A farmer has two fields, one for crops and the other for livestock. The field used for crops is 8 times
larger in area than the field used for livestock. Surprisingly, the fencing required to enclose the
smaller field (i.e. its perimeter) is twice the fencing required to enclose the larger field. Both fields
are rectangular and the fence around each is continuous (i.e. no gaps). The fencing is available only in
1 meter units. How many meters of fencing is required to enclose the two fields?

**ANSWER**:

*This puzzle has multiple solutions. Logically, we know that small field
(area-wise) will be a thin/skinny rectangle given it has twice the perimeter of the larger field despite
being so much smaller in terms of area. Here are a few solutions:*

Large Field: 17 x 264, Small Field: 1 x 561 (total fencing: 1686 m)

Large Field: 18 x 140, Small Field: 1 x 315 (total fencing: 948 m)

Large Field: 20 x 78, Small Field: 1 x 195 (total fencing: 588 m)

Large Field: 24 x 47, Small Field: 1 x 141 (total fencing: 426 m)

Large Field: 33 x 1024, Small Field: 2 x 2112 (total fencing: 1686 m)

Large Field: 63 x 64, Small Field: 2 x 252 (total fencing: 762 m)

Large Field: 49 x 2280, Small Field: 3 x 4655 (total fencing: 13974 m)

Large Field: 50 x 1164, Small Field: 3 x 2425 (total fencing: 7284 m)

Large Field: 52 x 606, Small Field: 3 x 1313 (total fencing: 3948 m)

Large Field: 56 x 327, Small Field: 3 x 763 (total fencing: 2298 m)

Large Field: 57 x 296, Small Field: 3 x 703 (total fencing: 2118 m)

Large Field: 79 x 120, Small Field: 3 x 395 (total fencing: 1194 m)

Large Field: 84 x 110, Small Field: 3 x 385 (total fencing: 1164 m)
**EXPLANATION**: This is from an old book of Mensa puzzles. The book listed a single
solution (the example with 762 m of fencing) without any explanation, but there are in fact multiple
(infinite) solutions. There are 2 equations:

**ab = 8cd** and

**(2a + 2b) × 2 =
2c + 2d**. The next step requires assigning a value to one of the 4 variables, for example, we can
consider a solution where C (the height of the thin/skinny field) = 1. To continue, an understanding of
derivatives is required. You can find an explanation of the solution in the answer posted

here.

Graphically, we're looking at pairs of rectangles where the second has twice the perimeter of the first
and then looking at all the different possible areas for each rectangle and trying to find an area in the
first rectangle that is eight times an area in the second rectangle. For example, in the 4th solution
listed above:

Do you have a

suggestion for this puzzle (e.g. something that should
be mentioned/clarified in the question or solution, bug, typo, etc.)?