Should i use lagrange for this question? Consider a rectangular parallelepiped with vertical, horizontal, and height x, y, z respectively. Let the sum of the lengths of the 12 sides be L, and the surface area be S. Changing the values ​​of x, y and z while keeping the value of L constant keeps x, y and z.
In case that value S takes the maximum value, 
answer the following:
a. As it is known, a, express surface area S with only x, y, L 
b. Find partial derivative of S with respect to x and y
c. Use the result of b above to express the values ​​of x, y, z and the maximum value of S when S becomes maximum, using L,
My question: Can I use lagrange multiplier for this kind of question?
$F(x,y,z)=\lambda g(x,y,z)$
Or should i find $S_xx, S_xy,S_x,S_y$ ?
Then using $S_x=0,S_y=0$, $S_xy^2-S_xx.S_yy<0$ ??
 A: What you want to do is to maximize
$$S=2(xy+xz+yz)\tag1$$ subject to the constraint $$L=4(x+y+z)\tag2$$ where $L$ is given.
So, you have two possibilities.
In the first one, eliminate for example $z$ from $(2)$ 
$$z=\frac{L}{4}-x-y \implies S=\frac{1}{2} \left(L x+L y-4 x^2-4 x y-4 y^2\right)$$ Differentiate to get
$$\frac {\partial S}{\partial x}=\frac{1}{2} (L-8 x-4 y)\tag 3$$
$$\frac {\partial S}{\partial y}=\frac{1}{2} (L-4 x-8 y)\tag 4$$ Set them equal to $0$ to get $x=y=\frac L{12}$ from which $z=\frac L{12}$.
The second possibily is the use of Lagrange multipliers, looking for the extremum of
$$F=2(xy+xz+yz)+\lambda (4(x+y+z)-L)$$ Compute the partial derivatives
$$\frac {\partial F}{\partial x}=4 \lambda +2 (y+z)\tag 5$$
$$\frac {\partial F}{\partial y}=4 \lambda +2 (x+z)\tag 6$$
$$\frac {\partial F}{\partial z}=4 \lambda +2 (x+y)\tag 7$$
$$\frac {\partial F}{\partial \lambda}=4 (x+y+z)-L\tag 8$$
and set them equal to $0$.
Use equations $(5,6,7)$ to express $x,y,z$ as functions of $\lambda$; this will give $x=y=z=-\lambda$; replace in $(8)$ to get $\lambda=-\frac L{12}$. Then $x=y=z=\frac L{12}$.
The first approach is simpler because you face a linear constraint which allows to eliminate one of the variables and then to reduce the dimension of the problem. But, think how complex the problem would become if we had to maximize $L$ for a given $S$.
