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 6

Solución:

Resolvemos el problema usando el método de Lagrange.

Así, buscamos los valores \(a,\) \(b,\) \(c\) que satisfagan: \begin{equation*} abc=64. \end{equation*} y \begin{equation*} f\left( a,b,c\right) =\dfrac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \end{equation*} tenga un máximo. Definimos la función \begin{equation*} g\left( a,b,c\right) =abc-64 \end{equation*} con lo cual la restricción se escribe como \begin{equation*} g\left( a,b,c\right) =0. \end{equation*} Escribimos \(f\) como: \begin{equation*} f\left( a,b,c\right) =\dfrac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{ 3abc}{ab+ac+bc}. \end{equation*} Calculamos los gradientes de \(f\) y \(g\): \begin{eqnarray*} \nabla f\left( a,b,c\right) & = & \left( \dfrac{\partial f}{\partial a},\dfrac{ \partial f}{\partial b},\dfrac{\partial f}{\partial c}\right) =\left( \frac{ 3b^{2}c^{2}}{\left( ab+ac+bc\right) ^{2}},\frac{3a^{2}c^{2}}{\left( ab+ac+bc\right) ^{2}},\frac{3a^{2}b^{2}}{\left( ab+ac+bc\right) ^{2}}\right) \\ \nabla g\left( a,b,c\right) & = & \left( \dfrac{\partial g}{\partial a},\dfrac{ \partial g}{\partial b},\dfrac{\partial g}{\partial c}\right) =\left( bc,ac,ab\right) . \end{eqnarray*} De acuerdo con el método de Lagrange, debemos resolver el sistema de ecuaciones \begin{eqnarray*} \nabla f\left( a,b,c\right) & = & \lambda \nabla g\left( a,b,c\right) \\ abc & = & 64, \end{eqnarray*} es decir \begin{eqnarray*} \frac{3b^{2}c^{2}}{\left( ab+ac+bc\right) ^{2}} & = & \lambda bc \\ \frac{3a^{2}c^{2}}{\left( ab+ac+bc\right) ^{2}} & = & \lambda ac \\ \frac{3a^{2}b^{2}}{\left( ab+ac+bc\right) ^{2}} & = & \lambda ab \\ abc & = & 64. \end{eqnarray*} De la cuarta ecuación tenemos que \(a,\) \(b,\) \(c\neq 0.\) De donde el sistema se escribe como \begin{eqnarray} \frac{3bc}{\left( ab+ac+bc\right) ^{2}} & = & \lambda \notag \\ \frac{3ac}{\left( ab+ac+bc\right) ^{2}} & = & \lambda \label{Prob6}\tag{1} \\ \frac{3ab}{\left( ab+ac+bc\right) ^{2}} & = & \lambda \notag \\ abc & = & 64. \notag \end{eqnarray} De las tres primeras ecuaciones de (\ref{Prob6}) tenemos \begin{equation*} bc=ac=ab, \end{equation*} de donde \begin{equation*} a=b=c. \end{equation*} Sustituyendo en la cuarta ecuación de (\ref{Prob6}) obtenemos \begin{equation*} a=b=c=4 \end{equation*} y sustituyendo en la primera ecuación de (\ref{Prob6}) tenemos: \begin{eqnarray*} \lambda & = & \dfrac{3bc}{\left( ab+ac+bc\right) ^{2}} \\ & = & \dfrac{3\left( 16\right) }{\left( 3\left( 16\right) \right) ^{2}} \\ & = & \dfrac{1}{3\left( 16\right) }=\dfrac{1}{48}. \end{eqnarray*} Definimos la función auxiliar \begin{eqnarray*} h\left( a,b,c\right) & = & f\left( a,b,c\right) -\lambda g\left( a,b,c\right) \\ & = & \frac{3abc}{ab+ac+bc}-\lambda \left( abc-64\right) , \end{eqnarray*} así \begin{eqnarray*} \dfrac{\partial h }{\partial a}\left( a,b,c\right) & = & \dfrac{3bc\left( ab+ac+bc\right) -3abc\left( b+c\right) }{\left( ab+ac+bc\right) ^{2}} -\lambda bc \\ & = & bc\dfrac{3\left( ab+ac+bc\right) -3a\left( b+c\right) }{\left( ab+ac+bc\right) ^{2}}-\lambda bc \\ & = & bc\dfrac{3a\left( b+c\right) +3bc-3a\left( b+c\right) }{\left( ab+ac+bc\right) ^{2}}-\lambda bc \\ & = & bc\dfrac{3bc}{\left( ab+ac+bc\right) ^{2}}-\lambda bc \\ & = & \dfrac{3b^{2}c^{2}}{\left( ab+ac+bc\right) ^{2}}-\lambda bc \end{eqnarray*} de la misma manera \begin{eqnarray*} \dfrac{\partial h }{\partial b}\left( a,b,c\right) & = & \dfrac{3a^{2}c^{2}}{ \left( ab+ac+bc\right)^{2}}-\lambda ac \\ \dfrac{\partial h }{\partial