Máximos y Mínimos de Funciones de Varias Variables Angel Carrillo Hoyo, Elena de Oteyza de Oteyza\(^2\), Carlos Hernández Garciadiego\(^1\), Emma Lam Osnaya\(^2\) \(^1\) Instituto de Matemáticas, UNAM; \(^2\) Facultad de Ciencias, UNAM

Ejemplo 15

Solución:

La ecuación de la tangente a la elipse \(\dfrac{x^{2}}{16}+\dfrac{y^{2}}{9 }=1\) en el punto \(P\left( x_{1},y_{1}\right) \) es \begin{equation*} \dfrac{x_{1}x}{16}+\dfrac{y_{1}y}{9}=1. \end{equation*} Encontramos la intersección de esta recta tangente con los ejes coordenados.

• Si \(x=0,\) entonces \begin{eqnarray*} \dfrac{y_{1}y}{9} & = & 1 \\ y & = & \dfrac{9}{y_{1}}. \end{eqnarray*}
• Si \(y=0,\) entonces \begin{eqnarray*} \dfrac{x_{1}x}{16} & = & 1 \\ x & = & \dfrac{16}{x_{1}}. \end{eqnarray*}

El área del triángulo que buscamos es \begin{equation*} A=\dfrac{1}{2}bh \end{equation*} y en este caso la base y la altura del triángulo corresponden a los catetos, entonces \begin{equation*} A=\dfrac{1}{2}\left( \dfrac{16}{\left\vert x_{1}\right\vert }\right) \dfrac{9 }{\left\vert y_{1}\right\vert }=\dfrac{72}{\left\vert x_{1}y_{1}\right\vert } . \end{equation*} Así consideramos la función \begin{equation*} f\left( x,y\right) =\dfrac{72}{\left\vert xy\right\vert } \end{equation*} y la restricción \begin{equation*} \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}-1=0. \end{equation*}

