Quadratic surface maximization and Hessians 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? 
My approach: I got the partial of the above function w.r.t. $l$ and then tried to evaluate it at $1/2$, but got stuck. It most likely will involve some analysis of Hessians.
 A: The exponential function does not actually matter, because it is strictly increasing. The maxima and minima of $\exp(f(w,l))$ are attained (or not attained) precisely at the same points as maxima and minima of $f(w,l)$ itself. One advantage of working with $f(w,l)=-(w^2+l^2/4 - l/2)$ is that its derivatives are simpler than for $e^f$.  
And another advantage that we don't even need calculus: it's a quadratic polynomial in which we can complete the square. 
$$-(w^2 + l^2/4 - l/2) =- (w^2 + (l-1)^2/4-1/4) \tag1$$
To maximize $-(\dots)$, we minimize the content of the parentheses. The smallest $w^2$ can be is $0$, at $w=0$. The smallest $(l-1)^2/4$ can be is $0$, at $l=1$. Therefore, at $w=0$, $l=1$ the global maximum of (1) is attained, and it is equal to $1/4$. 
Consequently, $e^f$ has maximum value $e^{1/4}$.
A: To maximize $\displaystyle\exp\left(-\left(w^2+\frac14\ell^2-\frac14w\ell\right)\right)$ is to minimize $w^2+\frac14\ell^2+\frac14w\ell$, and that is the same as minimizing $4w^2+\ell^2+w\ell$.  As a function of $\ell$, this is
$$
\ell^2+w\ell+4w^2
$$
$$
= \left(\ell^2+w\ell + \frac{w^2}{4}\right) - \frac{w^2}{4} + 4w^2\tag{completing the square}
$$
$$
= \left(\ell^2+w\ell + \frac{w^2}{4}\right) + \frac{15w^2}{4} = \left(\ell+\frac w2\right)^2 + \frac{15w^2}{4}.
$$
The value of $\ell$ that minimizes this is $-w/2$, since that makes the square equal to $0$.  the whole expression is then $15w^2/4$, and it is easy to find the value of $w$ that minimizes that.
