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I have three identical, vertical, cylindrical tanks each with a diameter of 1040mm. They need to be crated. What is the minumum internal area of the rectangular crate required ?

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Would you like to put all the three in one crate, or each crate will be put in a different crate? –  user21436 Jan 31 '12 at 21:40
    
@Kannappan: It seems pretty clear to me that all three are to go in one crate. The question wouldn't make much sense otherwise. –  TonyK Jan 31 '12 at 22:17
    
@Mike: Your question is needlessly complicated. Vertical tanks? 1040mm? What you want to know is, what is the smallest rectangle that can contain three non-intersecting disks of radius 1? –  TonyK Jan 31 '12 at 22:19
    
@TonyK It wouldn't as you mention. But it is possible that OP is lost somewhere even in such a case. –  user21436 Jan 31 '12 at 22:20

1 Answer 1

There are two configurations you might think about. You could put the three cylinders in a line, needing $1040 \times (3\cdot 1040)=3244800 \text{ mm}^2$ Alternately, you can put two next to each other and the third beside them. This puts the centers at the corners of an equilateral triangle of side $1040$. The two next to each other force the width of the rectangle to be $2\cdot 1040=2080$. The height is the height of the triangle, $1040 \frac{\sqrt 3}2$ plus $1040$, so the total area is $2080 \cdot 1040 (1+ \frac{\sqrt 3}2 )\approx 403659 \text{ mm}^2$

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There may be a third configuration. See www2.stetson.edu/~efriedma/ciranima (and then click on it to see it move). –  Gerry Myerson Jan 31 '12 at 23:13
    
@GerryMyerson: Good point. They are again on an equilateral triangle, but there looks to be a better bounding rectangle. –  Ross Millikan Jan 31 '12 at 23:27

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