1. Supongamos que \(x\) y \(y\) tienen el mismo signo, en ese caso: \begin{equation*} f\left( x,y\right) =\dfrac{72}{xy}. \end{equation*} Calculamos los gradientes de \(f\) y \(g\): \begin{eqnarray*} \nabla f\left( x,y,z\right) & = & \left( \dfrac{\partial f}{\partial x},\dfrac{ \partial f}{\partial y}\right) =\left( -\dfrac{72}{x^{2}y},-\dfrac{72}{xy^{2} }\right) \\ \nabla g\left( x,y,z\right) & = & \left( \dfrac{\partial g}{\partial x},\dfrac{ \partial g}{\partial y}\right) =\left( \dfrac{x}{8},\dfrac{2}{9}y\right) . \end{eqnarray*} De acuerdo con el método de Lagrange, debemos resolver el sistema de ecuaciones \begin{eqnarray*} \nabla f\left( x,y\right) & = & \lambda \nabla g\left( x,y\right) \\ \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9} & = & 1, \end{eqnarray*} es decir \begin{eqnarray} -\dfrac{72}{x^{2}y} & = & \dfrac{1}{8}x\lambda \label{15.0}\tag{1} \\ -\dfrac{72}{xy^{2}} & = & \dfrac{2}{9}y\lambda \notag \\ \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9} & = & 1. \notag \end{eqnarray} Considerando las dos primeras ecuaciones del sistema anterior, tenemos \begin{eqnarray*} -72 & = & \dfrac{x^{3}y}{8}\lambda \\ -72 & = & \dfrac{2}{9}xy^{3}\lambda , \end{eqnarray*} de donde \begin{eqnarray*} \dfrac{x^{3}y}{8}\lambda & = & \dfrac{2}{9}xy^{3}\lambda \\ \dfrac{x^{3}y}{8}\lambda -\dfrac{2}{9}xy^{3}\lambda & = & 0 \\ xy\lambda \left( \dfrac{x^{2}}{8}-\dfrac{2}{9}y^{2}\right) & = & 0. \end{eqnarray*} Como \(x,y,\lambda \neq 0,\) entonces \begin{eqnarray*} \dfrac{x^{2}}{8}-\dfrac{2}{9}y^{2} & = & 0 \\ \left( \dfrac{x}{\sqrt{8}}-\dfrac{\sqrt{2}}{3}y\right) \left( \dfrac{x}{ \sqrt{8}}+\dfrac{\sqrt{2}}{3}y\right) & = & 0, \end{eqnarray*} de donde \begin{equation*} \dfrac{x}{\sqrt{8}}-\dfrac{\sqrt{2}}{3}y=0 \quad \quad \quad \text{o} \quad \quad \quad \dfrac{x}{\sqrt{8}}+\dfrac{\sqrt{2}}{3}y=0. \end{equation*}
   • Si \(\dfrac{x}{\sqrt{8}}-\dfrac{\sqrt{2}}{3}y=0,\) entonces \begin{eqnarray*} \dfrac{x}{\sqrt{8}}-\dfrac{\sqrt{2}}{3}y & = & 0 \\ y & = & \dfrac{3x}{\sqrt{8}\sqrt{2}} \\ y & = & \dfrac{3}{4}x. \end{eqnarray*}
   • Si \(\dfrac{x}{\sqrt{8}}+\dfrac{\sqrt{2}}{3}y=0,\) entonces \begin{eqnarray*} \dfrac{x}{\sqrt{8}}+\dfrac{\sqrt{2}}{3}y & = & 0 \\ y & = & -\dfrac{3x}{\sqrt{8}\sqrt{2}} \\ y & = & -\dfrac{3}{4}x. \end{eqnarray*}
   Ahora sustituimos los valores obtenidos de \(y\) en la restricción.
   • Si \(y=\dfrac{3}{4}x,\) entonces \begin{eqnarray* \dfrac{x^{2}}{16}+\dfrac{\left( \dfrac{3}{4}x\right) ^{2}}{9} & = & 1 \\ \dfrac{x^{2}}{16}+\dfrac{x^{2}}{16} & = & 1 \\ \dfrac{x^{2}}{8} & = & 1 \\ x^{2} & = & 8 \\ \left\vert x\right\vert & = & \sqrt{8} \\ \left\vert x\right\vert & = & 2\sqrt{2}, \end{eqnarray*} así \begin{equation*} x=2\sqrt{2} \quad \quad \quad \text{o } \quad \quad \quad x=-2\sqrt{2}. \end{equation*}
     • Si \(x=2\sqrt{2},\) entonces \begin{equation*} y=\dfrac{3}{4}\left( 2\sqrt{2}\right) =\dfrac{3}{2}\sqrt{2} \end{equation*} y \(\lambda \) es \begin{eqnarray*} \lambda & = & -\dfrac{72\left( 8\right) }{x^{3}y} \\ & = & -\dfrac{72\left( 8\right) }{\left( 2\sqrt{2}\right) ^{3}\dfrac{3}{2}\sqrt{ 2}} \\ & = & -\dfrac{72\left( 8\right) }{8\left( 2\sqrt{2}\right) \dfrac{3}{2}\sqrt{2}} \\ & = & -12. \end{eqnarray*}
     • Si \(x=-2\sqrt{2},\) entonces \begin{equation*} y=\dfrac{3}{4}\left( -2\sqrt{2}\right) =-\dfrac{3}{2}\sqrt{2} \end{equation*} y \(\lambda \) es \begin{eqnarray*} \lambda & = & -\dfrac{72\left( 8\right) }{\left( -2\sqrt{2}\right) ^{3}\left( - \dfrac{3}{2}\sqrt{2}\right) } \\ & = & -\dfrac{72\left( 8\right) }{8\left( 2\sqrt{2}\right) \dfrac{3}{2}\sqrt{2}} \\ & = & -12. \end{eqnarray*}
   • Si \(y=-\dfrac{3}{4}x,\) entonces \begin{eqnarray*} \dfrac{x^{2}}{16}+\dfrac{\left( -\dfrac{3}{4}x\right) ^{2}}{9} & = & 1 \\ \dfrac{x^{2}}{16}+\dfrac{x^{2}}{16} & = & 1 \\ \dfrac{x^{2}}{8} & = & 1 \\ x^{2} & = & 8 \\ \left\vert x\right\vert & = & \sqrt{8} \\ \left\vert x\right\vert & = & 2\sqrt{2}, \end{eqnarray*} así \begin{equation*} x=2\sqrt{2} \quad \quad \quad \text{ o } \quad \quad \quad x=-2\sqrt{2}. \end{equation*}
     • Si \(x=2\sqrt{2},\) entonces \begin{equation*} y=-\dfrac{3}{4}\left( 2\sqrt{2}\right) =-\dfrac{3}{2}\sqrt{2}, \end{equation*} como \(x\) y \(y\) deben tener el mismo signo, desechamos este caso.
     • Si \(x=-2\sqrt{2},\) entonces \begin{equation*} y=-\dfrac{3}{4}\left( -2\sqrt{2}\right) =\dfrac{3}{2}\sqrt{2}, \end{equation*} como \(x\) y \(y\) deben tener el mismo signo, también desechamos este caso.

