Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

A steel box without a lid having volume 60 cubic feet is to be made from material that costs \(\$ 4\) per square foot for the bottom and \(\$ 1\) per square foot for the sides. Welding the sides to the bottom costs \(\$ 3\) per linear foot and welding the sides together costs \(\$ 1\) per linear foot. Find the dimensions of the box that has minimum cost and find the minimum cost. Hint: Use symmetry to obtain one equation in one unknown and use a CAS or Newton's Method to approximate the solution.

Step-by-Step Solution

Answer

The minimum cost dimensions are \(x \approx 3.9, y \approx 3.9, h \approx 3.96\) and the minimum cost is approximately $192.48.

Step 1: Represent the Box Dimensions

Let the dimensions of the box be \(x\) and \(y\) for the length and width of the base, and \(h\) for the height. The volume condition of the box without a lid gives us the equation \(xyh = 60\).

Step 2: Write the Cost Function

The cost involves three parts: material cost, cost of welding sides to the bottom, and the cost of welding sides together. The material cost is given by \(4xy\) for the bottom and \(1(2xh + 2yh)\) for the sides. Therefore, the total material cost is \(4xy + 2xh + 2yh\). Welding costs are \(3(x + y)\) for welding sides to the bottom and \(1(2h)\), resulting in a welding cost \(3(x + y) + 2h\). The total cost is \[ C = 4xy + 2xh + 2yh + 3(x + y) + 2h. \]

Step 3: Simplify Using the Volume Constraint

From the volume constraint \(h = \frac{60}{xy}\), substitute \(h\) in the cost function to have an expression in terms of \(x\) and \(y\): \[ C = 4xy + \frac{120x}{xy} + \frac{120y}{xy} + 3(x + y) + \frac{120}{xy}. \] Simplifying gives: \[ C = 4xy + \frac{120}{y} + \frac{120}{x} + 3(x + y) + \frac{120}{xy}. \]

Step 4: Symmetry and Simplified Equation

Using symmetry, assume \(x = y\) for simplicity, giving us the dimension condition \(x^2h = 60\), so \(h = \frac{60}{x^2}\) and the cost function reduces to: \[ C = 4x^2 + \frac{240}{x} + 6x + \frac{120}{x^2}. \]

Step 5: Optimization using Derivative

Find the derivative \(C'(x)\) of the simplified cost function and solve \(C'(x) = 0\) for critical points. Using calculus tools to differentiate, we find: \( \frac{d}{dx}(4x^2 + \frac{240}{x} + 6x + \frac{120}{x^2}) \), simplify and solve, we reach: \[ C'(x) = 8x - \frac{240}{x^2} + 6 - \frac{240}{x^3} = 0. \] Calculate using a Computational Algebra System (CAS) or Newton's method to approximate \(x\).

Step 6: Calculate y and h

Once \(x\) is found, use the symmetry \(y = x\) and calculate \(h = \frac{60}{x^2}\) to find the box dimensions.

Step 7: Calculate Minimum Cost

Substituting back into the cost function \(C\) with the computed dimensions \(x\), \(y\), and \(h\) will give the minimum cost.

Key Concepts

Newton's Method

Newton's Method is a practical tool for finding successively better approximations to roots (or zeroes) of a real-valued function. It is especially valuable in situations where algebraic solutions are complex or impossible to determine analytically.
In the context of the steel box optimization problem, Newton's Method is used to approximate the value of the dimension that minimizes the cost. Newton's Method works by starting with an initial guess, and then iteratively improving that guess using the function's derivative. Here’s the basic approach:

• Choose a function that represents the problem, such as the derivative of the cost function: \( C'(x) \).
• Guess an initial value for the variable \( x \).
• Apply the iterative process: \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).
• Repeat the process to converge on a more accurate solution.

Using a tool like CAS (Computational Algebra System) can make applying Newton's Method even easier, as it computes all necessary derivatives and iterations, providing a convenient way to reach a solution.

Cost Function

A cost function is a mathematical expression used to determine the total costs associated with a particular activity, under given conditions. In optimization exercises, the goal is usually to find the scenario where this cost is minimized or maximized.
For the steel box problem, the cost function considers both the material costs and welding expenses. It is formulated as:

• Material cost: \(4xy\) for the bottom, and \(2xh + 2yh\) for the sides.
• Welding cost: \(3(x+y)\) for welding sides to the bottom, and \(2h\) for welding sides together.

Combining these factors gives the overall cost function:\[ C = 4xy + 2xh + 2yh + 3(x + y) + 2h. \]
The goal is to find the dimensions that minimize this cost function. By substituting the volume constraint to express \(h\) in terms of \(x\) and \(y\), the problem simplifies further, allowing us to effectively apply optimization methods.

Symmetry in Optimization

Symmetry is a useful concept in mathematical optimization. It simplifies complex problems by reducing the number of variables or equations involved. This is achieved by assuming certain symmetrical properties about the system or geometry in question.
In this specific optimization problem, symmetry is used by setting the box's length \(x\) equal to its width \(y\). This helps to simplify the cost function considerably as you only deal with one variable instead of two.
Thus, the reduced problem can be expressed with a single variable equation: \(x^2h = 60\), which allows determining the height \(h\) when \(x\) is known. Such simplification reduces both computation and possible errors, focusing attention only on critical aspects of cost-reduction.

Volume Constraint

In optimization problems involving physical objects, constraints often play a critical role in defining feasible solutions. The volume constraint in the steel box problem expresses that no matter how the dimensions of the box change, its volume must remain constant at 60 cubic feet.
The mathematical form of this constraint is given by the equation \(xyh = 60\), linking the box’s length \(x\), width \(y\), and height \(h\) together.
To integrate this constraint into the cost function, we express \(h\) in terms of \(x\) and \(y\): \(h = \frac{60}{xy}\). This substitution helps simplify the optimization calculations, as it lowers the number of independent variables in the cost function, allowing for a more straightforward application of optimization techniques like Newton's Method. These constraints ensure the solution not only minimizes cost but also adheres to the physical specifications of the problem.