Si a partir de \(g\left( x,y\right) =c,\) podemos escribir una de las variables en términos de la otra aunque sólo sea en una vecindad de \(\overline{v_{0}}=\left( x_{0},y_{0}\right) \), entonces al hacer la sustitución correspondiente en \(f\) obtenemos una función \(F\) de una variable, por lo que para encontrar sus máximos y mínimos podemos utilizar los criterios existentes para ese caso.

Como \(\nabla g\left( \overline{v_{0}}\right) \neq \overline{0}\) entonces alguna de las dos derivadas parciales de \(g\) es distinta de \(0\) en \(\overline{v_{0}}.\)

Sólo probaremos el caso en que \(\dfrac{\partial g}{\partial y}\left( \overline{v_{0}}\right) \neq 0.\)

Como \(\dfrac{\partial g}{\partial y}\left( \overline{v_{0}}\right) \neq 0\) y \(g\) es derivable, a partir de la ecuación \(g\left( x,y\right) =c,\) podemos expresar a \(y\) como una función de la variable \(x\). De hecho, el Teorema de la Función Implícita nos dice que hay dos abiertos \(U_{0}\subset \mathbb{R}^{2}\) y \(V\subset \mathbb{R}\) y una función \(\varphi :V\longrightarrow \mathbb{R}\) tales que \(v_0 \in U_{0}\subset U\), \(x_{0}\in V\), \(\varphi \left( x_{0}\right) =y_{0}\) y \begin{equation*} \left\{ \left. \left( x,y\right) \in U_{0}\right\vert \,g\left( x,y\right) =c\right\} =\left\{ \left. \left( x,\varphi \left( x\right) \right) \right\vert \,x\in V\right\} \end{equation*} Además, para \(x\in V\) se cumple \begin{equation} \dfrac{d\varphi }{dx}\left( x\right) =-\dfrac{\dfrac{\partial g}{\partial x} \left( x,\varphi \left( x\right) \right) }{\dfrac{\partial g}{\partial y} \left( x,y\right) } \label{derphi1}\tag{1.1} \end{equation} Más aún, en este caso, por ser \(g\) de clase \(C^{2}\), podemos derivar \(\dfrac{d\varphi }{dx}\) respecto a \(x\): \begin{eqnarray} \dfrac{d\,^{2}\varphi }{dx^{2}} & = & -\dfrac{\dfrac{\partial g}{\partial y} \left( \dfrac{\partial ^{2}g}{\partial x^{2}}+\dfrac{\partial ^{2}g}{ \partial y\partial x}\dfrac{d\varphi }{dx}\right) -\dfrac{\partial g}{ \partial x}\left( \dfrac{\partial ^{2}g}{\partial y\partial x}+\dfrac{ \partial ^{2}g}{\partial y^{2}}\dfrac{d\varphi }{dx}\right) }{\left( \dfrac{ \partial g}{\partial y}\right) ^{2}} \notag \\ & = & -\dfrac{\dfrac{\partial g}{\partial y}\dfrac{\partial ^{2}g}{\partial x^{2}}+\dfrac{\partial ^{2}g}{\partial y\partial x}\dfrac{d\varphi }{dx} \dfrac{\partial g}{\partial y}-\dfrac{\partial g}{\partial x}\dfrac{\partial ^{2}g}{\partial y\partial x}-\dfrac{\partial ^{2}g}{\partial y^{2}}\dfrac{ \partial g}{\partial x}\dfrac{d\varphi }{dx}}{\left( \dfrac{\partial g}{ \partial y}\right) ^{2}} \label{derphi2}\tag{1.2} \\ & = & -\dfrac{1}{\dfrac{\partial g}{\partial y}}\left( \dfrac{\partial ^{2}g}{ \partial x^{2}}+\dfrac{\partial ^{2}g}{\partial y\partial x}\dfrac{d\varphi }{dx}+\dfrac{d\varphi }{dx}\dfrac{\partial ^{2}g}{\partial y\partial x}+ \dfrac{\partial ^{2}g}{\partial y^{2}}\left( \dfrac{d\varphi }{dx}\right) ^{2}\right) \notag \end{eqnarray} Consideramos la función \begin{equation*} h=f-\lambda g. \end{equation*}

Como \begin{eqnarray*} \nabla f\left( \overline{v_{0}}\right) & = & \lambda \nabla g\left( \overline{ v_{0}}\right) \\ g\left( \overline{v_{0}}\right) & = & c \end{eqnarray*} entonces \(\overline{v_{0}}=\left( x_{0},y_{0}\right) \) y \(\lambda \) satisfacen \begin{eqnarray*} \dfrac{\partial f}{\partial x}\left( \overline{v_{0}}\right) & = & \lambda \dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) \\ \dfrac{\partial f}{\partial y}\left( \overline{v_{0}}\right) & = & \lambda \dfrac{\partial g}{\partial y}\left( \overline{v_{0}}\right) \\ g\left( \overline{v_{0}}\right) & = & c \end{eqnarray*} y la función restricción \(\left. f\right\vert _{S}\) se escribe como \begin{equation*} f\left( x,y\right) =f\left( x,\varphi \left( x\right) \right) . \end{equation*}

Definamos \(F:V\longrightarrow \mathbb{R}\) como \begin{equation*} F\left( x\right) =f\left( x,\varphi \left( x\right) \right) \end{equation*}

Entonces \(x_{0}\) es un punto máximo relativo, mínimo relativo o punto silla de \(F\) si \(\overline{v_{0}}=\left( x_{0},y_{0}\right) \) es lo correspondiente para \(f.\)

Así, \(F\) depende únicamente de la variable \(x.\) Calculamos sus derivadas.

