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'Wrapping Gifts' printed from https://nrich.maths.org/
The solution here comes from Tom, Ella and James of Madras College,
St Andrews, Scotland. They prove that, to use the smallest square
of wrapping paper to wrap a cuboid shape, you should place the
cuboid symmetrically with its longest edges parallel to the
diagonal of the wrapping paper and with a face containing the two
longest edges on the paper. If $c$ is the longest edge of the
cuboid, $b$ is the middle one and $a$ is the shortest edge of the
cuboid then the length of the diagonal of the wrapping paper has to
be at least $2c + 2a$ so the square has to be $(c + a)\sqrt 2$ by
$(c + a)\sqrt 2$. In this case the edges of the wrapping paper will
only just meet.
The smallest square of wrapping paper to wrap a cube of edge length
$a$ is a square of side $2a\sqrt 2$.
Here is Tom, Ella and James' solution:
Let the square of wrapping paper be $x$ units by $x$ units. Using
simple trigonometry $D \cos \theta = x / 2$. To make the edges of
the paper meet, $D$ must have a length $c / 2$ to go under the
cuboid, $a$ to go up the side and another $c / 2$ to go over the
top. $$D = c / 2 + a + c / 2 = c + a$$ $$x = 2(c + a ) \cos
\theta$$ Looking at this it is simply a scaling of the cosine
graph.
Because of the symmetries involved we are only considering $0 \leq
\theta \leq \pi/4$. In this interval the cosine graph is always
stationary or decreasing so $\theta = \pi/4$ gives the minimum
value.
The smallest wrapping paper therefore has sides of length $$x = 2(c
+ a)\cos \pi/4 = \sqrt 2 (c + a)$$ where $c$ is the length of
longest edge of the cuboid and $a$ is the length of the shortest
edge.
Using this formula for a cube of side $a$ the wrapping paper to
minimize wastage is a square of size $2\sqrt 2 a$. The area of the
paper is $8a^2$ so the area of paper not actually used to cover the
cube is $2a^2$. If the cube is placed centrally on the wrapping
paper with edges parallel to the diagonals of the paper, when
wrapping the cube the edges of the paper will just meet at the top
of the cube and the 'wastage' will be a triangle of area $a^2/2$ at
each corner.