Cuboid Challenge

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#Geometry #CuboidChallenge

Note by Lew Sterling Jr
6 years, 2 months ago

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Comments

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=(L24Lh+4h2)h.dVdh=L28Lh+12h2=0       12h28hL+L2For  max  h=8L64248L224=L6.       x=L2L6=2L3.......x=2L3    h=L6.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 - 6 years, 2 months ago
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