Aplicando la regla de la cadena, tenemos \begin{eqnarray} \dfrac{dF}{dx} & = & \dfrac{\partial f}{\partial x}\dfrac{dx}{dx}+\dfrac{ \partial f}{\partial y}\dfrac{d\varphi }{dx} \notag \\ & = & \dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y}\dfrac{ d\varphi }{dx} \label{derf1}\tag{1.3} \end{eqnarray} y \begin{eqnarray} \dfrac{d\,^{2}F}{dx^{2}} &=&\dfrac{\partial ^{2}f}{\partial x^{2}}\dfrac{dx}{ dx}+\dfrac{\partial ^{2}f}{\partial y\partial x}\dfrac{d\varphi }{dx}+\left( \dfrac{\partial ^{2}f}{\partial x\partial y}\dfrac{dx}{dx}+\dfrac{\partial ^{2}f}{\partial y^{2}}\dfrac{d\varphi }{dx}\right) \dfrac{d\varphi }{dx}+ \dfrac{d\,^{2}\varphi }{dx^{2}}\dfrac{\partial f}{\partial y} \notag \\ & = & \dfrac{\partial ^{2}f}{\partial x^{2}}+\dfrac{\partial ^{2}f}{\partial y\partial x}\dfrac{d\varphi }{dx}+\dfrac{\partial ^{2}f}{\partial x\partial y }\dfrac{d\varphi }{dx}+\dfrac{\partial ^{2}f}{\partial y^{2}}\left( \dfrac{ d\varphi }{dx}\right) ^{2}+\dfrac{d\,^{2}\varphi }{dx^{2}}\dfrac{\partial f}{ \partial y} \notag \\ & = & \dfrac{\partial ^{2}f}{\partial x^{2}}+2\dfrac{\partial ^{2}f}{\partial y\partial x}\dfrac{d\varphi }{dx}+\dfrac{\partial ^{2}f}{\partial y^{2}} \left( \dfrac{d\varphi }{dx}\right) ^{2}+\dfrac{d\,^{2}\varphi }{dx^{2}} \dfrac{\partial f}{\partial y} \label{derf2}\tag{1.4} \end{eqnarray} Sustituyendo (\ref{derphi1}): \(\dfrac{d\varphi }{dx}=-\dfrac{\dfrac{ \partial g}{\partial x}}{\dfrac{\partial g}{\partial y}}\) en (\ref{derf1}) obtenemos : \begin{equation} \dfrac{dF}{dx}=\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y} \left( -\dfrac{\dfrac{\partial g}{\partial x}}{\dfrac{\partial g}{\partial y} }\right) =\dfrac{\partial f}{\partial x}-\dfrac{\dfrac{\partial g}{\partial x }\dfrac{\partial f}{\partial y}}{\dfrac{\partial g}{\partial y}}. \label{derf3}\tag{1.5} \end{equation} Sustituyendo (\ref{derphi1}) y (\ref{derphi2}) en (\ref{derf2}) obtenemos: \begin{eqnarray*} \dfrac{d\,^{2}F}{dx^{2}} &=&\dfrac{\partial ^{2}f}{\partial x^{2}}+2\dfrac{ \partial ^{2}f}{\partial y\partial x}\dfrac{d\varphi }{dx}+\dfrac{\partial ^{2}f}{\partial y^{2}}\left( \dfrac{d\varphi }{dx}\right) ^{2}+\dfrac{ d\,^{2}\varphi }{dx^{2}}\dfrac{\partial f}{\partial y} \\ & = &\dfrac{\partial ^{2}f}{\partial x^{2}}+2\dfrac{\partial ^{2}f}{\partial y\partial x}\left( -\dfrac{\dfrac{\partial g}{\partial x}}{\dfrac{\partial g }{\partial y}}\right) +\dfrac{\partial ^{2}f}{\partial y^{2}}\left( -\dfrac{ \dfrac{\partial g}{\partial x}}{\dfrac{\partial g}{\partial y}}\right) ^{2}+ \left( -\dfrac{1}{\dfrac{\partial g}{\partial y}}\left( \dfrac{\partial ^{2}g}{\partial x^{2}}+2\dfrac{\partial ^{2}g}{\partial y\partial x}\left( - \dfrac{\dfrac{\partial g}{\partial x}}{\dfrac{\partial g}{\partial y}} \right) +\dfrac{\partial ^{2}g}{\partial y^{2}}\left( -\dfrac{\dfrac{ \partial g}{\partial x}}{\dfrac{\partial g}{\partial y}}\right) ^{2}\right) \right) \dfrac{\partial f}{\partial y} \\ & = & \dfrac{\partial ^{2}f}{\partial x^{2}}-2\dfrac{\partial ^{2}f}{\partial y\partial x}\left( \dfrac{\dfrac{\partial g}{\partial x}}{\dfrac{\partial g}{ \partial y}}\right) +\dfrac{\partial ^{2}f}{\partial y^{2}}\left( \dfrac{ \dfrac{\partial g}{\partial x}}{\dfrac{\partial g}{\partial y}}\right) ^{2} -\dfrac{\dfrac{\partial f}{\partial y}}{\dfrac{\partial g}{\partial y}} \left( \dfrac{\partial ^{2}g}{\partial x^{2}}-2\dfrac{\partial ^{2}g}{ \partial y\partial x}\left( \dfrac{\dfrac{\partial g}{\partial x}}{\dfrac{ \partial g}{\partial y}}\right) +\dfrac{\partial ^{2}g}{\partial y^{2}} \left( \dfrac{\dfrac{\partial g}{\partial x}}{\dfrac{\partial g}{\partial y}} \right) ^{2}\right) \\ & = & \dfrac{\partial ^{2}f}{\partial x^{2}}-2\dfrac{\partial ^{2}f}{\partial y\partial x}\left( \dfrac{\dfrac{\partial g}{\partial x}}{\dfrac{\partial g}{ \partial y}}\right) +\dfrac{\partial ^{2}f}{\partial y^{2}}\left( \dfrac{ \dfrac{\partial g}{\partial x}}{\dfrac{\partial g}{\partial y}}\right) ^{2} -\dfrac{\dfrac{\partial f}{\partial y}}{\dfrac{\partial g}{\partial y}} \dfrac{\partial ^{2}g}{\partial x^{2}}-2\left( -\dfrac{\dfrac{\partial f}{ \partial y}}{\dfrac{\partial g}{\partial y}}\right) \dfrac{\partial ^{2}g}{ \partial y\partial x}\left( \dfrac{\dfrac{\partial g}{\partial x}}{\dfrac{ \partial g}{\partial y}}\right) +\left( -\dfrac{\dfrac{\partial f}{\partial y }}{\dfrac{\partial g}{\partial y}}\right) \dfrac{\partial ^{2}g}{\partial y^{2}}\left( \dfrac{\dfrac{\partial g}{\partial x}}{\dfrac{\partial g}{ \partial y}}\right) ^{2} \\ & = & \dfrac{\partial ^{2}f}{\partial x^{2}}-\dfrac{\dfrac{\partial f}{\partial y}}{\dfrac{\partial g}{\partial y}}\dfrac{\partial ^{2}g}{\partial x^{2}} -2\left( \dfrac{\partial ^{2}f}{\partial y\partial x}\left( \dfrac{\dfrac{ \partial g}{\partial x}}{\dfrac{\partial g}{\partial y}}\right) -\dfrac{ \dfrac{\partial f}{\partial y}}{\dfrac{\partial g}{\partial y}}\dfrac{ \partial ^{2}g}{\partial y\partial x}\left( \dfrac{\dfrac{\partial g}{ \partial x}}{\dfrac{\partial g}{\partial y}}\right) \right) +\dfrac{\partial ^{2}f}{\partial y^{2}}\left( \dfrac{\dfrac{\partial g}{\partial x}}{\dfrac{ \partial g}{\partial y}}\right) ^{2}-\dfrac{\dfrac{\partial f}{\partial y}}{ \dfrac{\partial g}{\partial y}}\dfrac{\partial ^{2}g}{\partial y^{2}}\left( \dfrac{\dfrac{\partial g}{\partial x}}{\dfrac{\partial g}{\partial y}} \right) ^{2} \\ & = & \dfrac{\partial ^{2}f}{\partial x^{2}}-\dfrac{\dfrac{\partial f}{\partial y}}{\dfrac{\partial g}{\partial y}}\dfrac{\partial ^{2}g}{\partial x^{2}} -2\left( \dfrac{\dfrac{\partial g}{\partial x}}{\dfrac{\partial g}{\partial y }}\right) \left( \dfrac{\partial ^{2}f}{\partial y\partial x}-\dfrac{ \dfrac{ \partial f}{\partial y}}{\dfrac{\partial g}{\partial y}}\dfrac{ \partial ^{2}g }{\partial y\partial