tenglama uchun kritik nuqtalarni olish uchun har birini ikkinchi tenglamaga ulashimiz mumkin.

\(y = - 2 : 4{x^2} - 16 = 0\,\,\,\,\, \to \,\,\,\,\,x = \pm 2\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {2, - 2} \right)\,\,\,\,{\mbox{and}}\,\,\,\left( { - 2, - 2} \right)\)

\(y = 2 : - 4{x^2} + 48 = 0\,\,\,\,\, \to \,\,\,\,\,x = \pm 2\sqrt 3 \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {2\sqrt 3 ,2} \right)\,\,\,\,{\mbox{and}}\,\,\,\left( { - 2\sqrt 3 ,2} \right)\)

\(x = 0 : 9{y^2} + 16y - 20 = 0\,\,\,\,\, \to \,\,\,\,\,y = \frac{{ - 16 \pm \sqrt {976} }}{{18}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {0,\frac{{ - 16 - \sqrt {976} }}{{18}}} \right)\,\,\,\,{\mbox{and}}\,\,\,\,\left( {0,\frac{{ - 16 + \sqrt {976} }}{{18}}} \right)\)

Ushbu bo'limdagi dastlabki ikkita muammoda ta'kidlaganimizdek, faqat yuqoridagi ishda topilgan haqiqiy echimlarni yozishdan ehtiyot bo'ling. Yuqoridagi uchta satrning har biridan \(x\) va \(y\) ning barcha mumkin bo'lgan kombinatsiyalarini yozmang. Agar siz ushbu muammo uchun shunday qilsangiz, siz tanqidiy nuqtalar bo'lmagan ko'p sonli nuqtalarga ega bo'lasiz.

Bundan tashqari, "tartibsizlik" dan hayajonlanmang (i.e. ildizlar) ba'zi muhim nuqtalarda ishtirok etadi. Vaqti-vaqti bilan bu muammolar bilan ular hayot haqiqati bo'ladi.

Xulosa qilib aytganda, bu funktsiya quyidagi oltita muhim nuqtaga ega.

\[\left( { - 2, - 2} \right),\,\,\left( {2, - 2} \right),\,\,\left( {2\sqrt 3 ,2} \right),\,\,\left( { - 2\sqrt 3 ,2} \right),\,\,\,\left( {0,\frac{{ - 16 - \sqrt {976} }}{{18}}} \right),\,\,\,\left( {0,\frac{{ - 16 + \sqrt {976} }}{{18}}} \right)\].

Keyinchalik, bizga quyidagilar kerak bo'ladi:

\[\begin{align*}D\left( {x,y} \right) & = {f_{x\,x}}{f_{y\,y}} - {\left[ {{f_{x\,y}}} \right]^2}\\ & = \left[ { - 2{y^2} + 8} \right]\left[ {18y - 2{x^2} + 16} \right] - {\left[ { - 4xy} \right]^2}\\ & = \left[ { - 2{y^2} + 8} \right]\left[ {18y - 2{x^2} + 16} \right] - 16{x^2}{y^2}\end{align*}\]

\(D\left( {x,y} \right)\) yordamida endi har bir muhim nuqtani quyidagicha tasniflashimiz mumkin.

\[\begin{align*} & \left( { - 2, - 2} \right) & : \hspace{0.25in} & D\left( { - 2, - 2} \right) = - 256 < 0 & \hspace{0.25in} & {\mbox{Saddle Point}}\\ & \left( {2, - 2} \right) & : \hspace{0.25in} & D\left( {2, - 2} \right) = - 256 < 0 & \hspace{0.25in} & {\mbox{Saddle Point}}\\ & \left( { - 2\sqrt 3 ,2} \right) & : \hspace{0.25in} & D\left( { - 2\sqrt 3 ,2} \right) = - 768 < 0 & \hspace{0.25in} & {\mbox{Saddle Point}}\\ & \left( {2\sqrt 3 ,2} \right) & : \hspace{0.25in} & D\left( {2\sqrt 3 ,2} \right) = - 768 < 0 & \hspace{0.25in} & {\mbox{Saddle Point}}\\ & \left( {0,\frac{{ - 16 - \sqrt {976} }}{{18}}} \right) & : \hspace{0.25in} & D\left( {0,\frac{{ - 16 - \sqrt {976} }}{{18}}} \right) = 180.4 > 0\,\,\,\,\,\,{f_{x\,x}}\left( {0,\frac{{ - 16 - \sqrt {976} }}{{18}}} \right) = - 5.8 < 0& \hspace{0.25in} & {\mbox{Relative Maximum}}\\ & \left( {0,\frac{{ - 16 + \sqrt {976} }}{{18}}} \right) & : \hspace{0.25in} & D\left( {0,\frac{{ - 16 + \sqrt {976} }}{{18}}} \right) = 205.1 > 0\,\,\,\,\,\,{f_{x\,x}}\left( {0,\frac{{ - 16 + \sqrt {976} }}{{18}}} \right) = 6.6 > 0 & \hspace{0.25in} & {\mbox{Relative Minimum}}\end{align*}\]

\(D\) ijobiy bo'lsa, \({f_{x\,x}}\) qiymatini tekshirishni unutmang, shunda biz to'g'ri tasnifni olamiz (i.e. maksimal yoki minimal) va shuningdek, salbiy \(D\) uchun bizga ikkinchi tekshirish kerak emasligini unutmang, chunki biz tanqidiy nuqta egar nuqtasi bo'lishini bilamiz.