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Cuboid Challenge

Note by Llewellyn Sterling
1 year, 6 months ago

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Let the square base have sides =x and height = h. V=x * x * h. But if paper is L * L, x=L-2h. \(So~~ V=(L^2-4Lh+4h^2) * h.\\\dfrac{dV}{dh}=L^2-8Lh+12h^2=0 ~~~\implies 12h^2-8hL+L^2 \\For~~max~~h=\dfrac{8L-\sqrt{64^2-48L^2}}{24}=\dfrac L 6.~~\implies~ x= L-2*\dfrac L 6=\dfrac {2L}{3}...\large....x=\dfrac {2L}{3} ~~~~h=\dfrac L 6.\) Niranjan Khanderia · 1 year, 6 months ago

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