Relative Minimums and Maximums — Question 2

Question 2 – Solution

Strategy. Use the coupled equations to reduce the possible coordinates, then apply the Hessian determinant separately at each solution.

Step 1: Critical points $$f_x=3x^2-3y,\qquad f_y=3y^2-3x.$$ Thus $y=x^2$ and $x=y^2$. Substitution gives $$x=x^4,\qquad x(x^3-1)=0.$$ The real possibilities are $x=0$ and $x=1$, giving $$\boxed{(0,0)\quad\text{and}\quad(1,1)}.$$

Step 2: Hessian test $$f_{xx}=6x,\qquad f_{yy}=6y,\qquad f_{xy}=-3,$$ so $D=36xy-9$. At $(0,0)$, $D=-9<0$, hence it is a saddle point. At $(1,1)$, $D=27>0$ and $f_{xx}=6>0$, so it is a strict relative minimum with $$f(1,1)=\boxed{-1}.$$

Step 3: Saddle verification At the origin the critical value is $0$. Along $y=0$, $$f(x,0)=x^3,$$ which is positive for small $x>0$ and negative for small $x<0$. Thus every neighborhood contains values above and below $0$, independently verifying the saddle classification.