   En resumen las soluciones del sistema (\ref{15.0}) son \begin{eqnarray*} x & = & 2\sqrt{2}, \quad \quad y=\dfrac{3}{2}\sqrt{2}, \quad \quad \lambda =-12 \\ x & = & -2\sqrt{2}, \quad \quad y=-\dfrac{3}{2}\sqrt{2}, \quad \quad \lambda =-12. \end{eqnarray*} Definimos la función auxiliar \begin{eqnarray*} h\left( x,y\right) & = & f\left( x,y\right) -\lambda g\left( x,y\right) \\ & = & \dfrac{72}{xy}-\lambda \left( \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}-1\right) . \end{eqnarray*} Calculamos las derivadas de \(h\) de orden \(1\). \begin{eqnarray*} \dfrac{\partial h }{\partial x}\left( x,y\right) & = & -\dfrac{72}{x^{2}y} -\lambda \dfrac{x}{8} \\ \dfrac{\partial h }{\partial y}\left( x,y\right) & = & -\dfrac{72}{xy^{2}} -\lambda \dfrac{2}{9}y. \end{eqnarray*} Las derivadas parciales de \(h\) de orden \(2\) son: \begin{eqnarray*} \dfrac{\partial ^{2}h }{\partial x^{2}}\left( x,y\right) & = & \dfrac{72\left( 2\right) }{x^{3}y}-\dfrac{1}{8}\lambda =\dfrac{144}{x^{3}y}-\dfrac{1}{8} \lambda \\ \dfrac{\partial ^{2}h }{\partial y^{2}}\left( x,y\right) & = & \dfrac{72\left( 2\right) }{xy^{3}}-\dfrac{2}{9}\lambda =\dfrac{144}{xy^{3}}-\dfrac{2}{9} \lambda \\ \dfrac{\partial ^{2}h }{\partial y\partial x}\left( x,y\right) & = & \dfrac{72}{ x^{2}y^{2}}=\dfrac{\partial ^{2}h\left( x,y\right) }{\partial x\partial y} \end{eqnarray*} El determinante hessiano limitado es \begin{eqnarray*} \left\vert \overline{H}\left( x,y\right) \right\vert & = & \left\vert \begin{array}{ccc} 0 & \dfrac{\partial g}{\partial x}\left( x,y\right) & \dfrac{\partial g}{ \partial y}\left( x,y\right) \\ & & \\ \dfrac{\partial g}{\partial x}\left( x,y\right) & \dfrac{\partial ^{2}h}{ \partial x^{2}}\left( x,y\right) & \dfrac{\partial ^{2}h}{\partial y\partial x}\left( x,y\right) \\ & & \\ \dfrac{\partial g}{\partial y}\left( x,y\right) & \dfrac{\partial ^{2}h}{ \partial x\partial y}\left( x,y\right) & \dfrac{\partial ^{2}h}{\partial y^{2}}\left( x,y\right) \end{array} \right\vert \\ & = & \left\vert \begin{array}{ccc} 0 & \dfrac{x}{8} & \dfrac{2}{9}y \\ & & \\ \dfrac{x}{8} & \dfrac{144}{x^{3}y}-\dfrac{1}{8}\lambda & \dfrac{72}{ x^{2}y^{2}} \\ & & \\ \dfrac{2}{9}y & \dfrac{72}{x^{2}y^{2}} & \dfrac{144}{xy^{3}}-\dfrac{2}{9} \lambda \end{array} \right\vert \end{eqnarray*}

