SOLUTION: I measured the diameter of the central core of a roll of wrapping paper to be 3.8cm. I cut it along the spiral join and counted 8.5 spirals as i roll out the net. I measured the ba

Algebra ->  Trigonometry-basics -> SOLUTION: I measured the diameter of the central core of a roll of wrapping paper to be 3.8cm. I cut it along the spiral join and counted 8.5 spirals as i roll out the net. I measured the ba      Log On


   

Question 742370: I measured the diameter of the central core of a roll of wrapping paper to be 3.8cm. I cut it along the spiral join and counted 8.5 spirals as i roll out the net. I measured the base angle to be 30 degrees.
I wondered how cardboard rolls were made. if i used a rectangular piece of cardboard to make one, i'd have some waste. What would be the smallest percentage of cardboard wasted if i made the core that i'd just cut up using a piece of rectangular cardboard?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
NOTE: Since you are not specifying some overlap, I am assuming this is not a real life problem. In classroom problems, we assume that the edges can be somehow glued together without overlap. I will calculate that way.

I imagine the factory that makes those rolls cuts those strips of cardboard in large numbers, one next to the other, on a slant, from a long roll of cardboard.
If you tried to cut just one strip it the same way (on a slant) from a rectangle of cardboard, there would be a lot of waste:
c=3.8%2Api is the circumference of the roll, but it does not matter for the percentage waste. L=8.5c%2Fcos%2830%5Eo%29

Your best bet for cutting just one strip would be to cut it along the length of a very long and thin cardboard rectangle, like this (drawing not to scale):
L should have been much, much longer, but I chose to make it not to scale.
The waste in this case would be the two triangles at the ends.
Their total area is x%2Ah.
The area of the green cardboard rectangle used is L%2Bx%29%2Ah
The fraction wasted is xh%2F%28%28L%2Bx%29%2Ah%29
and we can express all those lengths (L, x, c, and h) as a function of c.
We knew that L=8.5c%2Fcos%2830%5Eo%29.
We know that cos%2830%5Eo%29=sqrt%283%29%2F2 , but I do not want to mess with that square root unless I have to.
x=c%2Acos%2830%5Eo%29 and h=c%2Asin%2830%5Eo%29=0.5c are the legs of those right triangles with hypotenuse c.
L%2Bx=8.5c%2Fcos%2830%5Eo%29%2Bc%2Acos%2830%5Eo%29=c%288.5%2Fcos%2830%5Eo%29%2Bcos%2830%5Eo%29%29=c%2A%28%288.5%2Bcos%5E2%2830%5Eo%29%29%2Fcos%2830%5Eo%29%29
Since cos%5E2%2830%5Eo%29=%28sqrt%283%29%2F2%29%5E2=3%2F4=0.75,
L%2Bx=c%2A%28%288.5%2B0.75%29%2Fcos%2830%5Eo%29%29=9.25c%2Fcos%2830%5Eo%29
The wasted fraction is
xh%2F%28%28L%2Bx%29%2Ah%29=x%2F%28L%2Bx%29=c%2Acos%2830%5Eo%29%2F%28%289.25c%2Fcos%2830%5Eo%29%29%29=c%2Acos%5E2%2830%5Eo%29%2F9.25c=c%2A0.75%2F9.25c=0.75%2F9.25=3%2F37=0.081(rounded)
That means that 8.1% of the long and thin cardboard rectangle would be wasted.
I don't think we can reduce the waste any further, but you would need a cardboard rectangle measuring L%2Bx=9.25c%2Fcos%2830%5Eo%29 by h=0.5c.
With c=3.8%2Api=11.94cm (rounded), that would be a strip 6cm wide by 127.5cm long.