x}\right) +\left( \dfrac{\dfrac{\partial g}{\partial x} }{\dfrac{\partial g}{\partial y}}\right) ^{2}\left( \dfrac{ \partial ^{2}f}{ \partial y^{2}}-\dfrac{\dfrac{\partial f}{\partial y}}{ \dfrac{\partial g}{ \partial y}}\dfrac{\partial ^{2}g}{\partial y^{2}}\right) \\ & = & \dfrac{1}{\left( \dfrac{\partial g}{\partial y}\right) ^{2}}\left( \left( \dfrac{\partial ^{2}f}{\partial x^{2}}-\dfrac{\dfrac{\partial f}{\partial y} }{\dfrac{\partial g}{\partial y}}\dfrac{\partial ^{2}g}{\partial x^{2}} \right) \left( \dfrac{\partial g}{\partial y}\right) ^{2}-2\left( \dfrac{ \partial g}{\partial x}\right) \left( \dfrac{\partial g}{\partial y}\right) \left( \dfrac{\partial ^{2}f}{\partial y\partial x}-\dfrac{\dfrac{\partial f }{\partial y}}{\dfrac{\partial g}{\partial y}}\dfrac{\partial ^{2}g}{ \partial y\partial x}\right) + \left( \dfrac{\partial g}{\partial x}\right) ^{2}\left( \dfrac{ \partial ^{2}f}{\partial y^{2}}-\dfrac{\dfrac{\partial f}{\partial y}}{ \dfrac{\partial g}{\partial y}}\dfrac{\partial ^{2}g}{\partial y^{2}}\right) \right) \end{eqnarray*} De la igualdad (\ref{derf3}) tenemos \begin{eqnarray*} \dfrac{dF}{dx} & = & \dfrac{\partial f}{\partial x}-\dfrac{\dfrac{\partial g}{ \partial x}\dfrac{\partial f}{\partial y}}{\dfrac{\partial g}{\partial y}} \\ & = & \dfrac{\partial f}{\partial x}-\dfrac{\dfrac{\partial f}{\partial y}}{ \dfrac{\partial g}{\partial y}}\left( \dfrac{\partial g}{\partial x}\right) . \end{eqnarray*} Como \begin{eqnarray} \dfrac{\partial f}{\partial x}\left( \overline{v_{0}}\right) & = & \lambda \dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) \label{Lagrange1}\tag{1.6} \\ \dfrac{\partial f}{\partial y}\left( \overline{v_{0}}\right) & = & \lambda \dfrac{\partial g}{\partial y}\left( \overline{v_{0}}\right) \notag \end{eqnarray} Despejando \(\lambda \) de la segunda ecuación de (\ref{Lagrange1}), tenemos \begin{equation*} \lambda =\dfrac{\dfrac{\partial f}{\partial y}}{\dfrac{\partial g}{\partial y }} \end{equation*} Entonces \begin{eqnarray*} \dfrac{dF}{dx} & = & \dfrac{\partial f}{\partial x}-\lambda \left( \dfrac{ \partial g}{\partial x}\right) \\ \dfrac{d\,^{2}F}{dx^{2}} & = & \dfrac{1}{\left( \dfrac{\partial g}{\partial y} \right) ^{2}}\left( \left( \dfrac{\partial ^{2}f}{\partial x^{2}}-\lambda \dfrac{\partial ^{2}g}{\partial x^{2}}\right) \left( \dfrac{\partial g}{ \partial y}\right) ^{2}-2\left( \dfrac{\partial g}{\partial x}\right) \left( \dfrac{\partial g}{\partial y}\right) \left( \dfrac{\partial ^{2}f}{\partial y\partial x}-\lambda \dfrac{\partial ^{2}g}{\partial y\partial x}\right) +\left( \dfrac{\partial g}{\partial x}\right) ^{2}\left( \dfrac{\partial ^{2}f}{\partial y^{2}}-\lambda \dfrac{\partial ^{2}g}{\partial y^{2}}\right) \right) \end{eqnarray*} De la primera ecuación de (\ref{Lagrange1}) tenemos \begin{equation*} \dfrac{dF}{dx}=\dfrac{\partial f}{\partial x}-\lambda \left( \dfrac{\partial g}{\partial x}\right) =0. \end{equation*} Calculamos las derivadas parciales de \(h\): \begin{equation*} \dfrac{\partial h}{\partial x}=\dfrac{\partial f}{\partial x}-\lambda \dfrac{ \partial g}{\partial x},\quad \quad \quad \dfrac{\partial h}{\partial y}= \dfrac{\partial f}{\partial y}-\lambda \dfrac{\partial g}{\partial y} \end{equation*} y \begin{eqnarray*} \dfrac{\partial ^{2}h}{\partial x^{2}} & = & \dfrac{\partial ^{2}f}{\partial x^{2}}-\lambda \dfrac{\partial ^{2}g}{\partial x^{2}} \\ \dfrac{\partial ^{2}h}{\partial y^{2}} & = & \dfrac{\partial ^{2}f}{\partial y^{2}}-\lambda \dfrac{\partial ^{2}g}{\partial y^{2}} \\ \dfrac{\partial ^{2}h}{\partial y\partial x} & = & \dfrac{\partial ^{2}f}{ \partial y\partial x}-\lambda \dfrac{\partial ^{2}g}{\partial y\partial x}= \dfrac{\partial ^{2}h}{\partial x\partial y} \end{eqnarray*} Como \begin{eqnarray*} \dfrac{d\,^{2}F}{dx^{2}} & = & \dfrac{1}{\left( \dfrac{\partial g}{\partial y} \right) ^{2}}\left( \left( \dfrac{\partial ^{2}f}{\partial x^{2}}-\lambda \dfrac{\partial ^{2}g}{\partial x^{2}}\right) \left( \dfrac{\partial g}{ \partial y}\right) ^{2}-2\left( \dfrac{\partial g}{\partial x}\right) \left( \dfrac{\partial g}{\partial y}\right) \left( \dfrac{\partial ^{2}f}{\partial y\partial x}-\lambda \dfrac{\partial ^{2}g}{\partial y\partial x}\right) +\left( \dfrac{\partial g}{\partial x}\right) ^{2}\left( \dfrac{\partial ^{2}f}{\partial y^{2}}-\lambda \dfrac{\partial ^{2}g}{\partial y^{2}}\right) \right) \\ & = & \dfrac{1}{\left( \dfrac{\partial g}{\partial y}\right) ^{2}}\left( \dfrac{ \partial ^{2}h}{\partial x^{2}}\left( \dfrac{\partial g}{\partial y}\right) ^{2}-2\left( \dfrac{\partial g}{\partial x}\right) \left( \dfrac{\partial g}{ \partial y}\right) \dfrac{\partial ^{2}h}{\partial x\partial y}+\left( \dfrac{\partial g}{\partial x}\right) ^{2}\dfrac{\partial ^{2}h}{\partial y^{2}}\right) \end{eqnarray*} Observamos que esto último es igual a \begin{equation*} \dfrac{d\,^{2}F}{dx^{2}}\left( x_{0}\right) =\dfrac{-1}{\left( \dfrac{ \partial g}{\partial y}\left( \overline{v_{0}}\right) \right) ^{2}} \left\vert \begin{array}{ccc} 0 & \dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) & \dfrac{\partial g} {\partial y}\left( \overline{v_{0}}\right) \\ & & \\ \dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) & \dfrac{\partial ^{2}h} {\partial x^{2}}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial y\partial x}\left( \overline{v_{0}}\right) \\ & & \\ \dfrac{\partial g}{\partial y}\left( \overline{v_{0}}\right) & \dfrac{\partial ^{2}h}{\partial x\partial y}\left( \overline{v_{0}}\right) & \dfrac{\partial ^{2}h}{\partial y^{2}} \left( \overline{v_{0}}\right) \end{array} \right\vert \end{equation*} A partir del criterio de la segunda derivada para funciones de una variable aplicado a la función \(F,\), obtenemos:

