Problem 25 - Calculus and its applications

Question

Denney Construction is planning to build a warehouse whose interior volume is to be \(252,000 \mathrm{ft}^{3}\). Construction costs per square foot are estimated as follows: Walls: \(\quad \$ 3.00\) Floor: \(\quad \$ 4.00\) Ceiling: \(\quad \$ 3.00\) a) The total cost of the building is \(C(x, y, z),\) where \(x\) is the length, \(y\) is the width, and \(z\) is the height, all in feet. Find a formula for \(C(x, y, z)\) b) What dimensions of the building will minimize the total cost? What is the minimum cost?

Step-by-Step Solution

Answer

Minimum cost is approximately $160,187 with dimensions 60.07 ft x 60.07 ft x 70 ft.

Step 1: Express Volume in Terms of Dimensions

The warehouse's interior volume is given by the formula for the volume of a rectangular prism: \( V = x \cdot y \cdot z \). We set this equal to the given volume: \( x \cdot y \cdot z = 252,000 \). This is our constraint.

Step 2: Define Cost in Terms of Dimensions

The cost function combines the costs of the walls, floor, and ceiling. The floor area is \( x \cdot y \) and costs \(4 per square foot, the ceiling is the same area \( x \cdot y \) and costs \)3, and the walls consist of two \( x \cdot z \) walls and two \( y \cdot z \) walls, each costing $3 per square foot. The total cost function is: \[C(x, y, z) = 4xy + 3xy + 3(2xz + 2yz) = 7xy + 6xz + 6yz.\]

Step 3: Substitute the Constraint

Since \( xyz = 252,000 \), solve for \( z = \frac{252,000}{xy} \) and substitute this into the cost function:\[C(x, y, z) = 7xy + 6x\left(\frac{252,000}{xy}\right) + 6y\left(\frac{252,000}{xy}\right).\]

Step 4: Simplify the Cost Function

Simplify the substituted cost function:\[C(x, y, z) = 7xy + \frac{1,512,000}{y} + \frac{1,512,000}{x}.\] This expression is in terms of the dimensions \(x\) and \(y\).

Step 5: Use Calculus for Minimization

To find the minimum cost, we need to find the critical points of \(C(x, y)\). Set the partial derivatives \( \frac{\partial C}{\partial x} \) and \( \frac{\partial C}{\partial y} \) and find where they equal zero. This will require solving the following system of equations:1. \( \frac{\partial C}{\partial x} = 7y - \frac{1,512,000}{x^2} = 0 \)2. \( \frac{\partial C}{\partial y} = 7x - \frac{1,512,000}{y^2} = 0 \).

Step 6: Solve the System of Equations

Solve these two equations simultaneously to find values for \( x \) and \( y \). From the first equation:\[ x^2 = \frac{1,512,000}{7y} \] and from the second:\[ y^2 = \frac{1,512,000}{7x}. \] By symmetry, assume \(x = y\). This simplifies to solve for one dimension.

Step 7: Determine Values of x, y, z

Choosing \( x = y \) and substituting back to our volume constraint,\[ x^2z = 252,000. \] Using the equation \( x^3 = \frac{1,512,000}{7} \), we find: \[x = \sqrt[3]{\frac{1,512,000}{7}}.\] Calculate to find \(x = y = 60.07\) (approx). To find \( z \), substitute back: \[ z = \frac{252,000}{60.07^2}. \] Compute \(z \approx 70\).

Step 8: Calculate the Minimum Cost

Finally, substitute \(x = y = 60.07\) and \(z = 70\) into the cost function to find the minimum cost: \[C = 7(60.07)(60.07) + 6(60.07)(70) + 6(60.07)(70) = \text{approximately } \$160,187\].

Key Concepts

Cost Function

When tackling an optimization problem in calculus related to costs, a **Cost Function** is essential. The cost function represents the total expense involved in the construction of a warehouse in this scenario. It considers the various surfaces like the walls, floor, and ceiling, each incurring different costs per square foot.

• For instance, the floor is described by the area \(x \cdot y\), costing \\(4 per square foot.
• The ceiling, being the same size, costs \\)3 per square foot.
• The walls' areas comprise two \( x \cdot z \) walls and two \( y \cdot z \) walls, both sets costing \$3 per square foot each.

By integrating all these components, we obtain the total cost formula: \[ C(x, y, z) = 7xy + 6xz + 6yz. \] The cost function is critical since it helps us understand how different dimensions affect the total building cost. Optimize this function to minimize costs and achieve a budget-friendly solution.
Step-by-step, the derivatives of the cost with respect to each variable provide insights into how changing those variables would affect the overall cost. Understanding and utilizing this function allows you to find the most economical dimensions for constructing the warehouse.

Rectangular Prism Volume

The **Volume of a Rectangular Prism** is a foundational concept in this optimization problem. It allows us to understand how the dimensions of a prism (length, width, height) dictate its overall space or volume. The formula is straightforward:\[ V = x \cdot y \cdot z. \]In this exercise, the specified volume of the warehouse is \(252,000 \, \text{ft}^3\). This becomes a constraint, making the problem more structured. You know that the dimensions need to satisfy this volume constraint:\[ x \cdot y \cdot z = 252,000. \]By using this constraint, one can express one of the dimensions in terms of the others, which simplifies the optimization of the cost function. Redistributing the available volume among the length, width, and height efficiently is key to determining the dimensions that lead to the lowest possible building cost. Each adjustment of dimensions must respect the volume requirement while also aligning with cost considerations.

Critical Points in Calculus

Finding **Critical Points in Calculus** is a pivotal aspect of solving optimization problems. These points tell us where a function might reach its extremum, either a minimum or maximum. To locate critical points for the cost function, evaluate its derivative with respect to each variable.The process begins by calculating the partial derivatives:

• \( \frac{\partial C}{\partial x} = 7y - \frac{1,512,000}{x^2} \)
• \( \frac{\partial C}{\partial y} = 7x - \frac{1,512,000}{y^2} \)

Set these derivatives equal to zero, because critical points occur where the rate of change is zero:

• \( 7y = \frac{1,512,000}{x^2} \)
• \( 7x = \frac{1,512,000}{y^2} \)

Solving these equations simultaneously provides the values of \(x\) and \(y\) that optimize the cost, under the volume constraint. Through symmetry, assuming \(x = y\) simplifies solving these equations, helping you find the dimensions where the cost is minimized. From here, calculate the corresponding value of \(z\) using the volume constraint. The use of calculus in this way gives precise dimensions rapidly and effectively, illustrating how calculus is not just theoretical but also highly practical in real-world applications.