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 3

Consideremos el triángulo cuyos vértices son \(\overline{c} _{1}=\left( -2,-1,1\right) ,\overline{c}_{2}=\left( 1,0,1\right) ,\overline{ c }_{3}=\left( 2,3,-1\right) \) y la función \begin{equation*} f\left( \overline{x}\right) =\left\Vert \overline{x}-\overline{c} _{1}\right\Vert ^{2}+\left\Vert \overline{x}-\overline{c}_{2}\right\Vert ^{2}+\left\Vert \overline{x}-\overline{c}_{3}\right\Vert ^{2} \end{equation*} encuentra los máximos, mínimos relativos y puntos silla de la función.

Solución:

Tenemos \begin{eqnarray*} f\left( x,y,z\right) &=&\left\Vert \left( x,y,z\right) -\left( -2,-1,1\right) \right\Vert ^{2}+\left\Vert \left( x,y,z\right) -\left( 1,0,1\right) \right\Vert ^{2}+\left\Vert \left( x,y,z\right) -\left( 2,3,-1\right) \right\Vert ^{2} \\ &=&\left( x+2\right) ^{2}+\left( y+1\right) ^{2}+\left( z-1\right) ^{2}+\left( x-1\right) ^{2}+y^{2}+\left( z-1\right) ^{2}+\left( x-2\right) ^{2}+ \left( y-3\right) ^{2}+\left( z+1\right) ^{2}. \end{eqnarray*} Calculamos las derivadas de primer orden \begin{eqnarray*} \dfrac{\partial f}{\partial x}\left( x,y,z\right) & = & 2\left( x+2\right) +2\left( x-1\right) +2\left( x-2\right) \\ \dfrac{\partial f}{\partial y}\left( x,y,z\right) & = & 2\left( y+1\right) +2y+2\left( y-3\right) \\ \dfrac{\partial f}{\partial z}\left( x,y,z\right) & = & 2\left( z-1\right) +2\left( z-1\right) +2\left( z+1\right). \end{eqnarray*} Igualamos estas derivadas a cero \begin{eqnarray*} 2\left( x+2\right) +2\left( x-1\right) +2\left( x-2\right) & = & 0 \\ 2\left( y+1\right) +2y+2\left( y-3\right) & = & 0 \\ 2\left( z-1\right) +2\left( z-1\right) +2\left( z+1\right) & = & 0 \end{eqnarray*} simplificando tenemos \begin{eqnarray*} 3x-1 & = & 0 \\ 3y-2 & = & 0 \\ 3z-1 & = & 0 \end{eqnarray*} de donde \begin{equation*} x=\dfrac{1}{3},\quad \quad \quad y=\dfrac{2}{3},\quad \quad \quad z=\dfrac{1 }{3}. \end{equation*} Así, el único punto estacionario de \(f\) es \(\left( \dfrac{1}{3}, \dfrac{2}{3},\dfrac{1}{3}\right) .\)

Para determinar si el punto \(\left( \dfrac{1}{3},\dfrac{2}{3},\dfrac{1}{3} \right) \) es un máximo o un mínimo relativo estricto, calculamos las derivadas de segundo orden: \begin{equation*} \dfrac{\partial ^{2}f}{\partial x^{2}}\left( x,y,z\right) =6 \quad \quad \quad \dfrac{\partial ^{2}f}{\partial y^{2}}\left( x,y,z\right) =6 \quad \quad \quad \dfrac{\partial ^{2}f}{\partial z^{2}}\left( x,y,z\right) =6 \end{equation*} y \begin{eqnarray*} \dfrac{\partial ^{2}f}{\partial y\partial x}\left( x,y,z\right) &=&0=\dfrac{ \partial ^{2}f}{\partial x\partial y}\left( x,y,z\right) \\ \dfrac{\partial ^{2}f}{\partial z\partial x}\left( x,y,z\right) &=&0=\dfrac{ \partial ^{2}f}{\partial x\partial z}\left( x,y,z\right) \\ \dfrac{\partial ^{2}f}{\partial z\partial y}\left( x,y,z\right) &=&0=\dfrac{ \partial ^{2}f}{\partial y\partial z}\left( x,y,z\right). \end{eqnarray*} El determinante hessiano en \(\left( \dfrac{1}{3},\dfrac{2}{3},\dfrac{1}{3} \right) \) es \begin{equation*} \left\vert H_{3}\left( f,\left( \dfrac{1}{3},\dfrac{2}{3},\dfrac{1}{3} \right) \right) \right\vert =\left\vert \begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array} \right\vert \end{equation*} De donde,

• \(\dfrac{\partial ^{2}f}{\partial x^{2}}\left( \dfrac{1}{3},\dfrac{2}{3} ,\dfrac{1}{3}\right) =6>0.\)
• \begin{equation*} \left\vert H_{2}\left( f,\left( \dfrac{1}{3},\dfrac{2}{3},\dfrac{1}{3} \right) \right) \right\vert =\left\vert \begin{array}{cc} 6 & 0 \\ 0 & 6 \end{array} \right\vert =36>0. \end{equation*}

El valor de la función en \(\left( \dfrac{1}{3},\dfrac{2}{3},\dfrac{1}{3} \right) \) es \begin{array}{l} f\left( x,y,z\right) =\left\Vert \left( \dfrac{1}{3},\dfrac{2}{3},\dfrac{1}{ 3 }\right) -\left( -2,-1,1\right) \right\Vert ^{2}+\left\Vert \left( \dfrac{ 1 }{ 3},\dfrac{2}{3},\dfrac{1}{3}\right) -\left( 1,0,1\right) \right\Vert ^{2}+\left\Vert \left( \dfrac{1}{3},\dfrac{2}{3},\dfrac{1}{3}\right) -\left( 2,3,-1\right) \right\Vert ^{2} \\ \\ =\left\Vert \left( \dfrac{7}{3},\dfrac{5}{3},-\dfrac{2}{3}\right) \right\Vert ^{2}+\left\Vert \left( -\dfrac{2}{3},\dfrac{2}{3},-\dfrac{2}{3} \right) \right\Vert ^{2}+\left\Vert \left( -\dfrac{5}{3},-\dfrac{7}{3}, \dfrac{4}{3}\right) \right\Vert ^{2} \\ \\ =\dfrac{26}{3}+\dfrac{4}{3}+10 \\ \\ =20. \end{array}