Finding dimensions of a rectangular box 
Find the dimensions of a rectangular box without a top, of the maximum capacity with a surface area of $108 \, cm^2$.

This is my attempt at solving the problem : 
If $x,y,z$ are the dimensions of the box,
Surface Area : 
$$xy + 2xz + 2yz = 108\,cm^2$$
Volume : 
$$V = xyz$$
 A: HINT:
As $x,y,z>0$, using A.M, G.M  inequality 
 $$xy+2zx+2yz\ge  3(xy\cdot 2zx\cdot 2yz)^{\frac13}=3\cdot2^{\frac23}\cdot (xyz)^\frac23$$
Taking cube in either side,
  $$3^3\cdot 2^2\cdot (xyz)^2\le (xy+2zx+2yz)^3=(108)^3=2^6\cdot3^9$$
$$\implies  (xyz)^2\le2^4\cdot3^6\implies xyz\le 2^2\cdot3^3=108 $$
A: You want to maximize V= xyz with the constraint that xy+ 2xz+ 2yz= 108.  One way to do that is to use the constraint to remove one variable.  For example, 2xz+ 2yz= (2x+ 2y)z= 108- xy so z= (108- xy)/(2x+ 2y).  Replacing z in xyz by that, V= xy(108- xy)/(2x+ 2y).  Find the partial derivatives of that with respect to x and y, set them equal to 0 and solve for x and y.
A simpler method is to use the "Laplace multiplier" method.  Given that V= xyz, the gradient of V is $\nabla V= yz\vec{i}+ xz\vec{j}+ xy\vec{k}$.  The constraint, that A= xy+ 2xz+ 2yz be a constant, has gradient $\nabla A= (y+ 2z)\vec{i}+ (x+ 2z)\vec{j}+ (2x+ 2y)\vec{k}$.  
It can be shown, geometrically, that at a max or min, those two vectors must be parallel which means one is a multiple of the other: $yz\vec{i}+ xz\vec{j}+ xy\vec{k}= \lambda[(y+ 2z)\vec{i}+ (x+ 2z)\vec{j}+ (2x+ 2y)\vec{k}]$ for some number $\lambda$ (the "Laplace multiplier").
That is the same as saying that $yz= \lambda(y+ 2z)$, $xz= \lambda(x+ 2z)$, and $xy= \lambda(x+ 2y)$.  Since a specific value for $\lambda$ is not necessary for a solution, it is often a good idea to start by eliminating $\lambda$ by dividing one equation by another.  For example, dividing the first equation by the second, $\frac{yz}{xz}= \frac{y+ 2z}{x+ 2z}$, and dividing the second equation by the third, $\frac{xz}{xy}= \frac{x+ 2z}{x+ 2y}$.  Clearing the fractions from each equation, y(x+ 2z)= x(y+ 2z) and z(x+ 2y)= y(x+ 2z).  Those reduce to 2yz= 2xz, so y= x, and xz= xy so y= z.  That is,  x= y= z.  Since they must satisfy the constraint, $xy+ 2xz+ 2yz= x^2+ 2x^2+ 2x^2= 5x^2= 108$ so x^2= 108/5= 21.6.  Thus the maximum value of V, with this constraint is $(21.6)^3= 10077.696$. 
