Problem 53
Question
The Spice-of-Life Company is preparing to create shipping crates. The company wishes the volume of each crate to be 6 cubic feet, with the crate's base to be square between 1 and 2 feet on a side. Assume that the material for the bottom costs \(\$ 5\), the sides \(\$ 2\), and the top \(\$ 1\) per square foot. Find the dimensions that yield the minimum cost.
Step-by-Step Solution
Verified Answer
The optimal dimensions are base \( x \approx 1.5874 \) feet, height \( h \approx 2.379 \) feet, minimizing cost to \$44.90.
1Step 1: Understand the Problem
The goal is to find the least expensive dimensions for a crate with a square base and a total volume of 6 cubic feet. The sides of the square base should be between 1 and 2 feet. The costs of the materials are given per square foot.
2Step 2: Set Variables and Equations
Let's denote the side length of the square base as \( x \) and the height as \( h \). The volume equation is \( x^2 \times h = 6 \), which gives \( h = \frac{6}{x^2} \).
3Step 3: Determine Costs of Materials
Calculate the cost of each part of the box: the bottom costs \( 5x^2 \) (since area is \( x^2 \)), the top costs \( x^2 \), and the sides cost \( 4xh \) (area of four sides is \( 4xh \)). The total cost equation is: \[ \text{Total Cost} = 5x^2 + x^2 + 8 \times \frac{6}{x} = 6x^2 + \frac{48}{x} \]
4Step 4: Formulate Optimization Problem
Find the derivative of the total cost function with respect to \( x \). The derivative is: \[ \frac{dC}{dx} = 12x - \frac{48}{x^2} \]. Set the derivative to zero for critical points: \[ 12x - \frac{48}{x^2} = 0 \].
5Step 5: Solve for Critical Points
Multiply the equation by \( x^2 \) to eliminate the fraction: \[ 12x^3 = 48 \]. Solving for \( x \) gives \( x = \sqrt[3]{4} \approx 1.5874 \). Check if this lies between 1 and 2 feet.
6Step 6: Analyze Critical Points for Minimum Cost
Substitute \( x = 1 \), \( x \approx 1.5874 \), and \( x = 2 \) in the total cost equation to compare costs: - Cost at \( x = 1 \): \( 6(1)^2 + \frac{48}{1} = 54 \)- Cost at \( x \approx 1.5874 \): Solve for \( C(1.5874) \)- Cost at \( x = 2 \): \( 6(2)^2 + \frac{48}{2} = 48 \)
7Step 7: Calculate Final Cost for Critical Value of x
For \( x \approx 1.5874 \), we calculate: - \( h \approx \frac{6}{1.5874^2} \approx 2.379 \)- \( C(1.5874) = 6(1.5874)^2 + \frac{48}{1.5874} \approx 44.90 \).
8Step 8: Conclusion
The minimum cost is obtained when \( x \approx 1.5874 \) and \( h \approx 2.379 \) with a total cost of approximately \( \$44.90 \).
Key Concepts
Optimization ProblemsCritical PointsVolume and Cost Calculations
Optimization Problems
Optimization problems are about finding the best solution within a defined set of constraints.
In this context, we want the crate's dimensions to minimize construction costs while meeting specific size and volume requirements.
We have conditions that the base must be square and between one and two feet, and the total volume must be 6 cubic feet.
To optimize, you start by understanding what you want to minimize or maximize. - Here, the focus is to minimize cost.
In engineering and mathematics, this often involves setting up an equation to represent the thing you want to optimize against the constraints. - Once you have this equation, the next step is to find values that either give the highest or lowest possible result, depending on your goal.
This process revolves around finding the **critical points** and evaluating them to find the optimal solution.
Think of it like searching for the perfect balance where all conditions are perfectly met and the result is least costly or most efficient.
In this context, we want the crate's dimensions to minimize construction costs while meeting specific size and volume requirements.
We have conditions that the base must be square and between one and two feet, and the total volume must be 6 cubic feet.