   • Si \(x=2\sqrt{2},\) \(y=\dfrac{3}{2}\sqrt{2\), \(\lambda =-12,\) entonces \begin{eqnarray*} \left\vert \overline{H}\left( 2\sqrt{2},\dfrac{3}{2}\sqrt{2}\right) \right\vert & = & \left\vert \begin{array}{ccc} 0 & \dfrac{\sqrt{2}}{4} & \dfrac{\sqrt{2}}{3} \\ \dfrac{\sqrt{2}}{4} & \dfrac{144}{\left( 2\sqrt{2}\right) ^{3}\left( \dfrac{3 }{2}\sqrt{2}\right) }-\dfrac{1}{8}\left( -12\right) & \dfrac{72}{\left( 2 \sqrt{2}\right) ^{2}\left( \dfrac{3}{2}\sqrt{2}\right) ^{2}} \\ \dfrac{\sqrt{2}}{3} & \dfrac{72}{\left( 2\sqrt{2}\right) ^{2}\left( \dfrac{3 }{2}\sqrt{2}\right) ^{2}} & \dfrac{144}{2\sqrt{2}\left( \dfrac{3}{2}\sqrt{2} \right) ^{3}}-\dfrac{2}{9}\left( -12\right) \end{array} \right\vert \\ && \\ & = & \left\vert \begin{array}{ccc} 0 & \dfrac{\sqrt{2}}{4} & \dfrac{\sqrt{2}}{3} \\ \dfrac{\sqrt{2}}{4} & \dfrac{9}{2} & 2 \\ \dfrac{\sqrt{2}}{3} & 2 & 8 \end{array} \right\vert \\ & = & -\dfrac{4}{3} < 0. \end{eqnarray*}

     Por el inciso 2 del Teorema 1, hay un mínimo en \(x=2\sqrt{2},\) \(y=\dfrac{3}{2}\sqrt{2}\) y su valor es \begin{equation*} f\left( x,y\right) =\dfrac{72}{\left\vert xy\right\vert }=\dfrac{72}{6}=12. \end{equation*}

   • Si \(x=-2\sqrt{2},\) \(y=-\dfrac{3}{2}\sqrt{2},\) \(\lambda =-12,\) entonces \begin{eqnarray*} \left\vert \overline{H}\left( -2\sqrt{2},-\dfrac{3}{2}\sqrt{2}\right) \right\vert & = & \left\vert \begin{array}{ccc} 0 & -\dfrac{\sqrt{2}}{4} & -\dfrac{\sqrt{2}}{3} \\ & & \\ -\dfrac{\sqrt{2}}{4} & \dfrac{144}{\left( -2\sqrt{2}\right) ^{3}\left( - \dfrac{3}{2}\sqrt{2}\right) }-\dfrac{1}{8}\left( -12\right) & \dfrac{72}{ \left( -2\sqrt{2}\right) ^{2}\left( -\dfrac{3}{2}\sqrt{2}\right) ^{2}} \\ & & \\ -\dfrac{\sqrt{2}}{3} & \dfrac{72}{\left( -2\sqrt{2}\right) ^{2}\left( - \dfrac{3}{2}\sqrt{2}\right) ^{2}} & \dfrac{144}{-2\sqrt{2}\left( -\dfrac{3}{2 }\sqrt{2}\right) ^{3}}-\dfrac{2}{9}\left( -12\right) \end{array} \right\vert \\ & & \\ & = & \left\vert \begin{array}{ccc} 0 & -\dfrac{\sqrt{2}}{4} & -\dfrac{\sqrt{2}}{3} \\ & & \\ -\dfrac{\sqrt{2}}{4} & \dfrac{9}{2} & 2 \\ & & \\ -\dfrac{\sqrt{2}}{3} & 2 & 8 \end{array} \right\vert \\ & = & -\dfrac{4}{3} < 0. \end{eqnarray*}

