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.
As usual, I misread the problem and missed the word "any".
The minimum Area is 1 if you get to cherry pick the case.
But for any combination of sizes, we need to maximize the product of the 2 dimensions.
See Desmos graphic
https://www.desmos.com/calculator/vlxdxpmxvg
One square, the larger one, has dimensions x by x.
The other has dimensions √(1-x^2) by √(1-x^2).
To maximize the area,
Area = x*(x + √(1-x^2))
A = x^2 + x√(1-x^2)
A' = 2x + √(1-x^2) - 2x^2/(2√(1-x^2)) = 0
2x*(√(1-x^2)) + √(1-x^2)*√(1-x^2) = x^2
2x*(√(1-x^2)) + (1-x^2) = x^2
2x*(√(1-x^2)) = x^2 - (1-x^2)
4x^2*(1-x^2) = x^4 - 2x^2*(1-x^2) + x^4 - 2x^2 + 1
4x^2 - 4x^4 = x^4 - 2x^2 + 2x^4 + x^4 - 2x^2 + 1
8x^4 - 8x^2 + 1 = 0
one root is x= √(2 + √2) / 2 = approx 0.9238795
Max area is about 1.20710678
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Posted by Larry
on 2024-04-01 11:41:01 |
Second attempt
