Strangely, the wine for a large wedding reception is to be served in a stunning cut-glass receptacle with the interior dimensions of ${}^{40\times 40\times 30}$(height). The receptacle is to be initially filled to the top. The wine can be purchased in bottles of the sizes given in the following table. Purchasing a larger bottle instead of multiple smaller bottles decreases the overall cost of the wine. To minimize the cost, (a) which bottle sizes should be purchased and how many of each should be purchased and, once the receptacle is filled, how much wine is left over in terms of (b) standard bottles and (c) liters?

1 standard bottle

1 magnum = 2 standard bottles

1 jeroboam = 4 standard bottles

1 rehoboam = 6 standard bottles

1 methuselah = 8 standard bottles

1 salmanazar = 12 standard bottles

1 balthazar = 16 standard bottles = 11.356 L

1 nebuchadnezzar =20 standard bottles

1 magnum = 2 standard bottles

1 jeroboam = 4 standard bottles

1 rehoboam = 6 standard bottles

1 methuselah = 8 standard bottles

1 salmanazar = 12 standard bottles

1 balthazar = 16 standard bottles = 11.356 L

1 nebuchadnezzar =20 standard bottles

Using the given conversion factors, find which bottle sizes should be purchased and how many of each should be purchased.Nebuchadnezzar is the largest size bottle. It would reduce the number of bottles required and effectively reduce the cost.

The volume of the receptacle is $48000{\mathrm{cm}}^3=48\mathrm{L}$.

It shows that 48 L wine can be filled in the receptacle.

$$\begin{gathered} 16\mathrm{standard}\mathrm{bottles}=11.356\mathrm{L} \\ 1\mathrm{standard}\mathrm{bottle}=\frac{11.356}{16} \\ =0.70975\mathrm{L} \end{gathered}$$

1 nebuchadnezzar can be filled by 20 standard bottles.

So, 3 nebuchadnezzars can be filled by$3\times 20\times 0.70975=42.585\mathrm{L}$.

And remaining wine can be filled in 1 methuselah as

$$\begin{gathered} 1\mathrm{methuselah}=8\mathrm{standard}\mathrm{bottles} \\ =8\times 0.70975 \\ =5.678\mathrm{L} \end{gathered}$$

Thus, 3 nebuchadnezzars bottles and 1 methuselah bottle should be purchased.

Once the receptacle is filled, the wine leftover is,

48 - ( 42.585 + 5.678 ) = 0.26 L

But,

$$\begin{gathered} 1\mathrm{L}=\frac{1}{0.70975}\mathrm{standard}\mathrm{bottles} \\ =1.409\mathrm{standard}\mathrm{bottles} \end{gathered}$$

Calculate the amount of wine leftover.

$$\begin{gathered} 0.26\mathrm{L}=0.26\times 1.409 \\ =0.37\mathrm{standard}\mathrm{bottles} \end{gathered}$$

Thus, the wine leftover in terms of bottle is 0.37 of standard bottles.

Once the receptacle is filled, the wine leftover is

48 - ( 42.585 + 5.678 ) = 0.26 L

Thus, the amount of wine that is leftover is 0.26 L.