• \(\left\vert \overline{H}\left( \overline{v_{0}}\right) \right\vert > 0 \) implica \(\dfrac{d\,^{2}F}{dx^{2}}\left( x_{0}\right) < 0,\) y entonces \(\overline{v_{0}}\) es un punto máximo relativo de \(\left. f\right\vert _{S}.\)
• \(\left\vert \overline{H}\left( \overline{v_{0}}\right) \right\vert < 0 \) implica \(\dfrac{d\,^{2}F}{dx^{2}}\left( x\,_{0}\right) > 0,\) y entonces \(\overline{v_{0}}\) es un punto mínimo relativo de \(\left. f\right\vert _{S}. \)
• \(\left\vert \overline{H}\left( \overline{v_{0}}\right) \right\vert =0\) implica \(\dfrac{d\,^{2\,}F}{dx^{2}}\left( x_{0}\right) =0,\) y entonces el criterio no da información.

Si a partir de \(g\left( x,y,z\right) =c,\) podemos escribir una de las variables en términos de las otras dos aunque sólo sea en una vecindad, entonces al hacer la sustitución correspondiente en \(f\) obtenemos una función \(F\) de dos variables, por lo que para encontrar sus máximos y mínimos podemos utilizar los criterios existentes para ese caso.

Como \(\nabla g\left( \overline{v_{0}}\right) \neq \overline{0}\) entonces alguna de las tres derivadas parciales de \(g\) es distinta de \(0\) en \(\overline{v_{0}}.\)

Sólo probaremos el caso en que \(\dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) \neq 0.\) Los otros casos se prueban de manera similar.