     Por el inciso 2 del Teorema 1, hay un mínimo en \(x=-2\sqrt{2},y=-\dfrac{3}{2}\sqrt{2}\) y su valor es \begin{equation*} f\left( x,y\right) =\dfrac{72}{\left\vert xy\right\vert }=\dfrac{72}{6}=12. \end{equation*}

2. Si \(x\) y \(y\) tienen signos distintos, entonces \begin{equation*} f\left( x,y\right) =-\dfrac{72}{xy}. \end{equation*} Calculamos los gradientes de \(f\) y \(g\): \begin{eqnarray*} \nabla f\left( x,y,z\right) & = & \left( \dfrac{\partial f}{\partial x},\dfrac{ \partial f}{\partial y}\right) =\left( \dfrac{72}{x^{2}y},\dfrac{72}{xy^{2}} \right) \\ \nabla g\left( x,y,z\right) & = & \left( \dfrac{\partial g}{\partial x},\dfrac{ \partial g}{\partial y}\right) =\left( \dfrac{x}{8},\dfrac{2}{9}y\right) . \end{eqnarray*} De acuerdo con el método de Lagrange, debemos resolver el sistema de ecuaciones \begin{eqnarray*} \nabla f\left( x,y\right) & = & \lambda \nabla g\left( x,y\right) \\ \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9} & = & 1. \end{eqnarray*} es decir \begin{eqnarray} \dfrac{72}{x^{2}y} & = & \dfrac{x}{8}\lambda \label{15.1}\tag{2} \\ \dfrac{72}{xy^{2}} & = & \dfrac{2}{9}y\lambda \notag \\ \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9} & = & 1. \notag \end{eqnarray} Considerando las dos primeras ecuaciones del sistema anterior, tenemos \begin{eqnarray*} 72 & = & \dfrac{x^{3}y}{8}\lambda \\ 72 & = & \dfrac{2}{9}xy^{3}\lambda , \end{eqnarray*} de donde \begin{eqnarray*} \dfrac{x^{3}y}{8}\lambda & = & \dfrac{2}{9}xy^{3}\lambda \\ \dfrac{x^{3}y}{8}\lambda -\dfrac{2}{9}xy^{3}\lambda & = & 0 \\ xy\lambda \left( \dfrac{x^{2}}{8}-\dfrac{2}{9}y^{2}\right) & = & 0. \end{eqnarray*} Como \(x,y,\lambda \neq 0,\) entonces \begin{eqnarray*} \dfrac{x^{2}}{8}-\dfrac{2}{9}y^{2} & = & 0 \\ \left( \dfrac{x}{\sqrt{8}}-\dfrac{\sqrt{2}}{3}y\right) \left( \dfrac{x}{ \sqrt{8}}+\dfrac{\sqrt{2}}{3}y\right) & = & 0 \end{eqnarray*} de donde \begin{equation*} \dfrac{x}{\sqrt{8}}-\dfrac{\sqrt{2}}{3}y=0 \quad \quad \quad \text{o } \quad \quad \quad \dfrac{x}{\sqrt{8}}+\dfrac{\sqrt{2}}{3}y=0. \end{equation*}
   • Si \(\dfrac{x}{\sqrt{8}}-\dfrac{\sqrt{2}}{3}y=0,\) entonces \begin{eqnarray*} \dfrac{x}{\sqrt{8}}-\dfrac{\sqrt{2}}{3}y & = & 0 \\ y & = & \dfrac{3x}{\sqrt{8}\sqrt{2}} \\ y & = & \dfrac{3}{4}x. \end{eqnarray*}
   • Si \(\dfrac{x}{\sqrt{8}}+\dfrac{\sqrt{2}}{3}y=0,\) entonces \begin{eqnarray*} \dfrac{x}{\sqrt{8}}+\dfrac{\sqrt{2}}{3}y & = & 0 \\ y & = & -\dfrac{3x}{\sqrt{8}\sqrt{2}} \\ y & = & -\dfrac{3}{4}x. \end{eqnarray*}
   Ahora sustituimos los valores obtenidos de \(y\) en la restricción.
