In Exercises 31–52, find the relative maxima, relative minima, and saddle points for the given functions. Determine whether the function has an absolute maximum or absolute minimum as well.$f(x,y)=2x^2-2xy+4y^2+3x+9y-5$

A function,$f(x,y)=2x^2-2xy+4y^2+3x+9y-5$

$$\begin{gathered} \mathrm{The}\mathrm{first}-\mathrm{order}\mathrm{partial}\mathrm{derivatives}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{are}: \\ {f}_x(x,y)=\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=4x-2y+3\mathrm{and}{f}_y(x,y)=\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=-2x+8y+9 \\ \mathrm{Now},\mathrm{solve}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}:4x-2y+3=0\mathrm{and}-2x+8y+9=0,\mathrm{we}\mathrm{get}, \\ \Rightarrow x=-\frac{3}{2}\mathrm{and}y=-\frac{3}{2} \\ \mathrm{We}\mathrm{find}\mathrm{only}\mathrm{one}\mathrm{stationary}\mathrm{point}s\mathrm{of}f,\mathrm{namely}:\left(-\frac{3}{2},-\frac{3}{2}\right) \\ \mathrm{The}\mathrm{second}-\mathrm{order}\mathrm{partial}\mathrm{derivatives}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{are}: \\ {f}_{xx}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{x}}^2}=4,f_{yy}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{y}}^2}=8\mathrm{and}{f}_{xy}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial \mathrm{x}\partial \mathrm{y}}=-2 \\ {f}_{xx}\left(-\frac{3}{2},-\frac{3}{2}\right)=4,f_{yy}\left(-\frac{3}{2},-\frac{3}{2}\right)=8\mathrm{and}{f}_{xy}\left(-\frac{3}{2},-\frac{3}{2}\right)=-2 \\ \mathrm{The}\mathrm{determinant}\mathrm{of}\mathrm{the}\mathrm{Hessian}\mathrm{is}: \\ det\left(H_f\left(x,y\right)\right)=\left(\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{x}}^2}\right)\left(\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{y}}^2}\right)-{\left(\frac{{\partial }^2\mathrm{f}}{\partial \mathrm{x}\partial \mathrm{y}}\right)}^2 \\ det\left(H_f\left(-\frac{3}{2},-\frac{3}{2}\right)\right)=4\times 8-{\left(-2\right)}^2=28 \end{gathered}$$

$$\begin{gathered} \mathrm{If}f\mathrm{has}\mathrm{a}\mathrm{stationary}\mathrm{point}\mathrm{at}({\mathrm{x}}_0,{\mathrm{y}}_0),\mathrm{then} \\ \left(\mathrm{a}\right)f\mathrm{has}\mathrm{a}\mathrm{relative}\mathrm{maximum}\mathrm{at}(x_0,y_0)\mathrm{if}det\left(H_f\right(x_0,y_0\left)\right)>0\mathrm{with} \\ {f}_{xx}(x_0,y_0)<0or{f}_{yy}(x_0,y_0)<0. \\ \left(\mathrm{b}\right)f\mathrm{has}\mathrm{a}\mathrm{relative}\mathrm{minimum}\mathrm{at}(x_0,y_0)\mathrm{if}det\left(H_f\right(x_0,y_0\left)\right)>0\mathrm{with} \\ {f}_{xx}(x_0,y_0)>0\mathrm{or}{f}_{yy}(x_0,y_0)>0. \\ \left(\mathrm{c}\right)f\mathrm{has}\mathrm{a}\mathrm{saddle}\mathrm{point}\mathrm{at}(x_0,y_0)\mathrm{if}det\left(H_f\right(x_0,y_0\left)\right)<0. \\ \left(\mathrm{d}\right)\mathrm{If}det\left(H_f\right(x_0,y_0\left)\right)=0,\mathrm{no}\mathrm{conclusion}\mathrm{may}\mathrm{be}\mathrm{drawn}\mathrm{about}\mathrm{the}\mathrm{behavior}\mathrm{of} \\ f\mathrm{at}(x_0,y_0). \\ \mathrm{In}\mathrm{the}\mathrm{given}\mathrm{function},det\left(H_f\left(-\frac{3}{2},-\frac{3}{2}\right)\right)=28>0\mathrm{with}{f}_{xx}\left(-\frac{3}{2},-\frac{3}{2}\right)=4>0 \\ \mathrm{and}{f}_{yy}\left(-\frac{3}{2},-\frac{3}{2}\right)=8>0. \\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{has}\mathrm{minimum}\mathrm{at}\left(-\frac{3}{2},-\frac{3}{2}\right)\mathrm{with}\mathrm{minimum}\mathrm{value}, \\ f\left(-\frac{3}{2},-\frac{3}{2}\right)=2{\left(-\frac{3}{2}\right)}^2-2\left(-\frac{3}{2}\right)\left(-\frac{3}{2}\right)+4{\left(-\frac{3}{2}\right)}^2+3\left(-\frac{3}{2}\right)+9\left(-\frac{3}{2}\right)-5=-14 \end{gathered}$$

$$\begin{gathered} \mathrm{When}y=0,\underset{x\to \infty }{\lim }f(x,0)=\infty \mathrm{and}\underset{x\to -\infty }{\lim }f(x,0)=\infty \\ \mathrm{Therefore},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{has}\mathrm{absolute}\mathrm{minimum}\mathrm{at}\left(-\frac{3}{2},-\frac{3}{2}\right)\mathrm{with} \\ \mathrm{f}\left(-\frac{3}{2},-\frac{3}{2}\right)=-14 \end{gathered}$$