Relative Minimums and Maximums — Question 6

Question 6 – Solution

Strategy. Use the two first-derivative equations to reconstruct the parameters, then test the resulting quadratic form around $P$.

Step 1: Determine the parameters $$f_x=2x+ay-4,\qquad f_y=ax+2by+2.$$ At $(1,-1)$, the equations $f_x=f_y=0$ become $$-2-a=0,\qquad a-2b+2=0.$$ Therefore $$\boxed{a=-2,\qquad b=0}.$$

Step 2: Hessian classification For the resulting function, $$f_{xx}=2,\qquad f_{yy}=0,\qquad f_{xy}=-2.$$ Thus $$D=(2)(0)-(-2)^2=-4<0,$$ so $P$ is a saddle point.

Step 3: Direct verification Write $x=1+h$, $y=-1+k$. Since the function is quadratic and $P$ is critical, $$f(1+h,-1+k)-f(1,-1)=h^2-2hk.$$ Along $k=0$, the change is $h^2>0$. Along $k=h$, it is $-h^2<0$. Hence every neighborhood of $P$ contains values above and below $f(P)$, confirming $$\boxed{P=(1,-1)\text{ is a saddle point}.}$$