Partials and maximization If we have that the contours of a response surface are elliptical and the response is given by the following function:
$$\large \exp\left(-\left(w^2 + \frac{1}{4}l^2 -\frac{1}{4} \cdot w \cdot l\right)\right)$$
then if we maximize this function w.r.t $l$ holding $w$ fixed at $1/2$.
 And if we call the maximizer l-star, then holding l-star fixed, maximize over w. How to show that the overall max isn't achieved? 
Note: I got the wrong equation last time I asked a question on it, for which the answer turned out to be "easy"
 A: For a function $f(w,l)$ the regular maximization process is given by
$$df=\frac{\partial f}{\partial w}dw+\frac{\partial f}{\partial l}dl=0$$
which has the solution $l_M$ and $w_M$ by
$$\frac{\partial f}{\partial w}=g_1(w,l)=0$$
$$\frac{\partial f}{\partial l}=g_2(w,l)=0$$
In your case you first differentiate wrt $l$ by keeping $w$ fixed at $w_0$
$$\frac{d f(w_0,l)}{d l}=g_2(w_0,l)=0$$
and you find $l^*$ then
$$\frac{d f(l^*,w)}{d w}=g_1(w,l^*)=0$$
and you find $w^*$. 
As you see for global extremum the solution is
$$g_1(w_M,l_M)=0$$
$$g_2(w_M,l_M)=0$$
whereas for your method it is
$$g_2(w_0,l^*)=0$$
$$g_1(w^*,l^*)=0$$
If both solutions are equal it must be that $w_0=w_M$; so you need a good guess to fix $w$.
-------EDIT---------
As I say that $\frac{d f(w_0,l)}{d l}$ it means that $w$ is fixed $w_0$. If $f=w^2l^2-l$ then
$$\frac{d f(w_0,l)}{d l}=w_0^2\,2\,l-1$$
-------EDIT-2-------
For global maximum the total differential is
$$df=e^{lw/4-w^2-l^2/4}\bigg(\frac l4-2w\bigg)dw+e^{lw/4-w^2-l^2/4}\bigg(\frac w4-\frac l2\bigg)dl=0$$
$$=e^{lw/4-w^2-l^2/4}\bigg(\big(\frac l4-2w\big)dw+\big(\frac w4-\frac l2\big)dl\bigg)=0$$
and the system to be solved
$$g_1(w,l)=\frac l4-2w=0$$
$$g_2(w,l)=\frac w4-\frac l2=0$$
which has the solution for global maximum at $l_M=0$ and $w_M=0$
In your case first differentiate wrt to $l$ by keeping $w$ fixed at $\frac 12$ 
$$\frac{d f(\frac 12,l)}{d l}=e^{l/8-0.25-l^2/4}\bigg(\frac 18-\frac l2\bigg)=0$$
which has the solution $l^*=\frac 14$. Then you fix $l=l^*$ and differentiate wrt $w$
$$\frac{d f(w,\frac 14)}{d w}=e^{w/16-w^2-1/64}\bigg(\frac{1}{16}-2w\bigg)=0$$
and $w^*=\frac{1}{32}$
-------EDIT-3-------
When substitiuted into objective function
For global maximum
$$(w_M=0,l_M=0)\Rightarrow \large \exp(0)=1$$
For your solution
$$(w^*=\frac{1}{32},l^*=\frac 14)\Rightarrow \large \exp\left(-\left((\frac{1}{32})^2 + \frac{1}{4}(\frac 14)^2 -\frac{1}{4} \cdot \frac{1}{32} \cdot \frac 14\right)\right)=0.985$$
