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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

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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.

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    the figure....here it goes – 2017-02-16
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    Not too relevant for this question perhaps but you may google-"Overhang of Blocks"... – 2017-04-06

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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.

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    could not understand about the height of 1cc blocks – 2017-02-16
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    If you have two $1$ cc blocks with differing heights $m,m'$ the area total is $\frac 1m+\frac1{m'}$. You can get the same total area from two blocks of the same height with less total height. In essence, you are being asked to find the cheapest (in terms of height) way to get $48 cm^2$ in area with these volumes. – 2017-02-16
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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 $H={3\over4}$.