Minus' Solution to

That Ain't Hay!
Posted Weeks of Novmeber 16 - 29, 1997

Let's look at the minimum and maximum.
If the crop were all hay, Farmer Brown would have a maximum square of 4000 square meters and minimum of 3000 meters. If the crop were all clover, Farmer Brown would have a minimum square of 1200 square meters and maximum of 1600 square meters. But we know that ol' Farmer Brown has more clover than he does hay.

The minimum dimensions of the outer square would suffice if Farmer Brown were to sell as little hay as possible. So, whatever the side length of the clover square, the entire square will have a side length at least 1 meter longer.

You can solve this problem using algebra, or use trial and error --which is often easier when you have certain restraints.
Try a side length of 30 for the clover. That would mean the inner square would have an area of 900 square meters, resulting in sales of $225 for the clover. The larger square would then have a total area of 32(32) = 1024, meaning that 124 square meters, or 124 kg of hay would sell for $12.40. Unfortunately, $225 + $12.40 falls short of the $300 minimum.

If the area of clover were 34(34), then 1156 square meters of clover would sell for $289. The smallest area for the entire square would be 35(35) = 1225, which would yield 69 square meters of hay that would sell for $6.90. $289 + $6.90 still doesn't add up to $300.

Now what? Well, make the outer square 36 x 36. That will give you an area of 1296 square meters, or 140 square meters of hay worth $14. $289 + $14 = $303.

Thus, the minimum dimensions for the outer square are 36 meters x 36 meters.