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.
Answer: A= about 1.21
Solution::
If x and y are the sides of two squares with combined area 1,
then x^2 + y^2 = 1.
Suppose
without loss of generality that x ≥ y. Then the shorter side of a rectangle containing both
squares without overlap must be at least x, and the longer side must be at least x+y. Hence
the desired value of A is the maximum of x(x + y).
To find this maximum, we let x = cos θ, y = sin θ with θ ∈ [0, π/4].
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