c}\left( a,b,c\right) & = & \dfrac{3a^{2}b^{2}}{ \left( ab+ac+bc\right)^{2}}-\lambda ab. \end{eqnarray*} Las derivadas parciales de \(h\) de orden \(2\) son: \begin{eqnarray*} \dfrac{\partial ^{2}h }{\partial a^{2}}\left( a,b,c\right) & = & \dfrac{\left(ab+ac+bc\right) ^{2}\left( 0\right) -3b^{2}c^{2}\left( 2\left( ab+ac+bc\right) \left( b+c\right) \right) }{\left( ab+ac+bc\right) ^{4}} \\ & = & \dfrac{-3b^{2}c^{2}\left( 2\left( b+c\right) \right) }{\left( ab+ac+bc\right) ^{3}} \\ & = & \dfrac{-6b^{2}c^{2}\left( b+c\right) }{\left( ab+ac+bc\right) ^{3}}. \end{eqnarray*} De la misma manera \begin{eqnarray*} \dfrac{\partial ^{2}h}{\partial b^{2}}\left( a,b,c\right) & = & \frac{ -6a^{2}c^{2}\left( a+c\right) }{\left( ab+ac+bc\right) ^{3}} \\ \dfrac{\partial ^{2}h }{\partial c^{2}}\left( a,b,c\right) & = & \frac{ -6a^{2}b^{2}\left( a+b\right) }{\left( ab+ac+bc\right) ^{3}} \end{eqnarray*} y \begin{eqnarray*} \dfrac{\partial ^{2}h }{\partial b\partial a}\left( a,b,c\right) & = & \dfrac{ \left( ab+ac+bc\right) ^{2}\left( 6bc^{2}\right) -3b^{2}c^{2}\left( 2\left( ab+ac+bc\right) \left( a+c\right) \right) }{\left( ab+ac+bc\right) ^{4}} -\lambda c \\ & = & \dfrac{\left( ab+ac+bc\right) \left( 6bc^{2}\right) -3b^{2}c^{2}\left( 2\left( a+c\right) \right) }{\left( ab+ac+bc\right) ^{3}}-\lambda c \\ & = & \dfrac{\left( ab+ac+bc\right) \left( 6bc^{2}\right) -3b^{2}c^{2}\left( 2\left( a+c\right) \right) }{\left( ab+ac+bc\right) ^{3}}-\lambda c \\ & = & \dfrac{\left( ac+b\left( a+c\right) \right) \left( 6bc^{2}\right) -3b^{2}c^{2}\left( 2\left( a+c\right) \right) }{\left( ab+ac+bc\right) ^{3}} -\lambda c \\ & = & \dfrac{6b^{2}c^{2}\left( a+c\right) +6abc^{3}-6b^{2}c^{2}\left( \left( a+c\right) \right) }{\left( ab+ac+bc\right) ^{3}}-\lambda c \\ & = & \dfrac{6abc^{3}}{\left( ab+ac+bc\right) ^{3}}-\lambda c \\ & = & \dfrac{\partial ^{2}h\left( a,b,c\right) }{\partial a\partial b} \end{eqnarray*} de la misma manera \begin{eqnarray*} \dfrac{\partial ^{2}h }{\partial c\partial a}\left( a,b,c\right) & = & \dfrac{6ab^{3}c}{\left( ab+ac+bc\right) ^{3}}-\lambda b=\dfrac{\partial ^{2}h\left( a,b,c\right) }{\partial c\partial a} \\ \dfrac{\partial ^{2}h }{\partial b\partial c}\left( a,b,c\right) & = & \dfrac{6a^{3}bc}{\left( ab+ac+bc\right) ^{3}}-\lambda a=\dfrac{\partial ^{2}h\left( a,b,c\right) }{\partial c\partial b}. \end{eqnarray*} Evaluamos las derivadas parciales de \(h,\) de segundo orden en \(a=4,\) \(b=4,\) \( c=4,\) \(\lambda =\dfrac{1}{48}\) \begin{eqnarray*} \dfrac{\partial h^{2} }{\partial a^{2}}\left( 4,4,4\right) & = & \dfrac{ -6\left( 16\right) \left( 16\right) \left( 8\right) }{\left( 3\left( 16\right) \right) ^{3}}=-\dfrac{1}{9} \\ \dfrac{\partial h^{2} }{\partial b^{2}}\left( 4,4,4\right) & = & \dfrac{ -6\left( 16\right) \left( 16\right) \left( 8\right) }{\left( 3\left( 16\right) \right) ^{3}}=-\dfrac{1}{9} \\ \dfrac{\partial h^{2} }{\partial c^{2}}\left( 4,4,4\right) & = & \dfrac{ -6\left( 16\right) \left( 16\right) \left( 8\right) }{\left( 3\left( 16\right) \right) ^{3}}=-\dfrac{1}{9} \end{eqnarray*} y \begin{eqnarray*} \dfrac{\partial ^{2}h }{\partial b\partial a}\left( a,b,c\right) & = & \dfrac{ 6\left( 4\right) \left( 4\right) \left( 4\right) ^{3}}{\left( 3\left( 16\right) \right) ^{3}}-\frac{1}{48}\left( 4\right) =-\dfrac{1}{36}=\dfrac{ \partial ^{2}h }{\partial a\partial b}\left( a,b,c\right) \\ \dfrac{\partial ^{2}h }{\partial c\partial a}\left( a,b,c\right) & = & \dfrac{ 6\left( 4\right) \left( 4\right) ^{3}\left( 4\right) }{\left( 3\left( 16\right) \right) ^{3}}-\frac{1}{48}\left( 