In connection to my previous post, I suggest the following as a better configuration.
Continuing from the previous post, we have
$X_{v,u} =$ Extra length gained when stacking $S_v$ onto $S_u$
$= (T) – √[200(T) – 100^2]$; where $T = v + u$ = the sum of the two radii
$L = 2∑R_k–∑X_i$ (where $R_k$ is the radius of the $k$-th sphere and $i = 1, 2, …,20$.)
Then, $Min(L) = Min(2∑R_k –∑X_i)$ $= 2∑R_k– Max(∑X_i ) = 2∑R_k– Y$ (say)
Thus, in order to find Min(L), it is sufficient to find the largest of Y’s from different configurations.
Before we go on, we diverse a bit to point out that $F(T) = T – √(200T–100^2)$ is a decreasing function in the interval $[61, 99]$.
[The proof, which involves the showing of $F’(T) < 0$, is rather strict forward and is omitted.]
The decreasing function provides the following hints for the best stacking method:-
“Putting two smallest spheres adjacent to each other will give a largest gain in X.” ----(*)
“Putting two largest spheres adjacent to each other will give a shortest gain in X.” ----(#)
Let’s first take a look at the simplified situation of having only 3 spheres ($S_{30}$, $S_{49}$, and $S_{50}$)
We need to consider the following 3 cases only:-
The largest in the middle-----$Y(S_{30}, S_{50}, S_{49}) = 2.54538371$
The medium in the middle---$Y(S_{30}, S_{49}, S_{50}) = 2.8473196$
The smallest in the middle---$Y(S_{50}, S_{30}, S_{49}) = 5.38260202$
From this, our observation is
“Putting the smallest between the two largest will get a better gain in length.” ----(@)
This can also be verified mathematically as:-
$Y(S_{50}, S_{30}, S_{49}) – Y$(some other combinations)
$= Y(S_{50}, S_{30}, S_{49}) – Y(S_{30}, S_{49}, S_{50})$; [as an example]
$= (X_{49,30}+ X_{30,50}) – [X_{50,49} + X_{49,30}]$
$= X_{30,50} – X_{50,49}$
$= [80–√[200(80)–100^2] – [99–√[200(99)–100^2]$
$ > 0$ (since $F(T)$ is a decreasing function.)
Next, consider the case of having only 4 spheres (2 largest ($S_{50}$ & $S_{49})$ + 2 smallest $(S_{30}$ & $S_{31})$.
Case-1. Placing the largest two adjacent to each other
----$Y(S_{30} + S_{49} + S_{50} + S_{31}) = … = 5.10724084 =$ length gained is too small
----$Y(S_{50} + S_{49} + S_{31} + S_{30}) = … = 16.6412261$
Case-2. Pair up (One small + one large)
--- $Y(S_{31} + S_{49} + S_{30} + S_{50}) = … = 7.92293509 = $length gained is too small
--- $Y(S_{30} + S_{49} + S_{50} + S_{31}) = … =$ length gained is too small (as case1-(i))
Case-3. Placing the smallest two adjacent to each other
--- $Y(S_{50} + S_{31} + S_{30} + S_{49}) = … = 19.1980326$
--- $Y(S_{50} + S_{30} + S_{31} + S_{49}) = … = 19.176509$
--- $Y(S_{50} + S_{49} + S_{31} + S_{30}) = … = 16.xxxxx$ (as case1-(ii))
This shows that the $S_{50} + S_{31} + S_{30} + S_{49}$ combination has a better usage of the pipe. In fact, all these figures further verified the correctness of (*), (#) and also (@). And they also give us another hint:-
“The largest two spheres should be placed at the far ends (one in each end) of the pipe.” ----(%)
Summing these up, I propose the following configuration:-
$(S_{50}+S_{48}+ … +S_{40}+S_{38}+S_{36}+ … +S_{32})+S_{30}+[S_{31}+S_{33}+ … +S_{47}+S_{49}]$
The corresponding extra length gained $= 89.06688385$
$Min(L) = 1590.933116$, [a figure far less than the $S_{50}+S_{30}+S_{49}+S_{31}+ ...$ configuration.]