Assuming we have a box given that the sum of all intervals is $a$. What is the maximal surface area of the box? I know I need to use Lagrange multiplier but when I find the hessian matrix I get that it is not defined. I get the following matrix:
$$\left( \begin{array}{cccc} 0 & 2 & 2 & 4\\ 2 & 0 & 2 & 4\\ 2 & 2 & 0 & 4\\ 4 & 4 & 4 & 0 \end{array}\right)$$ But it is neither positive nor negative defined. So for $$\left\{ x=y=z=\frac{a}{12}\right\}$$ I get... nothing? I am curtain that I am wrong somewhere.
The Lagrange multiplier is
$$L=2(xy+xz+yz)+\lambda(a-(4x+4y+4z))$$
We can find the partial derivatives:
$$\frac{\partial L}{\partial x}=2y+2z-4\lambda\\ \frac{\partial L}{\partial y}=2x+2z-4\lambda\\ \frac{\partial L}{\partial z}=2x+2y-4\lambda\\ \frac{\partial L}{\partial \lambda}=a-(4x+4y+4z)\\ $$
You are correct that the critical point is $x=y=z=\frac{a}{12}$. So the Hessian matrix is \begin{bmatrix} 0 &4&4&4\\ 4&0&2&2\\ 4&2&0&2\\ 4&2&2&0 \end{bmatrix}
The determinants $H_1=-16, H_2=64, H_3=-192$, which guarantees a maximum.