To optimize, you start by understanding what you want to minimize or maximize. - Here, the focus is to minimize cost.
In engineering and mathematics, this often involves setting up an equation to represent the thing you want to optimize against the constraints. - Once you have this equation, the next step is to find values that either give the highest or lowest possible result, depending on your goal.
This process revolves around finding the **critical points** and evaluating them to find the optimal solution.
Think of it like searching for the perfect balance where all conditions are perfectly met and the result is least costly or most efficient.
Critical Points
Critical points are values where a function's derivative is zero or undefined.
It's where potential maxima or minima can be located, making them central to solving optimization problems in calculus.
For the crate problem, finding the critical points allows us to determine which base length gives the minimal cost.- First, derive the cost function with respect to the base's side length, denoted as \( x \).
From our total cost equation \( C(x) = 6x^2 + \frac{48}{x} \), we compute the derivative: \( \frac{dC}{dx} = 12x - \frac{48}{x^2} \).- Setting \( \frac{dC}{dx} = 0 \) provides us the critical points, which requires solving \( 12x - \frac{48}{x^2} = 0 \).
These are potential points where the cost is either minimized or maximized.
Critical points are important, as they tell you where to focus your investigation to find the best possible solutions.
In practical scenarios, evaluating each critical point will reveal which represents the optimal solution under the given constraints.
It's where potential maxima or minima can be located, making them central to solving optimization problems in calculus.
For the crate problem, finding the critical points allows us to determine which base length gives the minimal cost.- First, derive the cost function with respect to the base's side length, denoted as \( x \).
From our total cost equation \( C(x) = 6x^2 + \frac{48}{x} \), we compute the derivative: \( \frac{dC}{dx} = 12x - \frac{48}{x^2} \).- Setting \( \frac{dC}{dx} = 0 \) provides us the critical points, which requires solving \( 12x - \frac{48}{x^2} = 0 \).
These are potential points where the cost is either minimized or maximized.
Critical points are important, as they tell you where to focus your investigation to find the best possible solutions.
In practical scenarios, evaluating each critical point will reveal which represents the optimal solution under the given constraints.
Volume and Cost Calculations
To tackle any optimization problem involving physical structures, understanding how volume and cost calculations relate is crucial.
In the crate problem, the relationship between the dimensions and cost is dictated by the cost of materials and the crate's volume.- We start with establishing that the crate needs a volume of 6 cubic feet.
Thus, for a square base of side \( x \) and height \( h \), the formula is: \( x^2 \times h = 6 \), giving us \( h = \frac{6}{x^2} \).- Then, calculate the cost for each part of the crate.
The bottom costs \( 5x^2 \), the top \( x^2 \), and the four sides \( 4xh \).
The total cost combines all these into: \( 5x^2 + x^2 + 8 \times \frac{6}{x} = 6x^2 + \frac{48}{x} \).- This setup gives us a clear picture of how changes in the dimensions of the base will directly affect the overall material cost.
This is critical for solving not just this problem, but any similar everyday problem where costs or resources need minimizing while meeting specific requirements.
In the crate problem, the relationship between the dimensions and cost is dictated by the cost of materials and the crate's volume.- We start with establishing that the crate needs a volume of 6 cubic feet.
Thus, for a square base of side \( x \) and height \( h \), the formula is: \( x^2 \times h = 6 \), giving us \( h = \frac{6}{x^2} \).- Then, calculate the cost for each part of the crate.
The bottom costs \( 5x^2 \), the top \( x^2 \), and the four sides \( 4xh \).
The total cost combines all these into: \( 5x^2 + x^2 + 8 \times \frac{6}{x} = 6x^2 + \frac{48}{x} \).- This setup gives us a clear picture of how changes in the dimensions of the base will directly affect the overall material cost.
This is critical for solving not just this problem, but any similar everyday problem where costs or resources need minimizing while meeting specific requirements.
Other exercises in this chapter
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