Granite Block Cutting and Pavement Length Calculation

The problem asks us to first determine the maximum number of slabs of a specific size that can be cut from a given block of granite. Then, using these slabs, we need to find the maximum length of a pavement of a given width that can be constructed.

Dimensions of the Block and Slabs

• Granite Block Dimensions: 3.1 m × 2.2 m × 2.1 m
• Slab Dimensions: 3 m × 2 m × 4 cm
• Slab Thickness: 4 cm = 0.04 m

So, each slab is 3 m × 2 m × 0.04 m.

Calculating the Maximum Number of Slabs

To find the maximum number of slabs that can be cut from the block, we consider the volume of the block and the volume of a single slab. While the precise number of slabs cut depends on the orientation and cutting strategy within the block (which involves integer division of dimensions), in some problems of this type, the theoretical maximum number of slabs is estimated by dividing the volume of the block by the volume of a single slab and taking the floor of the result. Let's calculate this value.

Volume of the granite block ($V_{\text{block}}$):

\(V_{\text{block}} = \text{length} \times \text{width} \times \text{height}\)

\(V_{\text{block}} = 3.1 \text{ m} \times 2.2 \text{ m} \times 2.1 \text{ m}\)

\(V_{\text{block}} = 14.322 \text{ m}^3\)

Volume of one slab ($V_{\text{slab}}$):

\(V_{\text{slab}} = \text{length} \times \text{width} \times \text{thickness}\)

\(V_{\text{slab}} = 3 \text{ m} \times 2 \text{ m} \times 0.04 \text{ m}\)

\(V_{\text{slab}} = 0.24 \text{ m}^3\)

Theoretical maximum number of slabs:

\(N_{\text{slabs}} \approx \lfloor \frac{V_{\text{block}}}{V_{\text{slab}}} \rfloor\)

\(N_{\text{slabs}} \approx \lfloor \frac{14.322 \text{ m}^3}{0.24 \text{ m}^3} \rfloor\)

\(N_{\text{slabs}} \approx \lfloor 59.675 \rfloor\)

\(N_{\text{slabs}} = 59\)

Assuming 59 is the maximum number of slabs that can be obtained from the block, we proceed with this number for the pavement calculation.

Calculating the Total Area of Slabs

Each slab has dimensions 3 m × 2 m. The area of one slab ($A_{\text{slab}}$) is:

\(A_{\text{slab}} = \text{length} \times \text{width}\)

\(A_{\text{slab}} = 3 \text{ m} \times 2 \text{ m}\)

\(A_{\text{slab}} = 6 \text{ m}^2\)

The total area covered by 59 slabs ($A_{\text{total}}$) is:

\(A_{\text{total}} = N_{\text{slabs}} \times A_{\text{slab}}\)

\(A_{\text{total}} = 59 \times 6 \text{ m}^2\)

\(A_{\text{total}} = 354 \text{ m}^2\)

Calculating the Maximum Length of the Pavement

The slabs are used to make a pavement with a width of 1.5 m. The total area of the pavement is equal to the total area of the slabs used.

Pavement Area = Pavement Length × Pavement Width

\(A_{\text{total}} = L_{\text{pavement}} \times W_{\text{pavement}}\)

We have \(A_{\text{total}} = 354 \text{ m}^2\) and \(W_{\text{pavement}} = 1.5 \text{ m}\). We need to find \(L_{\text{pavement}}\).

\(354 \text{ m}^2 = L_{\text{pavement}} \times 1.5 \text{ m}\)

\(L_{\text{pavement}} = \frac{354 \text{ m}^2}{1.5 \text{ m}}\)

\(L_{\text{pavement}} = \frac{354}{1.5}\)

To simplify the division, we can write 1.5 as 3/2:

\(L_{\text{pavement}} = \frac{354}{3/2} = 354 \times \frac{2}{3}\)

\(L_{\text{pavement}} = 118 \times 2\)

\(L_{\text{pavement}} = 236 \text{ m}\)

Summary

By considering the maximum number of slabs obtained (59) and their total area, we calculated the maximum length of the pavement that can be made with a width of 1.5 m.

The maximum length of the pavement is 236 m.

Thus, the maximum length of the pavement is 236 meters.