SOLUTION: I need to pack round objects into a rectangular box. The box is a fixed length of 20" and fixed width of 10". If the part is a round cylinder of 3.5" diameter, how many can I g
Algebra.Com
Question 550880: I need to pack round objects into a rectangular box. The box is a fixed length of 20" and fixed width of 10". If the part is a round cylinder of 3.5" diameter, how many can I get in the box? If the part is 4.5", how many can I get in the box? I'm really asking: "What is the formula to determine this?"
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
need to pack round objects into a rectangular box.
The box is a fixed length of 20" and fixed width of 10".
If the part is a round cylinder of 3.5" diameter, how many can I get in the box? If the part is 4.5", how many can I get in the box?
:
Assuming you mean the cylinder has 3.5" diameter and is 4.5" long
Each part will occupy a rectangular area 3.5 by 4.5
If you place two rows of 5, side by side, a total of 10 can be placed in the box
RELATED QUESTIONS
The width of a rectangular box is 20%
of the length. If the perimeter is 192 cm, then... (answered by stanbon)
Miguel makes a boat display box for a school project. The display box is in the shape of... (answered by macston)
The height of a rectangular box is 4 times its length, and its width is 5ft
more than... (answered by josgarithmetic)
Problem: You wish to pack a cardboard box inside a wooden crate. In order to have room... (answered by scott8148)
The length of a rectangular box is 10 inches longer than the width. If the perimeter of (answered by rfer)
The length of a rectangular box is 4 times its width, and the height is 6ft more than its (answered by rfer)
A flat rate shipping box is in the shape of a rectangular prism. You estimate that the... (answered by Boreal)
Volume of a rectangular Chinese box: The volume of this box is 192 cu. units. Find its... (answered by Alan3354)
Use quadratic formula and a calculator to solve. Round decimal approximations to the... (answered by josgarithmetic)