Problem 29 - Precalculus Essentials

Question

A contractor is to build a warehouse whose rectangular floor will have an area of 4000 square feet. The warehouse will be separated into two rectangular rooms by an interior wall. The cost of the exterior walls is \(\$ 175\) per linear foot and the cost of the interior wall is \(\$ 125\) per linear foot. Express the contractor's cost for building the walls, \(C,\) as a function of one of the dimensions of the warehouse's rectangular floor, \(x\).

Step-by-Step Solution

Answer

The contractor's cost for building the walls, \(C\), as a function of one of the dimensions of the warehouse's rectangular floor, \(x\), is \(C = 175 \times (2x + 2(\frac{4000}{x})) + 125 \times \frac{4000}{x}\)

Step 1: Understand the Dimensions of the Warehouse

Assume that the warehouse forms a rectangle with dimensions \(x\) and \( \frac{4000}{x} \) since the area of a rectangle is given by \(Length \times Width\). \(x\) is one side and \( \frac{4000}{x} \) is the other side.

Step 2: Calculate Cost of Exterior Walls

The total cost of the exterior walls is the cost per linear foot times the total length of the exterior walls. For a rectangle, the total length of the exterior walls is given by the perimeter, which is \(2 \cdot Length + 2 \cdot Width\). Therefore, the cost of the exterior walls, \(C1\), is \(175 \times (2x + 2(4000/x))\).

Step 3: Calculate Cost of the Interior wall

The interior wall separates the two rooms, and its length is same as the width of the rectangle, which is \( \frac{4000}{x} \). So the cost of the interior wall, \(C2\), is \(125 \times \frac{4000}{x}\).

Step 4: Express total cost as function of \(x\)

The total cost, \(C\), is the sum of the costs of the exterior and interior walls. So, \(C = C1 + C2 = 175 \times (2x + 2(\frac{4000}{x})) + 125 \times \frac{4000}{x}\). This is the cost function in terms of \(x\).