Find the least number A such that for any two squares of combined area 1, a rectangle of area A exists such that the two squares can be packed in the rectangle (without interior
overlap).
You may assume that the sides of the squares are parallel to the sides of the
rectangle.
(In reply to
solution by Larry)
Case 3 the two squares have area 16/25 and 9/25. Then the rectangle they fit in is (4/5+3/5)*(4/5) = 28/25 > 1.
If A=1 then it is impossible to find a rectangle of that size to fit the 16/25 and 9/25 case; but the converse is true, it is trivial to find a rectangle of area 28/25 which fits the edge case of two squares of area 1/2.
Therefore A now must be at least 28/25.
re: solution