   • Si \(y=\dfrac{3}{4}x,\) entonces \begin{eqnarray*} \dfrac{x^{2}}{16}+\dfrac{\left( \dfrac{3}{4}x\right) ^{2}}{9} & = & 1 \\ \dfrac{x^{2}}{16}+\dfrac{x^{2}}{16} & = & 1 \\ \dfrac{x^{2}}{8} & = & 1 \\ x^{2} & = & 8 \\ \left\vert x\right\vert & = & \sqrt{8} \\ \left\vert x\right\vert & = & 2\sqrt{2} \end{eqnarray*} así \begin{equation*} x=2\sqrt{2} \quad \quad \quad \text{ o } \quad \quad \quad x=-2\sqrt{2}. \end{equation*}
     • Si \(x=2\sqrt{2},\) entonces \begin{equation*} y=\dfrac{3}{4}\left( 2\sqrt{2}\right) =\dfrac{3}{2}\sqrt{2}, \end{equation*} como \(x\) y \(y\) deben tener signos distintos, desechamos este caso.
     • Si \(x=-2\sqrt{2},\) entonces \begin{equation*} y=\dfrac{3}{4}\left( -2\sqrt{2}\right) =-\dfrac{3}{2}\sqrt{2}, \end{equation*} como \(x\) y \(y\) deben tener signos distintos, desechamos también este caso.
   • Si \(y=-\dfrac{3}{4}x,\) entonces \begin{eqnarray*} \dfrac{x^{2}}{16}+\dfrac{\left( -\dfrac{3}{4}x\right) ^{2}}{9} & = & 1 \\ \dfrac{x^{2}}{16}+\dfrac{x^{2}}{16} & = & 1 \\ \dfrac{x^{2}}{8} & = & 1 \\ x^{2} & = & 8 \\ \left\vert x\right\vert & = & \sqrt{8} \\ \left\vert x\right\vert & = & 2\sqrt{2} \end{eqnarray*} así \begin{equation*} x=2\sqrt{2} \quad \quad \quad \text{ o } \quad \quad \quad x=-2\sqrt{2}. \end{equation*}
     • Si \(x=2\sqrt{2},\) entonces \begin{equation*} y=-\dfrac{3}{4}\left( 2\sqrt{2}\right) =-\dfrac{3}{2}\sqrt{2} \end{equation*} y \(\lambda \) es \begin{eqnarray*} \lambda & = & \dfrac{72\left( 8\right) }{x^{3}y} \\ & = & \dfrac{72\left( 8\right) }{\left( 2\sqrt{2}\right) ^{3}\left( -\dfrac{3}{2 }\sqrt{2}\right) } \\ & = & -\dfrac{72\left( 8\right) }{8\left( 2\sqrt{2}\right) \dfrac{3}{2}\sqrt{2}} \\ & = & -12. \end{eqnarray*}
     • Si \(x=-2\sqrt{2},\) entonces \begin{equation*} y=-\dfrac{3}{4}\left( -2\sqrt{2}\right) =\dfrac{3}{2}\sqrt{2} \end{equation*} y \(\lambda \) es \begin{eqnarray*} \lambda & = & \dfrac{72\left( 8\right) }{x^{3}y} \\ & = & \dfrac{72\left( 8\right) }{\left( -2\sqrt{2}\right) ^{3}\dfrac{3}{2}\sqrt{ 2}} \\ & = & -\dfrac{72\left( 8\right) }{8\left( 2\sqrt{2}\right) \dfrac{3}{2}\sqrt{2}} \\ & = & -12. \end{eqnarray*}
   En resumen, las soluciones del sistema (\ref{15.1}) son \begin{eqnarray*} x & = & -2\sqrt{2}, \quad \quad y=\dfrac{3}{2}\sqrt{2}, \quad \quad \lambda =-12 \\ x & = & 2\sqrt{2}, \quad \quad y=-\dfrac{3}{2}\sqrt{2}, \quad \quad \lambda =-12. \end{eqnarray*} Definimos la función auxiliar \begin{eqnarray*} h\left( x,y\right) & = & f\left( x,y\right) -\lambda g\left( x,y\right) \\ & = & -\dfrac{72}{xy}-\lambda \left( \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9} -1\right) . \end{eqnarray*} Calculamos las derivadas de \(h\) de orden \(1\). \begin{eqnarray*} \dfrac{\partial h }{\partial