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.$f(x,y)=x^3+y^3-12x-3y+15$

A function,$f(x,y)=x^3+y^3-12x-3y+15$

$$\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}}=3{\mathrm{x}}^2-12\mathrm{and}{f}_y(x,y)=\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=3{\mathrm{y}}^2-3 \\ \mathrm{Now},\mathrm{solve}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}:3{\mathrm{x}}^2-12=0\mathrm{and}3{\mathrm{y}}^2-3=0,\mathrm{we}\mathrm{get}, \\ {\mathrm{x}}^2=4\mathrm{and}{\mathrm{y}}^2=1 \\ \Rightarrow x=\pm 2\mathrm{and}y=\pm 1 \\ \mathrm{We}\mathrm{find}\mathrm{the}\mathrm{four}\mathrm{stationary}\mathrm{point}s\mathrm{of}f,\mathrm{namely}:(2,1),(2,-1),(-2,1),(-2,-1) \\ \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}=6x,f_{yy}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial {\mathrm{y}}^2}=6y\mathrm{and}{f}_{xy}(x,y)=\frac{{\partial }^2\mathrm{f}}{\partial \mathrm{x}\partial \mathrm{y}}=0 \\ {f}_{xx}(2,1)=12,f_{yy}(2,1)=6\mathrm{and}{f}_{xy}(2,1)=0 \\ {f}_{xx}(2,-1)=12,f_{yy}(2,-1)=-6\mathrm{and}{f}_{xy}(2,-1)=0 \\ {f}_{xx}(-2,1)=-12,f_{yy}(-2,1)=6\mathrm{and}{f}_{xy}(-2,1)=0 \\ {f}_{xx}(-2,-1)=-12,f_{yy}(-2,-1)=-6\mathrm{and}{f}_{xy}(-2,-1)=0 \\ \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(2,1\right)\right)=12\times 6-0=72 \\ det\left(H_f\left(2,-1\right)\right)=12\times \left(-6\right)-0=-72 \\ det\left(H_f\left(-2,1\right)\right)=\left(-12\right)\times 6-0=-72 \\ det\left(H_f\left(-2,-1\right)\right)=\left(-12\right)\times \left(-6\right)-0=72 \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(2,1\right)\right)=72>0\mathrm{with}{f}_{xx}\left(2,1\right)=12>0\mathrm{and}{f}_{yy}\left(2,1\right)=6>0. \\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{has}\mathrm{minimum}\mathrm{at}\left(2,1\right)\mathrm{with}\mathrm{minimum}\mathrm{value}, \\ f\left(2,1\right)={\left(2\right)}^3+{\left(1\right)}^3-12\left(2\right)-3\left(1\right)+15=-3 \\ \mathrm{Also},det\left(H_f\left(2,-1\right)\right)=-72<0.\mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{has}\mathrm{a}\mathrm{saddle}\mathrm{point}\mathrm{at}\left(2,-1\right) \\ \mathrm{with}f(2,-1)={\left(2\right)}^3+{\left(-1\right)}^3-12\left(2\right)-3(-1)+15=1 \\ \mathrm{Also},det\left(H_f\left(-2,1\right)\right)=-72<0.\mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{has}\mathrm{a}\mathrm{saddle}\mathrm{point}\mathrm{at}\left(-2,1\right) \\ \mathrm{with}f(-2,1)={\left(-2\right)}^3+{\left(1\right)}^3-12(-2)-3\left(1\right)+15=29 \\ Also,det\left(H_f\left(-2,-1\right)\right)=72>0\mathrm{with}{f}_{xx}\left(-2,-1\right)=-12<0\mathrm{and}{f}_{yy}\left(-2,-1\right)=-6<0. \\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{has}\mathrm{maximum}\mathrm{at}\left(-2,-1\right)\mathrm{with}\mathrm{maximum}\mathrm{value}, \\ f\left(-2,-1\right)={\left(-2\right)}^3+{\left(-1\right)}^3-12(-2)-3(-1)+15=33 \end{gathered}$$

$$\begin{gathered} \mathrm{When}x=0,\underset{y\to \infty }{\lim }f(0,y)=\infty \mathrm{and}\underset{y\to -\infty }{\lim }f(0,y)=-\infty \\ \mathrm{Therefore},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{has}\mathrm{local}\mathrm{minimum}\mathrm{at}\left(2,1\right)\mathrm{and}\mathrm{local}\mathrm{maximum}\mathrm{at}(-2,-1) \end{gathered}$$