Relative Minimum (Calculus) For what values of the constant a will the function $f(x,y)=ax^2-0.5y^2+xy-x-y$ have a relative minimum at the critical point $(0,1)$? 
$(A) \space\space a>0$
$(B) \space\space a>0.5 $
$(C) \space\space a<-0.5 $
$(D) \space\space a<0 $
$(E) \space\space (A) \space$ or $\space (C) $
$(F)$ real values of a not precisely described by any of $(A)$ to $(E)$
$(G)$ no values of $a$
NOTE: After determining the second derivative of $f$ with respect to $x$, I got $a>0$ i.e option $(A)$ is correct. However, after solving the simultaneous equations resulting from the partial derivatives of $x$ and $y$, I got $a=-0.5$. Hence, I felt option $(F)$ was correct. I don't seem to know what I am getting wrong.
 A: $$f_x(x,y) = 2ax + y - 1$$
$$f_y(x,y) = -y + x - 1$$
so
$$f_{xx}(x,y) = 2a$$
$$f_{yy}(x,y) = -1$$
$$f_{xy}(x,y) = f_{yx}(x,y) = 1$$
Now, define:
$$D(0,1) = f_{xx}(0,1)f_{yy}(0,1) - f_{xy}(0,1)^2 = -2a - 1 = -(2a + 1)$$
There is a relative minimum if $D(0,1) > 0$ and $f_{xx}(0,1) > 0$
$D>0$ for:
$$-(2a+1) < 0$$
$$2a+1 > 0$$
$$a < -0.5$$
$f_{xx}(0,1) > 0$ for:
$$2a > 0$$
$$a > 0$$
So to have a relative minimum, we need $a < 0$ and $a > 0$, which is not possible.
This leaves $(G)$ as the only option
Note:
If $D(0,1) > 0$, then $f_{yy}$ and $f_{xx}$ have the same sign, so it would have worked just as well to say we need $D(0,1) > 0$ and $f_{yy} > 0$, which is clearly not possible since $f_{yy} = -1$
A: For $f$ to have a critical point at $(0,1)$, its gradient must vanish there. $\nabla f=\langle2ax+y-1,x-y-1\rangle$, which is $\langle0,-2\rangle$ at this point. It should be obvious that there’s no value of $a$ that will make this zero, i.e., $(0,1)$ is not a critical point of $f$ for any value of $a$, despite what the problem said. (Did you copy it down correctly?)