4\right) =-\dfrac{1}{36}=\dfrac{ \partial ^{2}h }{\partial c\partial a}\left( a,b,c\right) \\ \dfrac{\partial ^{2}h }{\partial b\partial c}\left( a,b,c\right) & = & \dfrac{ 6\left( 4\right) ^{3}\left( 4\right) \left( 4\right) }{\left( 3\left( 16\right) \right) ^{3}}-\frac{1}{48}\left( 4\right) =-\dfrac{1}{36}=\dfrac{ \partial ^{2}h }{\partial c\partial b}\left( a,b,c\right). \end{eqnarray*} Evaluamos el determinante hessiano limitado en \(\overline{v_{0}}=\left( 4,4,4\right) \) y \(\lambda =\dfrac{1}{48}\) \begin{equation*} \left\vert \overline{H}_{4}\left( \overline{v_{0}}\right) \right\vert =\left\vert \begin{array}{cccc} 0 & \dfrac{\partial g}{\partial a}\left( \overline{v_{0}}\right) & \dfrac{ \partial g}{\partial b}\left( \overline{v_{0}}\right) & \dfrac{\partial g}{ \partial c}\left( \overline{v_{0}}\right) \\ & & & \\ \dfrac{\partial g}{\partial a}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial a^{2}}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial b\partial a}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial c\partial a}\left( \overline{v_{0}}\right) \\ & & & \\ \dfrac{\partial g}{\partial b}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial a\partial b}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial b^{2}}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial c\partial b}\left( \overline{v_{0}}\right) \\ & & & \\ \dfrac{\partial g}{\partial c}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial a\partial c}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial b\partial c}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial c^{2}}\left( \overline{v_{0}}\right) \end{array} \right\vert \end{equation*} De donde, \begin{equation*} \left\vert \overline{H}_{4}\left( 4,4,4\right) \right\vert =\left\vert \begin{array}{cccc} 0 & 16 & 16 & 16 \\ & & & \\ 16 & -\dfrac{1}{9} & -\dfrac{1}{36} & -\dfrac{1}{36} \\ & & & \\ 16 & -\dfrac{1}{36} & -\dfrac{1}{9} & -\dfrac{1}{36} \\ & & & \\ 16 & -\dfrac{1}{36} & -\dfrac{1}{36} & -\dfrac{1}{9} \end{array} \right\vert =-\frac{16}{3}. \end{equation*} Calculamos también el determinante de la submatriz de \(3\times 3\) \begin{equation*} \left\vert \overline{H}_{3}\left( \overline{v_{0}}\right) \right\vert =\left\vert \begin{array}{ccc} 0 & \dfrac{\partial g}{\partial a}\left( \overline{v_{0}}\right) & \dfrac{ \partial g}{\partial b}\left( \overline{v_{0}}\right) \\ & & \\ \dfrac{\partial g}{\partial a}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial a^{2}}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial b\partial a}\left( \overline{v_{0}}\right) \\ & & \\ \dfrac{\partial g}{\partial b}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial a\partial b}\left( \overline{v_{0}}\right) & \dfrac{ \partial ^{2}h}{\partial b^{2}}\left( \overline{v_{0}}\right) \end{array} \right\vert \end{equation*} De donde, \begin{equation*} \left\vert \overline{H}_{3}\left( 4,4,4\right) \right\vert =\left\vert \begin{array}{ccc} 0 & 16 & 16 \\ & & \\ 16 & -\dfrac{1}{9} & -\dfrac{1}{36} \\ & & \\ 16 & -\dfrac{1}{36} & -\dfrac{1}{9} \end{array} \right\vert =\frac{128}{3}. \end{equation*} Puesto que \(\left\vert \overline{H}_{3}\left( 4,4,4\right) \right\vert >0\) y \(\left\vert \overline{H}_{4}\left( 4,4,4\right) \right\vert < 0 \), por el inciso 1 del Teorema 2, la función \(f\) tiene un máximo relativo. El valor que alcanza la función en el punto es

\begin{equation*} f\left( 4,4,4\right) =\dfrac{3}{\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}}=4. \end{equation*}