Relative Minimums and Maximums — Question 3

Question 3 – Solution

Strategy. The quadratic form itself records the local geometry; the Hessian test and carefully chosen diagonal paths should agree.

Step 1: Critical point $$f_x=2x+4y,\qquad f_y=4x+2y.$$ Solving $x+2y=0$ and $2x+y=0$ gives $x=y=0$. Therefore the origin is the unique critical point.

Step 2: Hessian test $$f_{xx}=2,\qquad f_{yy}=2,\qquad f_{xy}=4.$$ Hence $$D=(2)(2)-4^2=-12<0.$$ Therefore $$\boxed{(0,0)\text{ is a saddle point}.}$$

Step 3: Path verification Along $y=x$, $$f(x,x)=x^2+4x^2+x^2=6x^2>0$$ for $x\ne 0$. Along $y=-x$, $$f(x,-x)=x^2-4x^2+x^2=-2x^2<0.$$

Both paths approach the origin, whose value is zero, so arbitrarily nearby values occur on both sides of the critical value.