Definimos la función \begin{equation*} h=f-\lambda g. \end{equation*}

Por hipótesis, \begin{eqnarray*} \nabla f\left( \overline{v_{0}}\right) &=&\lambda \nabla g\left( \overline{ v_{0}}\right) \\ g\left( \overline{v_{0}}\right) &=&c. \end{eqnarray*} O sea, \(\overline{v_{0}}=\left( x_{0},y_{0},z_{0}\right) \) y \(\lambda \) satisfacen \begin{eqnarray} \dfrac{\partial f}{\partial x}\left( \overline{v_{0}}\right) &=&\lambda \dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) \label{flang}\tag{1.7} \\ \dfrac{\partial f}{\partial y}\left( \overline{v_{0}}\right) &=&\lambda \dfrac{\partial g}{\partial y}\left( \overline{v_{0}}\right) \notag \\ \dfrac{\partial f}{\partial z}\left( \overline{v_{0}}\right) &=&\lambda \dfrac{\partial g}{\partial z}\left( \overline{v_{0}}\right) \notag \\ g\left( \overline{v_{0}}\right) &=&c \notag \end{eqnarray} Como \(\dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) \neq 0\) y \(g\) es derivable, a partir de la ecuación \(g\left( x,y,z\right) =c,\) podemos expresar a \(x\) como una función de las variables \(y\) y \(z\). De hecho, el Teorema de la Función Implícita nos dice que hay dos abiertos \(U_{0}\subset U\) y \(V\subset \mathbb{R}^{2}\) y una función \(\varphi :V\longrightarrow \mathbb{R}\) tales que \(\overline{v_{0}}=\left( x_{0},y_{0},z_{0}\right) \in U_{0}\), \((y_0,z_0) \in V\), \(\varphi \left( y_{0},z_{0}\right) =x_{0}\) y \begin{equation*} \left\{ \left. \left( x,y,z\right) \in U_{0}\right\vert \,g\left( x,y,z\right) =c\right\} =\left\{ \left. \left( \varphi \left( y,z\right) ,y,z\right) \right\vert \,\left( y,z\right) \in V\right\} \end{equation*} Además, para \(\left( y,z\right) \in V\) se cumple \begin{eqnarray} \dfrac{\partial \varphi }{\partial y}\left( y,z\right) &=&-\dfrac{\dfrac{ \partial g}{\partial y}\left( \varphi \left( y,z\right) ,y,z\right) }{\dfrac{ \partial g}{\partial x}\left( \varphi \left( y,z\right) ,y,z\right) } \label{Parfi}\tag{1.8} \\ \dfrac{\partial \varphi }{\partial z}\left( y,z\right) &=&-\dfrac{\dfrac{ \partial g}{\partial z}\left( \varphi \left( y,z\right) ,y,z\right) }{\dfrac{ \partial g}{\partial x}\left( \varphi \left( y,z\right) ,y,z\right) } \notag \end{eqnarray} Más aún, en este caso, por ser \(g\) de clase \(C^{2}\), tenemos \begin{eqnarray} \dfrac{\partial ^{2}\varphi }{\partial y^{2}} & = & \dfrac{-\dfrac{\partial g}{ \partial x}\left( \dfrac{\partial ^{2}g}{\partial x\partial y}\dfrac{ \partial \varphi }{\partial y}+\dfrac{\partial ^{2}g}{\partial y^{2}}\right) +\dfrac{\partial g}{\partial y}\left( \dfrac{\partial ^{2}g}{\partial x^{2}} \dfrac{\partial \varphi }{\partial y}+\dfrac{\partial ^{2}g}{\partial y\partial x}\right) }{\left( \dfrac{\partial g}{\partial x}\right) ^{2}} \notag \\ & = & \dfrac{-\dfrac{\partial g}{\partial x}\dfrac{\partial ^{2}g}{\partial x\partial y}\dfrac{\partial \varphi }{\partial y}-\dfrac{\partial g}{ \partial x}\dfrac{\partial ^{2}g}{\partial y^{2}}+\dfrac{\partial g}{ \partial y}\dfrac{\partial ^{2}g}{\partial x^{2}}\dfrac{\partial \varphi }{ \partial y}+\dfrac{\partial g}{\partial y}\dfrac{\partial ^{2}g}{\partial y\partial x}}{\left( \dfrac{\partial g}{\partial x}\right) ^{2}} \notag \\ &=&\dfrac{1}{\left( \dfrac{\partial g}{\partial x}\right) ^{2}}\left( \dfrac{ \partial g}{\partial y}\dfrac{\partial ^{2}g}{\partial x^{2}}\left( -\dfrac{ \dfrac{\partial g}{\partial y}}{\dfrac{\partial g}{\partial x}}\right) + \dfrac{\partial g}{\partial y}\dfrac{\partial ^{2}g}{\partial y\partial x}- \dfrac{\partial g}{\partial x}\dfrac{\partial ^{2}g}{\partial y\partial x} \left( -\dfrac{\dfrac{\partial g}{\partial y}}{\dfrac{\partial g}{\partial x} }\right) -\dfrac{\partial g}{\partial x}\dfrac{\partial ^{2}g}{\partial y^{2} }\right) \label{SegpracialFiY}\tag{1.9} \\ & = & \dfrac{1}{\left( \dfrac{\partial g}{\partial x}\right) ^{2}}\left( - \dfrac{\left( \dfrac{\partial g}{\partial y}\right) ^{2}\dfrac{\partial ^{2}g }{\partial x^{2}}}{\dfrac{\partial g}{\partial x}}+\dfrac{\dfrac{ \partial g}{ \partial y}\dfrac{\partial ^{2}g}{\partial y\partial x}\dfrac{ \partial g}{ \partial x}}{\dfrac{\partial g}{\partial x}}+\dfrac{\dfrac{ \partial g}{ \partial x}\dfrac{\partial ^{2}g}{\partial y\partial x}\dfrac{ \partial g}{ \partial y}}{\dfrac{\partial g}{\partial x}}-\dfrac{\left( \dfrac{\partial g }{\partial x}\right) ^{2}\dfrac{\partial ^{2}g}{\partial y^{2}}}{\dfrac{ \partial g}{\partial x}}\right) \notag \\ &=&\dfrac{1}{\left( \dfrac{\partial g}{\partial x}\right) ^{3}}\left( 2 \dfrac{\partial g}{\partial y}\dfrac{\partial ^{2}g}{\partial y\partial x} \dfrac{\partial g}{\partial x}-\left( \dfrac{\partial g}{\partial y}\right) ^{2}\dfrac{\partial ^{2}g}{\partial x^{2}}-\left( \dfrac{\partial g}{ \partial x}\right) ^{2}\dfrac{\partial ^{2}g}{\partial y^{2}}\right) \notag \end{eqnarray} De modo similar obtenemos \begin{equation} \dfrac{\partial ^{2}\varphi }{\partial z\partial y}=\frac{1}{\left( \dfrac{ \partial g}{\partial x}\right) ^{3}}\left( \dfrac{\partial ^{2}g}{\partial z\partial x}\dfrac{\partial g}{\partial y}\dfrac{\partial g}{\partial x}- \dfrac{\partial ^{2}g}{\partial x^{2}}\dfrac{\partial g}{\partial z}\dfrac{ \partial g}{\partial y}+\dfrac{\partial ^{2}g}{\partial x\partial y}\dfrac{ \partial g}{\partial z}\dfrac{\partial g}{\partial x}-\dfrac{\partial ^{2}g}{ \partial z\partial y}\left( \dfrac{\partial g}{\partial x}\right) ^{2}\right) \label{SegpracialFiYZ}\tag{1.10} \end{equation} y \begin{equation} \dfrac{\partial ^{2}\varphi }{\partial z^{2}}=\frac{1}{\left( \dfrac{ \partial g}{\partial x}\right) ^{3}}\left( 2\dfrac{\partial ^{2}g}{\partial z\partial x}\dfrac{\partial g}{\partial z}\dfrac{\partial g}{\partial x}- \dfrac{\partial ^{2}g}{\partial x^{2}}\left( \dfrac{\partial g}{\partial z} \right) ^{2}-\dfrac{\partial ^{2}g}{\partial z^{2}}\left( \dfrac{\partial g}{ \partial x}\right) ^{2}\right) \label{SegpracialFiZ}\tag{1.11} \end{equation} Definamos \(F:V_{0}\longrightarrow \mathbb{R}\) como \begin{equation*} F\left( y,z\right) =f\left( \varphi \left( y,z\right) ,y,z\right) \end{equation*}

Entonces \(\left( y_{0},z_{0}\right) \) es un punto máximo relativo, mínimo relativo o punto silla de \(F\) si \(\overline{v_{0}}=\left( x_{0},y_{0},z_{0}\right) \) es lo correspondiente para \(f\)

Afirmamos

(a) \[\dfrac{\partial F}{\partial y}\left( y_{0},z_{0}\right) =0=\dfrac{ \partial F}{\partial z}\left( y_{0},z_{0}\right) \]

(b) \[\dfrac{\partial ^{2}F}{\partial y^{2}}\left( y_{0},z_{0}\right) =- \dfrac{1}{\left( \dfrac{\partial g}{\partial x}\left( \overline{v_{0}} \right) \right) ^{2}}\left\vert \overline{H}_{3}\left( \overline{v_{0}} \right) \right\vert ;\] de donde, el signo de \(\dfrac{\partial ^{2}F}{ \partial y^{2}}\left( y_{0},z_{0}\right) \) es menos el signo de \(\left\vert \overline{H}_{3}\left( \overline{v_{0}}\right) \right\vert \)

