5 blocks of volume 1 cc ,1 cc ,1 cc ,1 cc , and 4 cc are placed one above another ..... I came across this sum in an examination...
5 blocks of volume 1 cc
,1 cc
,1 cc
,1 cc
, and 4 cc are placed one above another to form a structure like this

Suppose the sum of surface areas of upper face of each block is $48 cm^2
$
.
Determine
the minimum possible height of the whole structure.
 A: It is "obvious" that all the $1$ cc blocks should be the same height, though you should justify this by showing that if two differ you can reduce the height by making them the same.  Let the $1$ cc blocks have height of $h$ and the $4$ cc block have height $k$.  The overall height is $4h+k$.  The total area is $48=\frac 4h +\frac 4k$.  Solve this for one variable, plug it into the total height of $4h+k$ getting an expression in one variable, differentiate, set to zero, solve.
A: For each such block $B_k$ we have $h_k={V_k\over A_k}$, where $A_k$ denotes the area of the top surface of $B_k$. Let $B_0$ be the large block. Then
$$H:=\sum_{k=0}^4 h_k=\sum_{k=0}^4{V_k\over A_k}=4\left({1\over A_0}+{1\over4}\sum_{k=1}^4{1\over A_k}\right)\ .$$
Since the function $x\mapsto{1\over x}$ $(x>0)$ is convex we obtain
$$H\geq4\left({1\over A_0}+{1\over{1\over4}\sum_{k=1}^4 A_k}\right)=4\left({1\over A_0}+{4\over 48-A_0}\right)\ ,\tag{1}$$
with equality iff all $A_k$ $(1\leq k\leq 4)$ are equal. The RHS goes to $\infty$ when $A_0\to0+$ or $A_0\to48-$, and takes its minimum at $A_0=16$. In this case $A_k=8$ $(1\leq k\leq 4)$, and $(1)$ gives $H={3\over4}$.
