Namo Buddhaya.
\section{Maksima, Minima, dan Pelana Kuda dari Fungsi Dua Peubah}
Misalkan ada sebuah fungsi $f \,:\, \mathbb{R}^2 \to \mathbb{R}$ yang kontinyu. Andaikan didefinisikan $\varphi := f(x, y)$. Andaikan pula, diketahui $\varphi = 0$. Oleh karena itu, pastilah $\varphi$ bergantung pada $x$ dan $y$. Kita akan mencari titik $(x, y)$ yang menyebabkan $\varphi$ bernilai stasioner. Mula-mula, $x$ dan $y$ dianggap bergantung pada $t \in \mathbb{R}$ sehingga $\varphi$ boleh dianggap bergantung pada $t$. Oleh karena itu, agar $\varphi$ bernilai stasioner, maka
\[ \frac{d\varphi}{dt} = \frac{\partial\varphi}{\partial x}\frac{dx}{dt} + \frac{\partial\varphi}{\partial y}\frac{dy}{dt} = 0 \]
untuk setiap $t$, sehingga syarat agar $\varphi$ bernilai stasioner adalah
\[ \frac{\partial\varphi}{\partial x} = 0 ~~~~~ \text{dan} ~~~~~ \frac{\partial\varphi}{\partial y} = 0. \]
Misalkan titik $(x, y)$ yang menyebabkan $\varphi$ bernilai stasioner adalah $(x_0, y_0)$. Titik $(x_0, y_0)$ menyebabkan $\varphi$ bernilai maksimum apabila $d^2\varphi/dt^2 < 0$ di titik $(x_0, y_0)$ sehingga
\[ \frac{d}{dt}\left(\frac{\partial\varphi}{\partial x}\frac{dx}{dt} + \frac{\partial\varphi}{\partial y}\frac{dy}{dt}\right) < 0 \]
alias
\[ \frac{d}{dt}\frac{\partial\varphi}{\partial x}\frac{dx}{dt} + \frac{\partial\varphi}{\partial x}\frac{d^2x}{dt^2} + \frac{d}{dt}\frac{\partial\varphi}{\partial y}\frac{dy}{dt} + \frac{\partial\varphi}{\partial y}\frac{d^2\varphi}{dt^2} < 0. \]
Karena sudah diketahui $\partial\varphi/\partial x = 0$ dan $\partial\varphi/\partial y = 0$ di titik $(x_0, y_0)$, maka diperoleh
\[ \left(\frac{\partial^2\varphi}{\partial x^2}\frac{dx}{dt} + \frac{\partial^2\varphi}{\partial x\partial y}\frac{dy}{dt}\right)\frac{dx}{dt} + \left(\frac{\partial^2\varphi}{\partial x\partial y}\frac{dx}{dt} + \frac{\partial^2\varphi}{\partial y^2}\frac{dy}{dt}\right)\frac{dy}{dt} < 0 \]
alias
\[ \frac{\partial^2\varphi}{\partial x^2}\left(\frac{dx}{dt}\right)^2 + 2\frac{\partial^2\varphi}{\partial x\partial y}\frac{dx}{dt}\frac{dy}{dt} + \frac{\partial^2\varphi}{\partial y^2}\left(\frac{dy}{dt}\right)^2 < 0 \]
alias
\[ \frac{\partial^2\varphi}{\partial x^2}\left[\left(\frac{dx}{dt}\right)^2 + 2\frac{\partial^2\varphi/\partial x\partial y}{\partial^2\varphi/\partial x^2}\frac{dx}{dt}\frac{dy}{dt} + \frac{\partial^2\varphi/\partial y^2}{\partial^2\varphi/\partial x^2}\left(\frac{dy}{dt}\right)^2\right] < 0 \]
alias
\[ \frac{\partial^2\varphi}{\partial x^2}\left[\left\{\frac{dx}{dt} + \frac{\partial^2\varphi/\partial x\partial y}{\partial^2\varphi/\partial x^2}\right\}^2 + \left\{\frac{\partial^2\varphi/\partial y^2}{\partial^2\varphi/\partial x^2} - \left(\frac{\partial^2\varphi/\partial x\partial y}{\partial^2\varphi/\partial x^2}\right)^2\right\}\left(\frac{dy}{dt}\right)^2\right] < 0 \]
alias $\partial^2\varphi/\partial x^2 < 0$ dan
\[ \frac{\partial^2\varphi}{\partial x^2}\left\{\frac{\partial^2\varphi/\partial y^2}{\partial^2\varphi/\partial x^2} + \left(\frac{\partial^2\varphi/\partial x\partial y}{\partial^2\varphi/\partial x^2}\right)^2\right\} < 0 \]
alias
\[ \frac{\partial^2\varphi}{\partial y^2} - \frac{(\partial^2\varphi/\partial x\partial y)^2}{\partial^2\varphi/\partial x^2} < 0 \]
alias
\[ \Delta := \frac{\partial^2\varphi}{\partial x^2}\frac{\partial^2\varphi}{\partial y^2} - \left(\frac{\partial^2\varphi}{\partial x\partial y}\right)^2 > 0. \]
Jadi, syarat agar titik $(x_0, y_0)$ menjadikan $\varphi$ bernilai maksimum adalah
\[ \Delta > 0 ~~~~~ \text{dan} ~~~~~ \partial^2\varphi/\partial x^2 < 0 ~~~~~ \text{atau} ~~~~~ \partial^2\varphi/\partial y^2 < 0 \]
di titik $(x_0, y_0)$. Dengan cara serupa, syarat agar titik $(x_0, y_0)$ menjadikan $\varphi$ bernilai minimum ($d^2\varphi/dt^2 >0$) adalah
\[ \Delta > 0 ~~~~~ \text{dan} ~~~~~ \partial^2\varphi/\partial x^2 > 0 ~~~~~ \text{atau} ~~~~~ \partial^2\varphi/\partial y^2 > 0 \]
di titik $(x_0, y_0)$. Apabila $\Delta < 0$ di titik $(x_0, y_0)$, maka titik $(x_0, y_0)$ merupakan titik pelana kuda. Apabila $\Delta = 0$ maka tidak diperoleh informasi apa-apa.
Alhamdulillah hirobbil alamin.