(c) \[H\left( y_{0},z_{0}\right) =\left\vert \begin{array}{ccc} \dfrac{\partial ^{2}F}{\partial x^{2}}\left( y_{0},z_{0}\right) & & \dfrac{ \partial ^{2}F}{\partial y\partial x}\left( y_{0},z_{0}\right) \\ & & \\ \dfrac{\partial ^{2}F}{\partial x\partial y}\left( y_{0},z_{0}\right) & & \dfrac{\partial ^{2}F}{\partial y^{2}}\left( y_{0},z_{0}\right) \end{array} \right\vert =-\dfrac{1}{\left( \dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) \right) ^{2}}\left\vert \overline{H}_{4}\left( \overline{v_{0}}\right) \right\vert ;\] de donde, el signo de \(H\) es menos el signo de \(\left\vert \overline{H}_{4}\left( \overline{v_{0}}\right) \right\vert .\)

Al aplicar a \(F\) el criterio de la segunda derivada para funciones en dos variables, tenemos que bajo las hipótesis de cada uno de los apartados (1) a (3) el punto \(\left( y_{0},z_{0}\right) \) es un punto máximo relativo, mínimo relativo o punto silla de \(F\), respectivamente y el teorema está probado.

Con relación al apartado (4) observamos que la función \(f\left( x,y,z\right) =x^{4}+y^{4}+z^{4}\) sujeta a la restricción \(x+y+z=0\) tiene un mínimo absoluto estricto en el origen y la función \(-f\left( x,y,z\right) \) sujeta a la restricción \(x+y+z=0\) tiene un máximo absoluto en el origen. En tanto que, la función \(g\left( x,y,z\right) =x^{3}+y^{3}+z^{3}\) sujeta a la restricción \(x+y+z=0\) tiene un punto silla en el origen. En los tres casos, el determinante hessiano limitado de \(f\) en el origen es \(0.\)

Pruebas de las afirmaciones

Aplicando la regla de la cadena a \begin{equation*} F\left( y,z\right) =f\left( \varphi \left( y,z\right) ,y,z\right) \end{equation*} tenemos \begin{eqnarray} \dfrac{\partial F}{\partial y} & = & \dfrac{\partial f}{\partial x}\dfrac{ \partial \varphi }{\partial y}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial y}+\dfrac{\partial f}{\partial z}\dfrac{\partial z}{\partial y} \label{ParF}\tag{1.12} \\ & = & \dfrac{\partial f}{\partial x}\dfrac{\partial \varphi }{\partial y}+ \dfrac{\partial f}{\partial y} \notag \\ \\ \dfrac{\partial F}{\partial z} & = & \dfrac{\partial f}{\partial x}\dfrac{ \partial \varphi }{\partial z}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial z}+\dfrac{\partial f}{\partial z}\dfrac{\partial z}{\partial z} \notag \\ & = & \dfrac{\partial f}{\partial x}\dfrac{\partial \varphi }{\partial z}+ \dfrac{\partial f}{\partial z} \notag \end{eqnarray} De nuevo por la regla de la cadena, las igualdades (\ref{Parfi}) y dado que \(f\left( \varphi \left( y_{0},z_{0}\right) ,y_{0},z_{0}\right) =\lambda g\left( \varphi \left( y_{0},z_{0}\right) ,y_{0},z_{0}\right) ,\) tenemos que las segundas parciales de \(F\) evaluadas en \(\left( y_{0},z_{0}\right) \) son \begin{eqnarray} \dfrac{\partial ^{2}F}{\partial y^{2}} & = & \left( \dfrac{\partial ^{2}f}{ \partial x^{2}}\dfrac{\partial \varphi }{\partial y}+\dfrac{\partial ^{2}f}{ \partial y\partial x}\right) \dfrac{\partial \varphi }{\partial y}+\dfrac{ \partial f}{\partial x}\dfrac{\partial ^{2}\varphi }{\partial y^{2}}+\dfrac{ \partial ^{2}f}{\partial x\partial y}\dfrac{\partial \varphi }{\partial y}+ \dfrac{\partial ^{2}f}{\partial y^{2}} \label{2aParFY}\tag{1.13} \\ & = & \dfrac{\partial ^{2}f}{\partial x^{2}}\left( \dfrac{\partial \varphi }{ \partial y}\right) ^{2}+2\dfrac{\partial ^{2}f}{\partial y\partial x}\dfrac{ \partial \varphi }{\partial y}+\dfrac{\partial f}{\partial x}\dfrac{\partial ^{2}\varphi }{\partial y^{2}}+\dfrac{\partial ^{2}f}{\partial y^{2}} \notag \\ & = & \dfrac{\partial ^{2}f}{\partial x^{2}}\left( -\dfrac{\dfrac{\partial g}{ \partial y}}{\dfrac{\partial g}{\partial x}}\right) ^{2}+2\dfrac{\partial ^{2}f}{\partial y\partial x}\left( -\dfrac{\dfrac{\partial g}{\partial y}}{ \dfrac{\partial g}{\partial x}}\right) +\lambda \dfrac{\partial g}{\partial x }\dfrac{\partial ^{2}\varphi }{\partial y^{2}}+\dfrac{\partial ^{2}f}{ \partial y^{2}} \notag \\ & = & \frac{1}{\left( \dfrac{\partial g}{\partial x}\right) ^{2}}\left[ \dfrac{ \partial ^{2}f}{\partial x^{2}}\left( \dfrac{\partial g}{\partial y}\right) ^{2}-2\dfrac{\partial ^{2}f}{\partial y\partial x}\dfrac{\partial g}{ \partial y}\dfrac{\partial g}{\partial x}+\lambda \left( \dfrac{\partial g}{ \partial x}\right) ^{3}\dfrac{\partial ^{2}\varphi }{\partial y^{2}}+\dfrac{ \partial ^{2}f}{\partial y^{2}}\left( \dfrac{\partial g}{\partial x}\right) ^{2}\right] , \notag \end{eqnarray} \begin{eqnarray} \dfrac{\partial ^{2}F}{\partial z\partial y} & = & \left( \dfrac{\partial ^{2}f }{\partial x^{2}}\dfrac{\partial \varphi }{\partial z}+\dfrac{\partial ^{2}f }{\partial z\partial x}\right) \dfrac{\partial \varphi }{\partial y}+\dfrac{ \partial f}{\partial x}\dfrac{\partial ^{2}\varphi }{\partial z\partial y}+ \dfrac{\partial ^{2}f}{\partial x\partial y}\dfrac{\partial \varphi }{ \partial z}+\dfrac{\partial ^{2}f}{\partial z\partial y} \notag \\ & = & \dfrac{\partial ^{2}f}{\partial x^{2}}\dfrac{\partial \varphi }{\partial y }\dfrac{\partial \varphi }{\partial z}+\dfrac{\partial ^{2}f}{\partial z\partial x}\dfrac{\partial \varphi }{\partial y}+\dfrac{\partial f}{ \partial x}\dfrac{\partial ^{2}\varphi }{\partial z\partial y}+\dfrac{ \partial ^{2}f}{\partial x\partial y}\dfrac{\partial \varphi }{\partial z}+ \dfrac{\partial ^{2}f}{\partial z\partial y} \label{2aParFYZ}\tag{1.14} \\ & = & \dfrac{\partial ^{2}f}{\partial x^{2}}\left( -\dfrac{\dfrac{\partial g}{ \partial y}}{\dfrac{\partial g}{\partial x}}\right) \left( -\dfrac{\dfrac{ \partial g}{\partial z}}{\dfrac{\partial g}{\partial x}}\right) +\dfrac{ \partial ^{2}f}{\partial z\partial x}\left( -\dfrac{\dfrac{\partial g}{ \partial y}}{\dfrac{\partial g}{\partial x}}\right) +\lambda \dfrac{\partial g}{\partial x}\dfrac{\partial ^{2}\varphi }{\partial z\partial y} +\dfrac{\partial ^{2}f}{\partial x\partial y}\left( -\dfrac{\dfrac{ \partial g}{\partial z}}{\dfrac{\partial g}{\partial x}}\right) +\dfrac{ \partial ^{2}f}{\partial z\partial y} \notag \\ &=&\frac{1}{\left( \dfrac{\partial g}{\partial x}\right) ^{2}}\left[ \dfrac{ \partial ^{2}f}{\partial x^{2}}\dfrac{\partial g}{\partial y}\dfrac{\partial g}{\partial z}-\dfrac{\partial ^{2}f}{\partial z\partial x}\dfrac{\partial g }{\partial y}\dfrac{\partial g}{\partial x}+\lambda \left( \dfrac{\partial g }{\partial x}\right) ^{3}\dfrac{\partial ^{2}\varphi }{\partial z\partial y} -\dfrac{\partial ^{2}f}{\partial x\partial y}\dfrac{\partial g}{ \partial z}\dfrac{\partial g}{\partial x}+\dfrac{\partial ^{2}f}{\partial z\partial y}\left( \dfrac{\partial g}{\partial x}\right) ^{2}\right] \notag \end{eqnarray} y \begin{eqnarray} \dfrac{\partial ^{2}F}{\partial z^{2}} & = &\left( \dfrac{\partial ^{2}f}{ \partial x^{2}}\dfrac{\partial \varphi }{\partial z}+\dfrac{\partial ^{2}f}{ \partial z\partial x}\right) \dfrac{\partial \varphi }{\partial z}+\dfrac{ \partial f}{\partial x}\dfrac{\partial ^{2}\varphi }{\partial z^{2}}+\dfrac{ \partial ^{2}f}{\partial x\partial z}\dfrac{\partial \varphi }{\partial z}+ \dfrac{\partial ^{2}f}{\partial z^{2}} \label{2aParFZ}\tag{1.15} \\ & = & \dfrac{\partial ^{2}f}{\partial x^{2}}\left( -\dfrac{\dfrac{\partial g}{ \partial z}}{\dfrac{\partial g}{\partial x}}\right) ^{2}+2\dfrac{\partial ^{2}f}{\partial z\partial x}\left( -\dfrac{\dfrac{\partial g}{\partial z}}{ \dfrac{\partial g}{\partial x}}\right) +\lambda \dfrac{\partial g}{\partial x }\dfrac{\partial ^{2}\varphi }{\partial z^{2}}+\dfrac{\partial ^{2}f}{ \partial z^{2}} \notag \\ & = & \frac{1}{\left( \dfrac{\partial g}{\partial x}\right) ^{2}}\left[ \dfrac{ \partial ^{2}f}{\partial x^{2}}\dfrac{\partial g}{\partial z}-2\dfrac{ \partial ^{2}f}{\partial z\partial x}\dfrac{\partial g}{\partial x}\dfrac{ \partial g}{\partial z}+\lambda \left( \dfrac{\partial g}{\partial x}\right) ^{3}\dfrac{\partial ^{2}\varphi }{\partial z^{2}}+\dfrac{\partial ^{2}f}{ \partial z^{2}}\left( \dfrac{\partial g}{\partial x}\right) ^{2}\right] \notag \end{eqnarray} Al sustituir \(\dfrac{\partial \varphi }{\partial y}\) y \(\dfrac{\partial \varphi }{\partial z}\) en (\ref{ParF}) de acuerdo con las igualdades (\ref {Parfi}), obtenemos \begin{eqnarray*} \dfrac{\partial F}{\partial y} & = & -\dfrac{\partial f}{\partial x}\dfrac{ \dfrac{\partial g}{\partial y}}{\dfrac{\partial g}{\partial x}}+\dfrac{ \partial f}{\partial y} \\ \dfrac{\partial F}{\partial z} & = & -\dfrac{\partial f}{\partial x}\dfrac{ \dfrac{\partial g}{\partial z}}{\dfrac{\partial g}{\partial x}}+\dfrac{ \partial f}{\partial z} \end{eqnarray*} Así, al evaluar las parciales en \(\left( y_{0},z_{0}\right) \) tenemos \begin{equation*} \dfrac{\partial F}{\partial y}\left( y_{0},z_{0}\right) =-\dfrac{\partial f}{ \partial x}\left( \overline{v_{0}}\right) \dfrac{\dfrac{\partial g}{\partial y}\left( \overline{v_{0}}\right) }{\dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) }+\dfrac{\partial f}{\partial y}\left( \overline{ v_{0}}\right) =-\lambda \dfrac{\partial g}{\partial y}\left( \overline{v_{0}} \right) +\dfrac{\partial f}{\partial y}\left( \overline{v_{0}}\right) =0 \end{equation*} De modo similar \begin{equation*} \dfrac{\partial F}{\partial z}\left( y_{0},z_{0}\right) =-\dfrac{\partial f}{ \partial x}\left( \overline{v_{0}}\right) \dfrac{\dfrac{\partial g}{\partial z}\left( \overline{v_{0}}\right) }{\dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) }+\dfrac{\partial f}{\partial z}\left( \overline{ v_{0}}\right) =-\lambda \dfrac{\partial g}{\partial z}\left( \overline{v_{0}} \right) +\dfrac{\partial f}{\partial z}\left( \overline{v_{0}}\right) =0 \end{equation*}

