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I have 2 times a square with side length 2, 2 times a square with side length 3, 1 times a square with side length 4 and 1 times a square with side length 5. I have to create the rectangle with the biggest area with some or all of the squares.

I have already found the solution (I think), which is 2 x square (side 2) + 2 x square (side 3) + square (side 4), but I can't really argument, why that is the solution. Can I please have some help?

*This is not homework.

Nick Podowalski
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1 Answers1

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If you are thinking of the rectangle with side lengths of 6 and 7, then I believe you have it right but the proofing is more difficult and I can only think of drawing it out as a proof.

Sorry if this doesn't help.

Kieran Mcfarlane
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