Math Problem Statement

How can i fit Board 1 Width: 122cms Length: 92cms

Board 2 Width: 144cms Length: 122cms

Board 3 Width: 68cms Length: 127cms

20 napkins: Width: 38cms Length: 38cms into a 600 length 120 width piece of material

Solution

To determine whether the boards and napkins can fit into a single material of 600 cm length and 120 cm width, we will perform a step-by-step calculation of the area and layout:


1. Material Dimensions

  • Material Area: 600cm×120cm=72,000cm2600 \, \text{cm} \times 120 \, \text{cm} = 72,000 \, \text{cm}^2
  • Material Width Limitation: Items cannot exceed 120 cm in width.

2. Item Areas

Board 1:

  • Dimensions: 122cm×92cm122 \, \text{cm} \times 92 \, \text{cm}
  • Area: 122×92=11,224cm2122 \times 92 = 11,224 \, \text{cm}^2

Board 2:

  • Dimensions: 144cm×122cm144 \, \text{cm} \times 122 \, \text{cm}
  • Area: 144×122=17,568cm2144 \times 122 = 17,568 \, \text{cm}^2

Board 3:

  • Dimensions: 68cm×127cm68 \, \text{cm} \times 127 \, \text{cm}
  • Area: 68×127=8,636cm268 \times 127 = 8,636 \, \text{cm}^2

Napkins (20 pieces):

  • Dimensions of One Napkin: 38cm×38cm38 \, \text{cm} \times 38 \, \text{cm}
  • Area of One Napkin: 38×38=1,444cm238 \times 38 = 1,444 \, \text{cm}^2
  • Total Area for 20 Napkins: 20×1,444=28,880cm220 \times 1,444 = 28,880 \, \text{cm}^2

3. Total Area of Items

  • Total Area: 11,224+17,568+8,636+28,880=56,308cm211,224 + 17,568 + 8,636 + 28,880 = 56,308 \, \text{cm}^2
  • Remaining Material Area: 72,00056,308=15,692cm272,000 - 56,308 = 15,692 \, \text{cm}^2 (material area is sufficient).

4. Layout Feasibility

Width Limitation Check:

  • Board 1: 122 cm width (fits within 120 cm, but must rotate)
  • Board 2: 144 cm width (does not fit within 120 cm, must rotate)
  • Board 3: 68 cm width (fits within 120 cm)

Boards 1 and 2 must be rotated to fit the width of 120 cm.

Length Considerations:

After rotating, lengths will occupy space along the 600 cm length of the material. The total used length would be:

  1. Board 1: Rotated to 92cm(length)92 \, \text{cm} \, \text{(length)}
  2. Board 2: Rotated to 122cm(length)122 \, \text{cm} \, \text{(length)}
  3. Board 3: 127cm(length)127 \, \text{cm} \, \text{(length)}
  4. Napkins: Arrange in 38cm×38cm38 \, \text{cm} \times 38 \, \text{cm} sections. They can stack efficiently, occupying approximately 760cm(length)760 \, \text{cm} \, \text{(length)}.

Total Length Required: 92+122+127+760=1,101cm92 + 122 + 127 + 760 = 1,101 \, \text{cm}. The material length is insufficient.


5. Conclusion

With the given dimensions, the material (600 cm × 120 cm) is insufficient to accommodate all items at once. Adjustments such as reducing napkins or using multiple pieces of material are necessary.

Would you like a suggested optimized layout or adjustments to prioritize specific items?


5 Related Questions:

  1. What specific priority should be given to boards or napkins in the layout?
  2. Could we split the napkins across different pieces of material?
  3. Should we consider reducing the quantity of napkins?
  4. Can any boards or napkins be resized to fit the material better?
  5. Do you want a diagram showing a potential optimized layout?

Tip: Always measure twice and plan a layout diagram before cutting material to minimize waste.

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Math Problem Analysis

Mathematical Concepts

Geometry
Area Calculation
Spatial Reasoning

Formulas

Area = Length × Width
Total Length Required = Sum of rotated lengths
Material Area = Material Length × Material Width
Remaining Material Area = Material Area - Total Item Area

Theorems

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Suitable Grade Level

Grades 8-10