en donde las últimas igualdades en los dos desplegados se dan por las segunda y tercera igualdades en (\ref{flang}). Queda probado el inciso (a).

Para facilitar un poco la comparación de resultados de los cálculos en las pruebas de los incisos (b) y (c) haremos los siguientes cambios de nombre \begin{equation*} \begin{array}{lllllll} \alpha & = & \dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) & & o & = & \dfrac{\partial ^{2}f}{\partial z\partial x}\left( \overline{ v_{0}}\right) \\ \beta & = & \dfrac{\partial g}{\partial y}\left( \overline{v_{0}}\right) & & p & = & \lambda \dfrac{\partial ^{2}g}{\partial z\partial x}\left( \overline{v_{0}}\right) \\ \gamma & = & \dfrac{\partial g}{\partial z}\left( \overline{v_{0}}\right) & & q & = & \dfrac{\partial ^{2}f}{\partial y^{2}}\left( \overline{v_{0}} \right) \\ k & = & \dfrac{\partial ^{2}f}{\partial x^{2}}\left( \overline{v_{0}}\right) & & r & = & \lambda \dfrac{\partial ^{2}g}{\partial y^{2}}\left( \overline{ v_{0}}\right) \\ l & = & \lambda \dfrac{\partial ^{2}g}{\partial x^{2}}\left( \overline{v_{0}} \right) & & s & = & \dfrac{\partial ^{2}f}{\partial z\partial y}\left( \overline{v_{0}}\right) \\ m & = & \dfrac{\partial ^{2}f}{\partial y\partial x}\left( \overline{v_{0}} \right) & & t & = & \lambda \dfrac{\partial ^{2}g}{\partial z\partial y} \left( \overline{v_{0}}\right) \\ n & = & \lambda \dfrac{\partial ^{2}g}{\partial y\partial x}\left( \overline{ v_{0}}\right) & & u & = & \dfrac{\partial ^{2}f}{\partial z^{2}}\left( \overline{v_{0}}\right) \\ & & & & v & = & \lambda \dfrac{\partial ^{2}g}{\partial z^{2}}\left( \overline{v_{0}}\right) \end{array} \end{equation*} Entonces las segundas derivadas parciales de \(h\) evaluadas en \(\overline{ v_{0}}\) se escriben como: \begin{equation*} \begin{array}{lllllll} a & = & \dfrac{\partial ^{2}h}{\partial x^{2}} & = & \dfrac{\partial ^{2}f}{ \partial x^{2}}-\lambda \dfrac{\partial ^{2}g}{\partial x^{2}} & = & k-l \\ b & = & \dfrac{\partial ^{2}h}{\partial x\partial y} & = & \dfrac{\partial ^{2}f}{\partial y\partial x}-\lambda \dfrac{\partial ^{2}g}{\partial y\partial x} & = & m-n \\ c & = & \dfrac{\partial ^{2}h}{\partial z\partial x} & = & \dfrac{\partial ^{2}f}{\partial z\partial x}-\lambda \dfrac{\partial ^{2}g}{\partial z\partial x} & = & o-p \\ d & = & \dfrac{\partial ^{2}h}{\partial y^{2}} & = & \dfrac{\partial ^{2}f}{ \partial y^{2}}-\lambda \dfrac{\partial ^{2}g}{\partial y^{2}} & = & q-r \\ e & = & \dfrac{\partial ^{2}h}{\partial y\partial z} & = & \dfrac{\partial ^{2}f}{\partial z\partial y}-\lambda \dfrac{\partial ^{2}g}{\partial z\partial y} & = & s-t \\ j & = & \dfrac{\partial ^{2}h}{\partial z^{2}} & = & \dfrac{\partial ^{2}f}{ \partial z^{2}}-\lambda \dfrac{\partial ^{2}g}{\partial z^{2}} & = & u-v \end{array} \end{equation*} Entonces las igualdades (\ref{SegpracialFiY}), (\ref{SegpracialFiYZ}) y (\ref {SegpracialFiZ}) se escriben como \begin{eqnarray*} \left( \dfrac{\partial g}{\partial x}\right) ^{3}\dfrac{\partial ^{2}\varphi }{\partial y^{2}} &=&2\dfrac{\partial ^{2}g}{\partial y\partial x}\dfrac{ \partial g}{\partial y}\dfrac{\partial g}{\partial x}-\dfrac{\partial ^{2}g}{ \partial x^{2}}\left( \dfrac{\partial g}{\partial y}\right) ^{2}-\dfrac{ \partial ^{2}g}{\partial y^{2}}\left( \dfrac{\partial g}{\partial x}\right) ^{2}=\dfrac{1}{\lambda }\left( 2n\beta \alpha -l\beta ^{2}-r\alpha ^{2}\right) \\ \\ \left( \dfrac{\partial g}{\partial x}\right) ^{3}\dfrac{\partial ^{2}\varphi }{\partial z\partial y} &=&\dfrac{\partial ^{2}g}{\partial z\partial x} \dfrac{\partial g}{\partial y}\dfrac{\partial g}{\partial x}-\dfrac{\partial ^{2}g}{\partial x^{2}}\dfrac{\partial g}{\partial z}\dfrac{\partial g}{ \partial y}+\dfrac{\partial ^{2}g}{\partial x\partial y}\dfrac{\partial g}{ \partial z}\dfrac{\partial g}{\partial x}-\dfrac{\partial ^{2}g}{\partial z\partial y}\left( \dfrac{\partial g}{\partial x}\right) ^{2} =\dfrac{1}{\lambda }\left( p\beta \alpha -l\beta \gamma +n\gamma \alpha -t\alpha ^{2}\right) \\ \\ \left( \dfrac{\partial g}{\partial x}\right) ^{3}\dfrac{\partial ^{2}\varphi }{\partial z^{2}} &=&2\dfrac{\partial ^{2}g}{\partial z\partial x}\dfrac{ \partial g}{\partial z}\dfrac{\partial g}{\partial x}-\dfrac{\partial ^{2}g}{ \partial x^{2}}\left( \dfrac{\partial g}{\partial z}\right) ^{2}-\dfrac{ \partial ^{2}g}{\partial z^{2}}\left( \dfrac{\partial g}{\partial x}\right) ^{2}=\dfrac{1}{\lambda }\left( 2p\gamma \alpha -l\gamma ^{2}-v\alpha ^{2}\right) \end{eqnarray*} Y las segundas derivadas parciales de \(F\) evaluadas en \(\left( y_{0},z_{0}\right) \) \begin{eqnarray*} \dfrac{\partial ^{2}F}{\partial y^{2}} & = & \frac{1}{\left( \dfrac{\partial g}{ \partial x}\right) ^{2}}\left[ \dfrac{\partial ^{2}f}{\partial x^{2}}\left( \dfrac{\partial g}{\partial y}\right) ^{2}-2\dfrac{\partial ^{2}f}{\partial y\partial x}\dfrac{\partial g}{\partial y}\dfrac{\partial g}{\partial x} +\lambda \left( \dfrac{\partial g}{\partial x}\right) ^{3}\dfrac{\partial ^{2}\varphi }{\partial y^{2}}+\dfrac{\partial ^{2}f}{\partial y^{2}}\left( \dfrac{\partial g}{\partial x}\right) ^{2}\right] \\ & = & \frac{1}{\left( \dfrac{\partial g}{\partial x}\right) ^{2}}\left[ \dfrac{ \partial ^{2}f}{\partial x^{2}}\left( \dfrac{\partial g}{\partial y}\right) ^{2}-2\dfrac{\partial ^{2}f}{\partial y\partial x}\dfrac{\partial g}{ \partial y}\dfrac{\partial g}{\partial x}+2n\beta \alpha -l\beta ^{2}-r\alpha ^{2}+\dfrac{\partial ^{2}f}{\partial y^{2}}\left( \dfrac{ \partial g}{\partial x}\right) ^{2}\right] \\ & = & \frac{k\beta ^{2}-2m\beta \alpha +2n\beta \alpha -l\beta ^{2}-r\alpha ^{2}+q\alpha ^{2}}{\alpha ^{2}} \\ & = & \frac{\alpha ^{2}\left( q-r\right) +\beta ^{2}\left( k-l\right) -2\beta \alpha \left( m-n\right) }{\alpha ^{2}} \\ & = & \frac{\alpha ^{2}d+\beta ^{2}a-2\beta \alpha b}{\alpha ^{2}}, \end{eqnarray*}

\begin{eqnarray*} \dfrac{\partial ^{2}F}{\partial z\partial y} & = & \left( \dfrac{\partial ^{2}f }{\partial x^{2}}\dfrac{\partial \varphi }{\partial z}+\dfrac{\partial ^{2}f }{\partial z\partial x}\right) \dfrac{\partial \varphi }{\partial y}+\dfrac{ \partial f}{\partial x}\dfrac{\partial ^{2}\varphi }{\partial z\partial y}+ \dfrac{\partial ^{2}f}{\partial x\partial y}\dfrac{\partial \varphi }{ \partial z}+\dfrac{\partial ^{2}f}{\partial z\partial y} \\ & = & \dfrac{\partial ^{2}f}{\partial x^{2}}\dfrac{\dfrac{\partial g}{\partial y }}{\dfrac{\partial g}{\partial x}}\dfrac{\dfrac{\partial g}{\partial z}}{ \dfrac{\partial g}{\partial x}}-\dfrac{\partial ^{2}f}{\partial z\partial x} \dfrac{\dfrac{\partial g}{\partial y}}{\dfrac{\partial g}{\partial x}} +\lambda \dfrac{\partial g}{\partial x}\dfrac{\partial ^{2}\varphi }{ \partial z\partial y}-\dfrac{\partial ^{2}f}{\partial x\partial y}\dfrac{ \dfrac{\partial g}{\partial z}}{\dfrac{\partial g}{\partial x}}+\dfrac{ \partial ^{2}f}{\partial z\partial y} \\ & = & \frac{1}{\left( \dfrac{\partial g}{\partial x}\right) ^{2}}\left[ \dfrac{ \partial ^{2}f}{\partial x^{2}}\dfrac{\partial g}{\partial y}\dfrac{\partial g}{\partial z}-\dfrac{\partial ^{2}f}{\partial z\partial x}\dfrac{\partial g }{\partial y}\dfrac{\partial g}{\partial x}+\lambda \left( \dfrac{\partial g }{\partial x}\right) ^{3}\dfrac{\partial ^{2}\varphi }{\partial z\partial y} - \dfrac{\partial ^{2}f}{\partial x\partial y}\dfrac{\partial g}{\partial x} \dfrac{\partial g}{\partial z}+\dfrac{\partial ^{2}f}{\partial z\partial y} \left( \dfrac{\partial g}{\partial x}\right) ^{2}\right] \\ & = & \frac{k\beta \gamma -o\beta \alpha +p\beta \alpha -\beta \gamma l+n\gamma \alpha -t\alpha ^{2}-m\alpha \gamma +s\alpha ^{2}}{\alpha ^{2}} \\ &=&\frac{\beta \gamma \left( k-l\right) -\beta \alpha \left( o-p\right) -\alpha \gamma \left( m-n\right) +\alpha ^{2}\left( s-t\right) }{\alpha ^{2}} \\ & = & \frac{\beta \gamma a-\beta \alpha c-\alpha \gamma b+\alpha ^{2}e}{\alpha ^{2}} \end{eqnarray*} y \begin{eqnarray*} \dfrac{\partial ^{2}F}{\partial z^{2}} & = & \frac{1}{\left( \dfrac{\partial g}{ \partial x}\right) ^{2}}\left[ \dfrac{\partial ^{2}f}{\partial x^{2}}\left( \dfrac{\partial g}{\partial z}\right) ^{2}-2\dfrac{\partial ^{2}f}{\partial z\partial x}\dfrac{\partial g}{\partial x}\dfrac{\partial g}{\partial z} +\lambda \left( \dfrac{\partial g}{\partial x}\right) ^{3}\dfrac{\partial ^{2}\varphi }{\partial z^{2}}+\dfrac{\partial ^{2}f}{\partial z^{2}}\left( \dfrac{\partial g}{\partial x}\right) ^{2}\right] \\ & = & \frac{1}{\left( \dfrac{\partial g}{\partial x}\right) ^{2}}\left[ \dfrac{ \partial ^{2}f}{\partial x^{2}}\left( \dfrac{\partial g}{\partial z}\right) ^{2}-2\dfrac{\partial ^{2}f}{\partial z\partial x}\dfrac{\partial g}{ \partial x}\dfrac{\partial g}{\partial z}+2p\gamma \alpha -l\gamma ^{2}-v\alpha ^{2}+\dfrac{\partial ^{2}f}{\partial z^{2}}\left( \dfrac{ \partial g}{\partial x}\right) ^{2}\right] \\ & = & \frac{k\gamma ^{2}-2o\gamma \alpha +2p\gamma \alpha -l\gamma ^{2}-v\alpha ^{2}+u\alpha ^{2}}{\alpha ^{2}} \\ &=&\frac{\gamma ^{2}\left( k-l\right) -2\gamma \alpha \left( o-p\right) +\alpha ^{2}\left( u-v\right) }{\alpha ^{2}} \\ & = & \frac{\gamma ^{2}a-2\gamma \alpha c+\alpha ^{2}j}{\alpha ^{2}} \end{eqnarray*} En tanto que \begin{eqnarray*} \left\vert \overline{H}_{3}\left( \overline{v_{0}}\right) \right\vert & = & \left\vert \begin{array}{ccc} 0 & \dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) & \dfrac{ \partial g}{\partial y}\left( \overline{v_{0}}\right) \\ & & \\ \dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial x^{2}}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial x\partial y}\left( \overline{v_{0}}\right) \\ & & \\ \dfrac{\partial g}{\partial y}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial x\partial y}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial y^{2}}\left( \overline{v_{0}}\right) \end{array} \right\vert =\left\vert \begin{array}{ccc} 0 & \alpha & \beta \\ \alpha & a & b \\ \beta & b & d \end{array} \right\vert \\ \\ &=& -d\alpha ^{2}+2b\alpha \beta -a\beta ^{2} \\ & = & -\left( d\alpha ^{2}-2b\alpha \beta +a\beta ^{2}\right) \end{eqnarray*} Por tanto, \begin{equation*} \dfrac{\partial ^{2}F}{\partial y^{2}}=\frac{\alpha ^{2}d+\beta ^{2}a-2\beta \alpha b}{\alpha ^{2}}=-\frac{1}{\left( \dfrac{\partial g}{\partial x} \right) ^{2}}H_{3}\left( \overline{v_{0}}\right) \end{equation*} y queda probada la afirmación (b).

