Problem 14 - Essential Calculus Early Transcendentals

Question

A rectangular storage container with an open top is to have a volume of 10 \(\mathrm{m}^{3} .\) The length of its base is twice the width. Material for the base costs \(\$ 10\) per square meter. Material for the sides costs \(\$ 6\) per square meter. Find the cost of materials for the cheapest such container.

Step-by-Step Solution

Answer

The minimum cost of materials is approximately \(\$120.00\).

Step 1: Define Variables

Let's define the variables first. Let the width of the base be \( w \) meters, the length be \( 2w \) meters (since it's twice the width), and the height be \( h \) meters. The volume of the container is given by the formula:\[ V = \text{length} \times \text{width} \times \text{height} = 2w \times w \times h = 2w^2h \]The volume must be 10 cubic meters:\[ 2w^2h = 10 \]We'll use this equation to express \( h \) in terms of \( w \).

Step 2: Express Height in Terms of Width

Rearrange the volume equation to solve for \( h \):\[ h = \frac{10}{2w^2} = \frac{5}{w^2} \]

Step 3: Calculate Surface Areas for Cost

Now, calculate the surface areas for cost calculation:- Base area: \( A_{base} = 2w \times w = 2w^2 \)- Four sides: Two pairs of identical sides. - Two sides of dimensions \( w \times h \): Total area = \( 2wh \) - Two sides of dimensions \( 2w \times h \): Total area = \( 4wh \)Thus, the total side area is \( 6wh \).

Step 4: Express Total Cost Function

The total material cost consists of the cost for the base and the sides:- Cost of the base: \( C_{base} = 2w^2 \times 10 = 20w^2 \) dollars- Cost of the sides: \( C_{sides} = 6wh \times 6 = 36wh \) dollarsSubstitute \( h = \frac{5}{w^2} \) from Step 2:\[ C_{sides} = 36w \times \frac{5}{w^2} = \frac{180}{w} \]

Step 5: Formulate Total Cost Function

The total cost function \( C(w) \) is given by:\[ C(w) = 20w^2 + \frac{180}{w} \]

Step 6: Find Minimum Cost Using Calculus

To find the minimum cost, differentiate \( C(w) \) with respect to \( w \) and set it to zero:\[ \frac{dC}{dw} = 40w - \frac{180}{w^2} \]Set the derivative to zero to find critical points:\[ 40w - \frac{180}{w^2} = 0 \]\[ 40w^3 = 180 \]\[ w^3 = \frac{180}{40} = 4.5 \]\[ w = \sqrt[3]{4.5} \]

Step 7: Compute Corresponding Dimensions

Calculate \( w \), \( l \), and \( h \):- \( w = \sqrt[3]{4.5} \)- \( l = 2w = 2 \times \sqrt[3]{4.5} \)- \( h = \frac{5}{w^2} = \frac{5}{(\sqrt[3]{4.5})^2} \)

Step 8: Calculate Minimum Total Cost

Substitute \( w = \sqrt[3]{4.5} \) into the cost function:\[ C(w) = 20(\sqrt[3]{4.5})^2 + \frac{180}{\sqrt[3]{4.5}} \]Calculate this value for the minimum cost.

Key Concepts

Volume of Rectangular Prism

Understanding the volume of a rectangular prism is crucial in many real-world applications, such as determining how much space an object can hold. Here, we are considering a rectangular storage container with specific constraints. To calculate the volume of a rectangular prism, we apply the formula:

• Volume = length × width × height

In this exercise, the length of the base is twice the width, which leads to the expression for volume as follows:

• Volume = 2w × w × h = 2w²h

Given that the volume must be 10 cubic meters, we can set up the equation:

• 2w²h = 10

This equation allows us to define any one dimension in terms of the others. Solving for the height, we find:

• h = \( \frac{5}{w^2} \)

This relationship between height and width will be vital for optimizing the dimensions of the container for cost efficiency.

Surface Area Calculation

Calculating the surface area of a container is essential when you need to determine the amount of material required to construct it. In this specific problem, the container's open-top design means we focus on the base and the sides:

• Base area is calculated as: \( A_{base} = 2w² \)
• For the sides, there are two smaller sides with the area: \( 2wh \) and two larger sides with the area: \( 4wh \)

Together, the total area for the sides becomes:

• Total side area = 2wh + 4wh = 6wh

By expressing the height in terms of width \( h = \frac{5}{w^2} \), we can simplify the surface area calculations, which will come into play in managing costs.

Cost Minimization

Cost minimization involves finding the most economical way to construct the container, considering the cost of materials for both the base and the sides:

• Base cost: \( C_{base} = 2w² imes 10 = 20w² \) dollars.
• Sides cost: \( C_{sides} = 6wh imes 6 = 36wh \) dollars.

For the sides, substituting the expression for height from earlier, we have:

• \( C_{sides} = 36w \times \frac{5}{w^2} = \frac{180}{w} \)

The total cost function combines these:

• Total cost \( C(w) = 20w² + \frac{180}{w} \)

To find the minimum cost, we use calculus. By taking the derivative \( \frac{dC}{dw} = 40w - \frac{180}{w^2} \) and setting it to zero, we solve for \( w \):

• \( 40w^3 = 180 \)
• \( w^3 = 4.5 \)
• \( w = \sqrt[3]{4.5} \)

These calculations give us the width needed to minimize costs, while the length and height can be derived similarly, ensuring we achieve the most cost-effective dimensions.