Math Problem Statement
What is the answer
Solution
The problem is asking to find a formula for the amount of packing material, , needed to surround a cardboard box with given dimensions, allowing for space between the box and a wooden crate.
The dimensions of the box are:
- Length: feet
- Width: feet
- Depth: feet
There must be:
- A 0.5-foot space on the front, back, and sides.
- A 1-foot space on the top and bottom.
Step 1: Dimensions of the crate
The crate must be larger than the box to accommodate the packing material. So, we add the required space around each dimension:
-
Length of the crate: Since there is a 0.5 ft space on the front and back, the total added space is 1 foot. Thus, the length of the crate is:
-
Width of the crate: Similar to the length, there is a 0.5 ft space on both sides, so the width of the crate is:
-
Depth of the crate: For the depth, we account for the 1-foot space at the top and bottom, so the depth of the crate is:
Step 2: Volume of the crate and the box
The volume of the crate and the box can now be calculated:
-
Volume of the crate:
-
Volume of the box:
Step 3: Formula for the packing material
The amount of packing material is the difference between the volume of the crate and the volume of the box:
Substituting the expressions for the volumes:
This is the formula for the amount of packing material needed in terms of .
Would you like further details or a simplification of this formula?
Related Questions:
- How would the formula change if the spaces around the box were different sizes?
- What is the packing material required if the box length is 4 feet?
- How does the volume of the box affect the overall packing material?
- Can you derive the expanded form of the packing material formula?
- What happens to the packing material required as the box size approaches zero?
Tip: Remember that when calculating volumes, ensure the units match across all dimensions.
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Math Problem Analysis
Mathematical Concepts
Volume
Geometry
Algebra
Formulas
V_box = x(x + 2)(x - 1)
V_crate = (x + 1)(x + 3)(x + 1)
M(x) = V_crate - V_box
Theorems
Volume of Rectangular Prism
Suitable Grade Level
Grades 10-12