Finalmente \begin{eqnarray*} \left\vert \overline{H}_{4}\left( \overline{v_{0}}\right) \right\vert & = & \left\vert \begin{array}{cccc} 0 & \dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) & \dfrac{ \partial g}{\partial y}\left( \overline{v_{0}}\right) & \dfrac{\partial g}{ \partial z}\left( \overline{v_{0}}\right) \\ & & & \\ \dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial x^{2}}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial y\partial x}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial z\partial x}\left( \overline{v_{0}}\right) \\ & & & \\ \dfrac{\partial g}{\partial y}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial x\partial y}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial y^{2}}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial z\partial y}\left( \overline{v_{0}}\right) \\ & & & \\ \dfrac{\partial g}{\partial z}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial x\partial z}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial y\partial z}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial z^{2}}\left( \overline{v_{0}}\right) \end{array} \right\vert = \left\vert \begin{array}{cccc} 0 & \alpha & \beta & \gamma \\ \alpha & a & b & c \\ \beta & b & d & e \\ \gamma & c & e & j \end{array} \right\vert \\ \\ & = & \alpha ^{2}e^{2}+b^{2}\gamma ^{2}+c^{2}\beta ^{2}-ad\gamma ^{2}-aj\beta ^{2}-dj\alpha ^{2} +2a\beta \gamma e-2b\alpha \gamma e-2c\alpha \beta e-2bc\beta \gamma +2cd\alpha \gamma +2bj\alpha \beta \end{eqnarray*} Por tanto \begin{eqnarray*} -\alpha ^{2}\left\vert \overline{H}_{4}\left( \left( \overline{v_{0}}\right) \right) \right\vert &=&dj\alpha ^{4}-\alpha ^{4}e^{2}-b^{2}\alpha ^{2}\gamma ^{2}-c^{2}\alpha ^{2}\beta ^{2}+ad\alpha ^{2}\gamma ^{2}+aj\alpha ^{2}\beta ^{2}+2b\alpha ^{3}\gamma e +2c\alpha ^{3}\beta e-2cd\alpha ^{3}\gamma -2bj\alpha ^{3}\beta -2a\alpha ^{2}\beta \gamma e+2bc\alpha ^{2}\beta \gamma \end{eqnarray*} En tanto que \begin{eqnarray*} \left\vert H\left( y_{0},z_{0}\right) \right\vert & = & \left\vert \begin{array}{cc} \dfrac{\partial ^{2}F}{\partial y^{2}} & \dfrac{\partial ^{2}F}{\partial y\partial z} \\ & \\ \dfrac{\partial ^{2}F}{\partial y\partial z} & \dfrac{\partial ^{2}F}{ \partial z^{2}} \end{array} \right\vert =\dfrac{\partial ^{2}F}{\partial y^{2}}\dfrac{\partial ^{2}F}{ \partial z^{2}}-\left( \dfrac{\partial ^{2}F}{\partial y\partial z}\right) ^{2} \\ \\ & = & \frac{\alpha ^{2}d+\beta ^{2}a-2\beta \alpha b}{\alpha ^{2}}\times \frac{ \gamma ^{2}a-2\gamma \alpha c+\alpha ^{2}j}{\alpha ^{2}}-\left( \frac{\beta \gamma a-\beta \alpha c-\alpha \gamma b+\alpha ^{2}e}{\alpha ^{2}}\right) ^{2} \\ & = & \frac{1}{\alpha ^{4}}\left[ dj\alpha ^{4}-\alpha ^{4}e^{2}-b^{2}\alpha ^{2}\gamma ^{2}-c^{2}\alpha ^{2}\beta ^{2}+ad\alpha ^{2}\gamma ^{2}+aj\alpha ^{2}\beta ^{2}+2b\alpha ^{3}\gamma e +2c\alpha ^{3}\beta e-2cd\alpha ^{3}\gamma -2bj\alpha ^{3}\beta -2a\alpha ^{2}\beta \gamma e+2bc\alpha ^{2}\beta \gamma \right] \end{eqnarray*} De donde, \begin{eqnarray*} t\alpha ^{4}\left\vert H\left( \left( y_{0},z_{0}\right) \right) \right\vert & = & dj\alpha ^{4}-\alpha ^{4}e^{2}-b^{2}\alpha ^{2}\gamma ^{2}-c^{2}\alpha ^{2}\beta ^{2}+ad\alpha ^{2}\gamma ^{2}+aj\alpha ^{2}\beta ^{2}+2b\alpha ^{3}\gamma e +2c\alpha ^{3}\beta e-2cd\alpha ^{3}\gamma -2bj\alpha ^{3}\beta -2a\alpha ^{2}\beta \gamma e+2bc\alpha ^{2}\beta \gamma \end{eqnarray*} Así, \begin{eqnarray*} \alpha ^{4}\left\vert H\left( y_{0},z_{0}\right) \right\vert & = & -\alpha ^{2}\left\vert \overline{H}_{4}\left( \overline{v_{0}}\right) \right\vert \\ \left\vert H\left( y_{0},z_{0}\right) \right\vert & = & -\frac{1}{\alpha ^{2}} \left\vert \overline{H}_{4}\left( \overline{v_{0}}\right) \right\vert =- \frac{1}{\left( \dfrac{\partial g}{\partial x}\left( \overline{v_{0}}\right) \right) ^{2}}\left\vert \overline{H}_{4}\left( \overline{v_{0}}\right) \right\vert \end{eqnarray*} y queda probada la afirmación (c).