What is the area of the smallest rectangle that can fit 10 rectangles with areas 1 to 10, inclusive? Rectangles must have integer sides and cannot overlap.
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asked Jan 27 at 2:31
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2 Answers
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This answer seems a bit too trivial to be the intended one, but:
The answer is 55.
The minimum area must be 1+2+3+4+5+6+7+8+9+10=55
This can be done by taking rectangles of dimensions 1$\times$1, 2$\times$1,..., 10$\times$1 and arranging them in a straight line to get a rectangle with dimensions 55$\times$1
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2$\begingroup$ Rot13(5k11 vf nyfb rkgerzryl fvzcyr (ebjf bs 10+1, 9+2, 8+3 rgp)) $\endgroup$– JafeJan 27 at 3:48
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$\begingroup$ Oh that's right, then it's just a poorly framed question $\endgroup$ Jan 27 at 3:52
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$\begingroup$ of course you are right. I completely stuffed up this question. It should have been a square... $\endgroup$ Jan 27 at 4:28
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One example of smallest perimeter rectangle
Rectangle 8x7, with one empty 1x1 square, which is the nerest to an 8x8 square
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$\begingroup$ that's very nice! This is the question I should have asked. $\endgroup$ Jan 27 at 12:04