91Ó°ÊÓ

A function,$g(x,y)=4x^2−8xy+y^2+3y+5$

$$\begin{gathered} \mathrm{The}\mathrm{first}-\mathrm{order}\mathrm{partial}\mathrm{derivatives}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{are}: \\ {g}_x(x,y)=\frac{∂g}{∂\mathrm{x}}=8x-8y\mathrm{and}{g}_y(x,y)=\frac{∂g}{∂\mathrm{y}}=-8x+2y+3 \\ \mathrm{Now},\mathrm{solve}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}:8x-8y=0\mathrm{and}-8x+2y+3=0,\mathrm{we}\mathrm{get}, \\ x=y\mathrm{and}-8x+2x+3=0 \\ ⇒x=\frac{1}{2}=y \\ \mathrm{We}\mathrm{find}\mathrm{only}\mathrm{one}\mathrm{stationary}\mathrm{point}s\mathrm{of}g,\mathrm{namely}:\left(\frac{1}{2},\frac{1}{2}\right) \\ \mathrm{The}\mathrm{second}-\mathrm{order}\mathrm{partial}\mathrm{derivatives}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{are}: \\ {g}_{xx}(x,y)=\frac{{∂}^{2}g}{∂{\mathrm{x}}^2}=8,g_{yy}(x,y)=\frac{{∂}^{2}g}{∂{\mathrm{y}}^2}=2\mathrm{and}{g}_{xy}(x,y)=\frac{{∂}^{2}g}{∂\mathrm{x}∂\mathrm{y}}=-8 \\ {g}_{xx}\left(\frac{1}{2},\frac{1}{2}\right)=8,g_{yy}\left(\frac{1}{2},\frac{1}{2}\right)=2and{g}_{xy}\left(\frac{1}{2},\frac{1}{2}\right)=-8 \\ \mathrm{The}\mathrm{determinant}\mathrm{of}\mathrm{the}\mathrm{Hessian}\mathrm{is}: \\ det\left(H_g\left(x,y\right)\right)=\left(\frac{{∂}^{2}g}{∂{\mathrm{x}}^2}\right)\left(\frac{{∂}^{2}g}{∂{\mathrm{y}}^2}\right)-{\left(\frac{{∂}^{2}g}{∂\mathrm{x}∂\mathrm{y}}\right)}^2 \\ det\left(H_g\left(\frac{1}{2},\frac{1}{2}\right)\right)=8×2-{\left(-8\right)}^2 \\ =16-64 \\ =-48 \end{gathered}$$

$$\begin{gathered} \mathrm{If}g\mathrm{has}\mathrm{a}\mathrm{stationary}\mathrm{point}\mathrm{at}({\mathrm{x}}_0,{\mathrm{y}}_0),\mathrm{then} \\ \left(\mathrm{a}\right)g\mathrm{has}\mathrm{a}\mathrm{relative}\mathrm{maximum}\mathrm{at}(x_0,y_0)\mathrm{if}det\left(H_g\right(x_0,y_0\left)\right)>0\mathrm{with} \\ {g}_{xx}(x_0,y_0)<0or{g}_{yy}(x_0,y_0)<0. \\ \left(\mathrm{b}\right)g\mathrm{has}\mathrm{a}\mathrm{relative}\mathrm{minimum}\mathrm{at}(x_0,y_0)\mathrm{if}det\left(H_g\right(x_0,y_0\left)\right)>0\mathrm{with} \\ {g}_{xx}(x_0,y_0)>0\mathrm{or}{g}_{yy}(x_0,y_0)>0. \\ \left(\mathrm{c}\right)g\mathrm{has}\mathrm{a}\mathrm{saddle}\mathrm{point}\mathrm{at}(x_0,y_0)\mathrm{if}det\left(H_g\right(x_0,y_0\left)\right)<0. \\ \left(\mathrm{d}\right)\mathrm{If}det\left(H_g\right(x_0,y_0\left)\right)=0,\mathrm{no}\mathrm{conclusion}\mathrm{may}\mathrm{be}\mathrm{drawn}\mathrm{about}\mathrm{the}\mathrm{behavior}\mathrm{of} \\ g\mathrm{at}(x_0,y_0). \\ \mathrm{In}\mathrm{the}\mathrm{given}\mathrm{function},det\left(H_g\left(\frac{1}{2},\frac{1}{2}\right)\right)=-48<0.\mathrm{Hence},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{saddle}\mathrm{point} \\ \mathrm{at}\left(\frac{1}{2},\frac{1}{2}\right)\mathrm{with}g\left(\frac{1}{2},\frac{1}{2}\right)=4{\left(\frac{1}{2}\right)}^2-8\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+{\left(\frac{1}{2}\right)}^2+3\left(\frac{1}{2}\right)+5=\frac{23}{4} \end{gathered}$$

There are no maximum and minimum points.