Saumya has pencil box of volume `60 cm^(3)`. What can be the maximum length of a pencil that can be accommodate in the box. Given that all the sides are integral ( in cm ) and different from each other ?

Options

A. `7 sqrt(2) cm `
B. `sqrt(905) cm `
C. `sqrt(170) cm `
D. `sqrt(3602) cm `

Answer

To find the maximum length of a pencil that can be accommodated in Saumya's pencil box with a volume of 60 cm³, we need to consider the dimensions of the box as a cuboid. The length of the pencil will be the diagonal of the cuboid, which can be calculated using the formula:

\[ d = \sqrt{L^2 + B^2 + H^2} \]

where \(L\), \(B\), and \(H\) are the length, breadth, and height of the cuboid, respectively.

### Step 1: Factor the Volume
First, we need to find the integral dimensions of the cuboid that multiply to give the volume of 60 cm³. We can start by finding the prime factorization of 60:

\[ 60 = 2^2 \times 3^1 \times 5^1 \]

### Step 2: Find Possible Dimensions
Next, we need to find sets of three different integers (length, breadth, height) that multiply to 60. The combinations of factors of 60 that are different from each other are:

1. \(1, 5, 12\) (1 × 5 × 12 = 60)
2. \(2, 3, 10\) (2 × 3 × 10 = 60)
3. \(2, 5, 6\) (2 × 5 × 6 = 60)

### Step 3: Calculate the Diagonal for Each Combination
Now, we will calculate the diagonal for each combination to find which one gives the maximum length.

1. For \(L = 1\), \(B = 5\), \(H = 12\):
\[ d = \sqrt{1^2 + 5^2 + 12^2} = \sqrt{1 + 25 + 144} = \sqrt{170} \]

2. For \(L = 2\), \(B = 3\), \(H = 10\):
\[ d = \sqrt{2^2 + 3^2 + 10^2} = \sqrt{4 + 9 + 100} = \sqrt{113} \]

3. For \(L = 2\), \(B = 5\), \(H = 6\):
\[ d = \sqrt{2^2 + 5^2 + 6^2} = \sqrt{4 + 25 + 36} = \sqrt{65} \]

### Step 4: Compare the Diagonal Lengths
Now we compare the diagonal lengths calculated:
- \(\sqrt{170} \approx 13.04\)
- \(\sqrt{113} \approx 10.63\)
- \(\sqrt{65} \approx 8.06\)

The maximum diagonal length is \(\sqrt{170}\).

### Conclusion
Thus, the maximum length of the pencil that can be accommodated in the box is:

\[ \text{Maximum length of pencil} = \sqrt{170} \text{ cm} \]