Show, with calculations, which OPTION of the two options (Option 1 or Option 2) you would advise Franz to use…

Question

Show, with calculations, which OPTION of the two options (Option 1 or Option 2) you would advise Franz to use so that he can pack the maximum number of boxes on the floor of the trailer.

ScanSolve AI Tutor · Accepted Answer

To determine which packing option allows for the maximum number of boxes on the floor of the trailer, we need to calculate how many boxes fit for each option.

Box dimensions:
- Length ( $L_b$ ) = $0.3 \text{ m}$
- Breadth ( $B_b$ ) = $0.215 \text{ m}$
- Height ( $H_b$ ) = $0.235 \text{ m}$
Trailer dimensions:
- Length ( $L_t$ ) = $394 \text{ inches}$
- Breadth ( $B_t$ ) = $119 \text{ inches}$
- Height ( $H_t$ ) = $94.5 \text{ inches}$
Conversion factor: $1 \text{ inch} = 2.54 \text{ cm}$

Convert all dimensions to a consistent unit (centimeters).
$L_b = 0.3 \text{ m} \times 100 \text{ cm/m} = 30 \text{ cm}$ $B_b = 0.215 \text{ m} \times 100 \text{ cm/m} = 21.5 \text{ cm}$ $L_t = 394 \text{ inches} \times 2.54 \text{ cm/inch} = 1000.76 \text{ cm}$ $B_t = 119 \text{ inches} \times 2.54 \text{ cm/inch} = 302.26 \text{ cm}$
Calculate the number of boxes that fit on the floor for Option 1.
In Option 1, the box's length ( $L_b$ ) aligns with the trailer's length ( $L_t$ ), and the box's breadth ( $B_b$ ) aligns with the trailer's breadth ( $B_t$ ).
$\text{Number of boxes along trailer length} = \left\lfloor \frac{L_t}{L_b} \right\rfloor = \left\lfloor \frac{1000.76 \text{ cm}}{30 \text{ cm}} \right\rfloor = \left\lfloor 33.358... \right\rfloor = 33$ $\text{Number of boxes along trailer breadth} = \left\lfloor \frac{B_t}{B_b} \right\rfloor = \left\lfloor \frac{302.26 \text{ cm}}{21.5 \text{ cm}} \right\rfloor = \left\lfloor 14.058... \right\rfloor = 14$ $\text{Total boxes for Option 1} = 33 \times 14 = 462 \text{ boxes}$
Calculate the number of boxes that fit on the floor for Option 2.
In Option 2, the box's breadth ( $B_b$ ) aligns with the trailer's length ( $L_t$ ), and the box's length ( $L_b$ ) aligns with the trailer's breadth ( $B_t$ ).
$\text{Number of boxes along trailer length} = \left\lfloor \frac{L_t}{B_b} \right\rfloor = \left\lfloor \frac{1000.76 \text{ cm}}{21.5 \text{ cm}} \right\rfloor = \left\lfloor 46.547... \right\rfloor = 46$ $\text{Number of boxes along trailer breadth} = \left\lfloor \frac{B_t}{L_b} \right\rfloor = \left\lfloor \frac{302.26 \text{ cm}}{30 \text{ cm}} \right\rfloor = \left\lfloor 10.075... \right\rfloor = 10$ $\text{Total boxes for Option 2} = 46 \times 10 = 460 \text{ boxes}$
Compare the total number of boxes for both options.
Option 1 allows 462 boxes, while Option 2 allows 460 boxes. Therefore, Option 1 allows for the maximum number of boxes on the floor.

Franz should use $\text{Option 1}$ .

Show, with calculations, which OPTION of the two options (Option 1 or Option 2) you would advise Franz to use so that he can pack the maximum number of boxes on the floor of the trailer.

Answer

Step-by-step solution

Convert all dimensions to a consistent unit (centimeters).

Calculate the number of boxes that fit on the floor for Option 1.

Calculate the number of boxes that fit on the floor for Option 2.

Compare the total number of boxes for both options.

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