Ch. 4 - Applications of the Derivative

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 4 - Applications of the Derivative

Problem 4.5.21

Chapter 4, Problem 4.5.21

Shipping crates A square-based, box-shaped shipping crate is designed to have a volume of 16 ft³. The material used to make the base costs twice as much (per square foot) as the material in the sides, and the material used to make the top costs half as much (per square foot) as the material in the sides. What are the dimensions of the crate that minimize the cost of materials?

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Define the variables: Let x be the side length of the square base and h be the height of the crate. The volume constraint is given by the equation V = x^2 * h = 16 ft³.

Express the height h in terms of x using the volume constraint: h = 16 / x^2.

Set up the cost function: Let C_base, C_side, and C_top be the costs per square foot for the base, sides, and top, respectively. Given that C_base = 2 * C_side and C_top = 0.5 * C_side, the total cost function is C = 2 * C_side * x^2 + 4 * C_side * x * h + 0.5 * C_side * x^2.

Substitute h from step 2 into the cost function: C = 2 * C_side * x^2 + 4 * C_side * x * (16 / x^2) + 0.5 * C_side * x^2. Simplify this expression to find C in terms of x.

Differentiate the cost function with respect to x, set the derivative equal to zero, and solve for x to find the critical points. Use the second derivative test or analyze the behavior of the cost function to determine which critical point minimizes the cost.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Volume of a Box

The volume of a box is calculated by multiplying its length, width, and height. For a square-based crate, the volume formula simplifies to V = x²h, where x is the length of a side of the base and h is the height. Understanding how to express the volume in terms of the dimensions is crucial for setting up the problem correctly.

Cost Function

The cost function represents the total cost of materials used in constructing the crate. It is derived from the areas of the base, sides, and top, each multiplied by their respective costs per square foot. Formulating this function accurately is essential for determining the dimensions that minimize the overall cost.

Optimization Techniques

Optimization techniques, such as finding critical points using derivatives, are used to minimize or maximize functions. In this context, applying calculus to the cost function allows us to find the dimensions of the crate that yield the lowest cost. Understanding how to differentiate and analyze these functions is key to solving the problem.

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