Problem 353 - Calculus Volume 1

Question

Set up, but do not evaluate, each optimization problem. You are constructing a box for your cat to sleep in. The plush material for the square bottom of the box costs $$\$ 5 / \mathrm{ft}^{2}$$ and the material for the sides costs $$\$ 2 / \mathrm{ft}^{2}$$. You need a box with volume \(4 \mathrm{ft}^{2}\). Find the dimensions of the box that minimize cost. Use \(x\) to represent the length of the side of the box.

Step-by-Step Solution

Answer

Minimize \( C(x) = 5x^2 + \frac{32}{x} \).

Step 1: Define Variables

Let the side of the square base of the box be \( x \), and let the height of the box be \( h \). Then the volume of the box is given by the formula \( V = x^2 \cdot h \). Since the volume needs to be \( 4 \) ft³, we have \( x^2 \cdot h = 4 \).

Step 2: Formulate the Cost Function

The cost of the material for the bottom of the box is \$5 per ft², hence the cost for the base is \( 5x^2 \). The cost for the sides depends on the perimeter of the base and the height, so it is \( 2 \times 4x \times h \). Thus, the cost of the sides is \( 8xh \). The total cost \( C \) is therefore given by \( C = 5x^2 + 8xh \).

Step 3: Use the Volume Constraint

From \( x^2 \cdot h = 4 \), solve for \( h \) in terms of \( x \): \( h = \frac{4}{x^2} \). Substitute this expression into the cost function to get \( C = 5x^2 + 8x \left( \frac{4}{x^2} \right) \).

Step 4: Simplify the Cost Function

Substitute \( h = \frac{4}{x^2} \) into \( C = 5x^2 + 8xh \). This gives \( C = 5x^2 + 8 \times \frac{4}{x} = 5x^2 + \frac{32}{x} \). This is the function to minimize.

Step 5: Summary of Optimization Problem

We need to find the value of \( x \) that minimizes the cost function \( C(x) = 5x^2 + \frac{32}{x} \), subject to the constraint of the volume being \( 4 \) ft³.

Key Concepts

Cost Function

A cost function represents the total expense associated with making a product or providing a service. In the context of this optimization problem, we are calculating the cost associated with the construction of a cat box. The cost is dependent on the materials used for the box's bottom and sides.

• The bottom, being made of plush, incurs a higher cost of \(5 per square foot, making it crucial to factor in its area when calculating overall costs.
• The sides, made from a different material, cost \)2 per square foot. Here, both the perimeter of the base and the height of the box contribute to the cost calculation.

The formula for the total cost is summatively expressed as: \( C = 5x^2 + 8xh \), where:

• \( x \) is the side length of the square base.
• \( h \) is the height of the box.
• The function is derived from adding both the costs of the bottom and sides.

Volume Constraint

Volume constraint ensures that the box can hold a specific minimum volume, in this case, \( 4 \text{ ft}^3 \). This constraint is important as it affects the dimensions and resulting cost of the box. By employing the formula for the volume of a box, \( V = x^2 \cdot h \), we ensure the box holds exactly the required volume.
To adhere to the constraint, solve the volume equation for height \( h \) in terms of \( x \):

• \( x^2 \cdot h = 4 \)
• \( h = \frac{4}{x^2} \) ensures the box will hold the necessary volume.

This substitution is vital as it allows integration into the cost function for optimization purposes.

Minimization

Minimization in optimization is finding the best possible low value for a function, often to reduce costs or increase efficiency. For this problem, the objective is to minimize the cost function: \( C(x) = 5x^2 + \frac{32}{x} \). Through this function, the goal is to determine the side length \( x \) that results in the lowest possible construction cost while meeting the volume criteria.
Minimization involves calculus tools such as setting the derivative of the cost function to zero to find critical points. These points help in identifying potential minima or maxima:

• Differentiation helps trace the cost's behavior with relation to \( x \).
• Analyzing the derivative assists in pinpointing the precise \( x \) value that lowers the cost.

Variable Definition

Clear variable definitions are foundational in solving optimization problems. They help understand and formulate the equations effectively. In this task, the two main variables are \( x \) and \( h \):

• \( x \) represents the length of each side of the square bottom of the box.
• \( h \) is the height of the box, determined by the volume constraint equation.

By defining these variables, you set a basis for forming the expressions of the cost function and constraints. Defining \( x \) upfront also enables substitutions and simplifications, especially when combining it with the provided volume constraint.