Find the critical point of the function Consider the following function.
$g(x, y)=e^{−8x^2−6y^2+24y}$
(a) Find the critical point of g.
(b) Using your critical point in (a), find the value of D(a, b) from the Second Partials test that is used to classify the critical point.
(c) Use the Second Partials test to classify the critical point from (a).
For C, the options are either
Saddle Point, Relative Minimum, Relative Maximum, or Inconclusive  
I could really used some help finding the critical point. I separated them into $f_x$ and $f_y$ and set them equal to 0 and got (24,0). I'm not sure if this answer is right and what I am supposed to do to find b.
 A: Without derivatives:
$g(x,y)=\exp(-8x^2-6(y-2)^2+24 ) \le \exp(24)=g(0,2)$
A: Your answer is not right ! We have 
$g_x(x,y)=g(x,y)(-16x)=0 \iff x=0$
and
$g_y(x,y)=g(x,y)(-12y+24)=0 \iff y=2$.
A: If we write this function as
$$ g(x,y) \ \ = \ \ e^{−8x^2−6y^2+24y} \ \ = \ \ e^{−(8x^2+6y^2-24y)} \ \ = \ \ e^{−[8x^2 \ + \ 6(y-2)^2 \ + 24]} \ \  = \ \ e^{−8x^2}  ·  e^{−6(y-2)^2}  ·  e^{24} \ \ , $$
one thing we can see is that the level curves are ellipses  $ \ 8x^2 \ + \ 6(y-2)^2 \ = \ C \ \ , \ \ C \ \ge \ 0 \ $ centered on $ \ (0,2) \ \ , $ using "completing the square" as Fred shows.  Since we can express $ \ g \ $ as an "exponentially-decaying" function, we would expect there to be a maximum value for the function $ \ g(0,2) \ = \ e^{24} \ . $  [The graph is only worth plotting with extreme vertical scale compression, since $   \ e^{24} \ = \ (e^3)^8 \ \approx \ 20^8 \ \approx \ 2.6 · 10^{10} \ .  $ ]
You didn't show what you found for the first partial derivatives, so it isn't possible to see why you found the result you did.  We work out the relevant partial derivatives here to see up the Hessian matrix:
$$ f_x \ \ = \ \  e^{24} \ · \ (-16x) · e^{−8x^2} \ · \ e^{−6(y-2)^2} \ \ , \ \  f_y \ \ = \ \  e^{24} \ · \  e^{−8x^2} \ · \ [-6·2·(y-2)]·e^{−6(y-2)^2} \ \ ; $$
[the second partial derivatives require the application of the Product Rule]
$$ f_{xx} \ \ = \ \  e^{24} \ · \ \left[ \ (-16) · e^{−8x^2} \ + \ (-16x) · (-16x) · e^{−8x^2} \ \right] \ · \ e^{−6(y-2)^2} \ \ , $$
$$ f_{yy} \ \ = \ \  e^{24} \ · \ (-16) · e^{−8x^2}   \ · \ \left[ \ (-12) · e^{−6(y-2)^2} \ + \ [-6·2·(y-2)]^2·e^{−6(y-2)^2} \right]\ \ , $$
$$ f_{xy} \ \ = \ \  e^{24} \ · \  (-16) · e^{−8x^2}  \ · \ [-12·(y-2)]·e^{−6(y-2)^2} \ . $$
Simplifying the expressions, the Hessian matrix is then
$$ \mathcal{H} \ \ = \ \ \left| \begin{array}{cc} -16·(1-16x^2) & 16·12·(y-2) \\
16·12·(y-2) & 12^2·(y-2)^2 \ - \ 12 \end{array}  \right| \ · \ e^{−8x^2}  ·  e^{−6(y-2)^2}  ·  e^{24}  $$
[there is no compelling need to "pretty up" the entries, as it is easier to see what the evaluation will be in this form] .
At the critical point,
$$ \mathcal{H} \ |_{(0,2)} \ \ = \ \ \left| \begin{array}{cc} -16·(1-0) & 16·12·(2-2) \\
16·12·(2-2) & 12^2·(2-2)^2 \ - \ 12 \end{array}  \right| \ · \ e^0  ·  e^0  ·  e^{24}  $$
$$  = \ \ \left| \begin{array}{cc} -16 & 0 \\
0 &  -12 \end{array}  \right|  · \  e^{24} \ \ > \ \ 0 \ \ .  $$
This shows that the critical point is a local extremum.  Since
$$ f_{xx} \ |_{(0,2)} \ \ = \ \  e^{24} \ · \ \left[ \ (-16) · e^0 \ + \ 0 · 0 · e^0 \ \right] \ · \ e^0 \ \ = \ \  (-16) · e^{24} \ \ < \ \ 0 , $$
it is a local maximum, as we anticipated.