x}\left( x,y\right) & = & \dfrac{72}{x^{2}y} -\lambda \dfrac{x}{8} \\ \dfrac{\partial h }{\partial y}\left( x,y\right) & = & \dfrac{72}{xy^{2}} -\lambda \dfrac{2}{9}y. \end{eqnarray*} Las derivadas parciales de \(h\) de orden \(2\) son: \begin{eqnarray*} \dfrac{\partial ^{2}h }{\partial x^{2}}\left( x,y\right) & = & -\dfrac{72\left( 2\right) }{x^{3}y}-\dfrac{1}{8}\lambda =-\dfrac{144}{x^{3}y}-\dfrac{1}{8} \lambda \\ \dfrac{\partial ^{2}h }{\partial y^{2}}\left( x,y\right) & = & -\dfrac{72\left( 2\right) }{xy^{3}}-\dfrac{2}{9}\lambda =-\dfrac{144}{xy^{3}}-\dfrac{2}{9} \lambda \\ \dfrac{\partial ^{2}h }{\partial y\partial x}\left( x,y\right) & = & -\dfrac{72 }{x^{2}y^{2}}=\dfrac{\partial ^{2}h\left( x,y\right) }{\partial x\partial y}. \end{eqnarray*} El determinante hessiano limitado es \begin{eqnarray*} \left\vert \overline{H}\left( x,y\right) \right\vert & = & \left\vert \begin{array}{ccc} 0 & \dfrac{\partial g}{\partial x}\left( x,y\right) & \dfrac{\partial g}{ \partial y}\left( x,y\right) \\ & & \\ \dfrac{\partial g}{\partial x}\left( x,y\right) & \dfrac{\partial ^{2}h}{ \partial x^{2}}\left( x,y\right) & \dfrac{\partial ^{2}h}{\partial y\partial x}\left( x,y\right) \\ & & \\ \dfrac{\partial g}{\partial y}\left( x,y\right) & \dfrac{\partial ^{2}h}{ \partial x\partial y}\left( x,y\right) & \dfrac{\partial ^{2}h}{\partial y^{2}}\left( x,y\right) \end{array} \right\vert \\ \\ & = & \left\vert \begin{array}{ccc} 0 & \dfrac{x}{8} & \dfrac{2}{9}y \\ & & \\ \dfrac{x}{8} & -\dfrac{144}{x^{3}y}-\dfrac{1}{8}\lambda & -\dfrac{72}{ x^{2}y^{2}} \\ & & \\ \dfrac{2}{9}y & -\dfrac{72}{x^{2}y^{2}} & -\dfrac{144}{xy^{3}}-\dfrac{2}{9} \lambda \end{array} \right\vert \end{eqnarray*}
   • Si \(x=-2\sqrt{2},\) \(y=\dfrac{3}{2}\sqrt{2},\) \(\lambda =12,\) entonces: \begin{eqnarray*} \left\vert \overline{H}\left( -2\sqrt{2},\dfrac{3}{2}\sqrt{2}\right) \right\vert & = & \left\vert \begin{array}{ccc} 0 & -\dfrac{\sqrt{2}}{4} & \dfrac{\sqrt{2}}{3} \\ & & \\ -\dfrac{\sqrt{2}}{4} & -\dfrac{144}{\left( -2\sqrt{2}\right) ^{3}\left( \dfrac{3}{2}\sqrt{2}\right) }-\dfrac{1}{8}\left( -12\right) & -\dfrac{72}{ \left( -2\sqrt{2}\right) ^{2}\left( \dfrac{3}{2}\sqrt{2}\right) ^{2}} \\ & & \\ \dfrac{\sqrt{2}}{3} & -\dfrac{72}{\left( -2\sqrt{2}\right) ^{2}\left( \dfrac{ 3}{2}\sqrt{2}\right) ^{2}} & -\dfrac{144}{-2\sqrt{2}\left( \dfrac{3}{2}\sqrt{ 2}\right) ^{3}}-\dfrac{2}{9}\left( -12\right) \end{array} \right\vert \\ & & \\ & = & \left\vert \begin{array}{ccc} 0 & -\dfrac{\sqrt{2}}{4} & \dfrac{\sqrt{2}}{3} \\ -\dfrac{\sqrt{2}}{4} & \dfrac{9}{2} & -2 \\ \dfrac{\sqrt{2}}{3} & -2 & 8 \end{array} \right\vert \\ & = & -\dfrac{4}{3} < 0. \end{eqnarray*}

