## Calculus with Applications (10th Edition)

The critical points are $(\frac{9}{2},\frac{3}{2})$ and (0,0). There is a saddle point at (0,0). There is a relative maximum at $(\frac{9}{2},\frac{3}{2})$
$f(x,y)=x^{2}+4y^{3}-6xy-1$ $f_x(x,y)=2x-6y=0 \rightarrow x=3y$ $f_y(x,y)=12y^{2}-6x=0$ Use the substitution method to solve the system of equations $f_y(x,y)=12y^{2}-6x=0=0 \rightarrow 12y^{2}-6(3y)=0$ $12y^{2}-18y=0$ $y=\frac{3}{2} \rightarrow x=\frac{9}{2}$ or $y=0 \rightarrow x=0$ The critical points are $(\frac{9}{2},\frac{3}{2})$ and (0,0) To identify any extrema: $f_{xx}(x,y)=2$ $f_{yy}(x,y)=24y$ $f_{xy}(x,y)=-6$ For (0,0): $f_{xx}(0,0)=2$ $f_{yy}(0,0)=0$ $f_{xy}(0,0)=-6$ $D=2\times0 - (-6)^{2}=-36\lt0 \rightarrow$ There is a saddle point at (0,0) For $(\frac{9}{2},\frac{3}{2})$: $f_{xx}(\frac{9}{2},\frac{3}{2})=2$ $f_{yy}(\frac{9}{2},\frac{3}{2})=36$ $f_{xy}(\frac{9}{2},\frac{3}{2})=-6$ $=2\times 36- (-6)^{2}=36\gt0$ and $f_{xx}(\frac{9}{2},\frac{3}{2})=2\gt0 \rightarrow$ There is a relative maximum at $(\frac{9}{2},\frac{3}{2})$