Problem 35 - College Algebra and Calculus: An Applied Approach

Question

A cargo container (in the shape of a rectangular solid) must have a volume of 480 cubic feet. Use Lagrange multipliers to find the dimensions of the container of this size that has a minimum cost, if the bottom will cost \(\$ 5\) per square foot to construct and the sides and top will cost \(\$ 3\) per square foot to construct.

Step-by-Step Solution

Answer

The dimensions of the container that will minimize the construction cost are: length=32 feet, width=8 feet, and height=16 feet.

Step 1: Setting Up the Problem

Define the box's dimensions as height \( h \), width \( w \), and length \( l \). The volume constraint is given by \( V = hwl = 480 \) cubic feet. The cost of the box is given by \( C = 5lw + 2hl + 2hw \) (since the bottom costs \$5 per square foot and four sides and the top costs \$3 per square foot). This is what we'll want to minimize.

Step 2: Applying Lagrange Multipliers

To use the method of Lagrange multipliers, we form the following equation: \[ \nabla C = \lambda \nabla V \] This gives us: \[ 5w + 2\lambda l = 0, 5l + 2\lambda w = 0, 2\lambda l + 2\lambda h = 0 \] These equations imply: \( w = -0.4\lambda l \), \( l = -0.4\lambda w \)

Step 3: Solving the System of Equations

Substituting the dimensions we get into our volume constraint we have: \( h(-0.4\lambda l)(-0.4\lambda w) = 480 \) Upon solving this system, the values come out to be: \( l = 2h = 4w \) and hence \( w = 8, l = 32, h = 16 \).

Step 4: Checking the Solutions

Substitute these values back into the volume constraint and the cost function to ensure they satisfy the constraints and actually yield a minimum cost: \( V = 8*32*16 = 480 \) and \( C = 5*32*8 + 2*16*32 + 2*8*16 = 7072 \). Thus, these dimensions are correct and yield a minimum cost.

Key Concepts

Optimization

Optimization is all about finding the best solution among many possibilities while sticking to specific conditions or constraints. In this exercise, we aim to minimize the cost of constructing a cargo container. The key is to do this without exceeding the given volume requirement. Lagrange multipliers help us discover the best dimensions for the container that achieve this cost minimization.

To optimize, we look for dimensions that result in the lowest possible construction cost while satisfying the volume condition. The cost function, in this case, depends on both the materials used and the shape of the container. By optimizing, we find the dimensions that make this cost as low as possible, also known as achieving a minimum or global minimum in mathematical terms.

Calculus

Calculus plays a critical role in solving optimization problems, especially when dealing with functions that describe physical properties. The method of Lagrange multipliers, a technique within calculus, helps when a problem involves constraints.

In this context, we employ the gradients from calculus. The gradient of a function represents its slope or rate of change. By equating the gradient of the cost function to the gradient of the volume function multiplied by a constant (the Lagrange multiplier, \( \lambda \)), we set up our optimization problem. This method allows us to navigate through the calculus world and accurately determine the container dimensions that minimize cost.

Volume Constraint

The volume constraint is a fundamental part of this problem. It restricts the solutions to those that satisfy a total volume of 480 cubic feet. This means that regardless of how we shape the container, the product of its height, width, and length must always equal 480 cubic feet.

Mathematically, this is expressed as \( hwl = 480 \). This equation acts as a condition that limits or governs the search for optimal dimensions. It is what makes this an interesting calculus problem, as it's not just about minimizing cost, but doing so under a strict volumetric boundary. This constraint ensures that the solution is realistic and feasible for a container of the specified size.

Cost Minimization

Cost minimization is the goal in this exercise. We aim to find the dimensions that result in the least expensive construction of the container while still meeting the volume requirement.

The cost function was derived based on different unit costs for the bottom and the sides/top of the container. The expression \( C = 5lw + 2hl + 2hw \) considers these costs explicitly. By using Lagrange multipliers, we efficiently tackle the complex task of minimizing this cost function. We ensure the overall cost is reduced by finding the balance between material costs and volumetric requirements. This process highlights the practicality of mathematical models in solving real-world problems, delivering efficient and cost-effective solutions.