     Por el inciso 2 del Teorema 1, hay un mínimo en \(x=-2\sqrt{2}, y=\dfrac{3}{2}\sqrt{2}\) y su valor es \begin{equation*} f\left( x,y\right) =\dfrac{72}{\left\vert xy\right\vert }=\dfrac{72}{6}=12. \end{equation*}

   • Si \(x=2\sqrt{2},\) \(y=-\dfrac{3}{2}\sqrt{2},\) \(\lambda =-12.\) \begin{eqnarray*} \left\vert \overline{H}\left( 2\sqrt{2},-\dfrac{3}{2}\sqrt{2}\right) \right\vert & = & \left\vert \begin{array}{ccc} 0 & \dfrac{\sqrt{2}}{4} & -\dfrac{\sqrt{2}}{3} \\ & & \\ \dfrac{\sqrt{2}}{4} & -\dfrac{144}{\left( 2\sqrt{2}\right) ^{3}\left( - \dfrac{3}{2}\sqrt{2}\right) }-\dfrac{1}{8}\left( -12\right) & -\dfrac{72}{ \left( 2\sqrt{2}\right) ^{2}\left( -\dfrac{3}{2}\sqrt{2}\right) ^{2}} \\ & & \\ -\dfrac{\sqrt{2}}{3} & -\dfrac{72}{\left( 2\sqrt{2}\right) ^{2}\left( - \dfrac{3}{2}\sqrt{2}\right) ^{2}} & -\dfrac{144}{2\sqrt{2}\left( -\dfrac{3}{2 }\sqrt{2}\right) ^{3}}-\dfrac{2}{9}\left( -12\right) \end{array} \right\vert \\ && \\ & = & \left\vert \begin{array}{ccc} 0 & \dfrac{\sqrt{2}}{4} & -\dfrac{\sqrt{2}}{3} \\ \dfrac{\sqrt{2}}{4} & \dfrac{9}{2} & -2 \\ -\dfrac{\sqrt{2}}{3} & -2 & 8 \end{array} \right\vert \\ & = & -\dfrac{4}{3} < 0. \end{eqnarray*}

     Por el inciso 2 del Teorema 1, hay un mínimo en \(x=2\sqrt{2},y=-\dfrac{3}{2}\sqrt{2}\) y su valor es \begin{equation*} f\left( x,y\right) =\dfrac{72}{\left\vert xy\right\vert }=\dfrac{72}{6}=12. \end{equation*}