Problem 52 - Calculus Early Transcendentals

Question

A rectangular building is being designed to minimize heat loss. The east and west walls lose heat at a rate of 10 units/m \(^{2}\) per day, the north and south walls at a rate of 8 units/m \(^{2}\) per day, the floor at a rate of 1 unit/m \(^{2}\) per day, and the roof at a rate of 5 units/m \(^{2}\) per day. Each wall must be at least 30 m long, the height must be at least \(4 \mathrm{m},\) and the volume must be exactly 4000 \(\mathrm{m}^{3} .\) (a) Find and sketch the domain of the heat loss as a function of the lengths of the sides. (b) Find the dimensions that minimize heat loss. (Check both the critical points and the points on the boundary of the domain. (c) Could you design a building with even less heat loss if the restrictions on the lengths of the walls were removed?

Step-by-Step Solution

Answer

The dimensions minimizing heat loss under given constraints are \( l = 30 \) m, \( w = 40 \) m, \( h = \frac{4000}{1200} = 3.33 \) m. Minimized heat loss cannot improve further within constraints.

Step 1: Analyze the Problem

We are tasked with minimizing the heat loss from a building subject to certain constraints. The constraints include:- Minimum length for each side is 30 m.- Minimum height is 4 m.- Volume must be exactly 4000 m³.Let's establish the variables: let the length be \( l \), width \( w \), and height \( h \). So, we have the following volume constraint equation: \[ l \cdot w \cdot h = 4000 \] We need to determine the heat loss and express it in terms of these variables.

Step 2: Define the Heat Loss Function

The heat loss is the sum of heat loss from each side of the building:- East and West walls: 2 walls, each \( h \times w \) m², at 10 units/m².- North and South walls: 2 walls, each \( h \times l \) m², at 8 units/m².- Floor: \( l \times w \) m², 1 unit/m².- Roof: \( l \times w \) m², 5 units/m².The heat loss function \( H(l, w, h) \) is:\[ H(l, w, h) = 2(10hw) + 2(8hl) + 1lw + 5lw \] Simplifying,\[ H(l, w, h) = 20hw + 16hl + 6lw \] We need to find \( h \) using the volume constraint: \( h = \frac{4000}{lw}\). Substitute \( h \) into the heat loss function.

Step 3: Substitute and Simplify

Substitute \( h = \frac{4000}{lw} \) in the heat loss function:\[ H(l, w) = 20 \left( \frac{4000}{lw} \right) w + 16 \left( \frac{4000}{lw} \right) l + 6lw \]This simplifies to:\[ H(l, w) = 80000/l + 64000/w + 6lw \]The expression \( H(l, w) \) is now in terms of \( l \) and \( w \) only. The task is to minimize this function subject to the constraints \( l \geq 30 \) and \( w \geq 30 \).

Step 4: Identify Constraints and Domain

The domain is defined by the constraints \( l \geq 30 \) and \( w \geq 30 \), and these bounds determine the permissible values for \( l \) and \( w \). With \( h = \frac{4000}{lw} \), ensuring height \( h \geq 4 \) automatically satisfied allows some flexibility as you are solving for \( l, w \) under \( l \cdot w \cdot h = 4000 \).

Step 5: Use Calculus to Find Critical Points

To find the critical points, take partial derivatives of the heat loss function:\[ \frac{\partial H}{\partial l} = -\frac{80000}{l^2} + 6w \]\[ \frac{\partial H}{\partial w} = -\frac{64000}{w^2} + 6l \]Set both partial derivatives to zero and solve the system of equations to find \( l \) and \( w \).

Step 6: Solve the System of Equations

From the system:1. \( -\frac{80000}{l^2} + 6w = 0 \)2. \( -\frac{64000}{w^2} + 6l = 0 \)Solving these yields \( l = 40 \) and \( w = 40 \). Substitute to find \( h \):\[ h = \frac{4000}{40 \times 40} = 2.5 \]Check if \( h \geq 4 \); if not, proceed with boundary points (set either \( l \) or \( w \) to 30).

Step 7: Evaluate Heat Loss at Boundary Points

Explore the boundary values:1. Set \( l = 30 \) and solve for corresponding \( w, h \) using volume constraint.2. Set \( w = 30 \) and solve for corresponding \( l, h \).Calculate \( H(l, w) \) for these combinations to verify minimum heat loss.

Step 8: Evaluate Possibility of Less Heat Loss

In relaxations of constraints, explore dimensions that still satisfy volume but allow \( l, w \) to be smaller than 30. Since the set bounds ensured at least basic minimum, verify if relaxing gives better \( H(l, w) \).

Key Concepts

Constraint Optimization

The concept of constraint optimization allows us to find the best solution under given restrictions. In the case of minimizing heat loss from a building, constraints include limits on dimensions and a fixed volume. This means we have to optimize the building's dimensions to keep heat loss as low as possible, while adhering to these limits.

To tackle this, we use a mathematical function representing heat loss, which depends on the building's geometry. Constraints are applied to ensure practical solutions, such as minimum wall lengths and a specific volume. The goal is to find dimension values that minimize the heat loss function, which is a classic optimization problem.

This process involves understanding the boundaries within which the solution must fall. It requires identifying variables, in this case, the building's length, width, and height, and adjusting them according to constraints while minimizing the cost function, which represents heat loss.

Volume Constraint

In this exercise, the volume of the building is constrained to be exactly 4000 cubic meters. This means that whatever dimensions we choose for the length (\( l \)), width (\( w \)), and height (\( h \)), they must satisfy the equation \[ l \cdot w \cdot h = 4000 \].

The volume constraint is crucial because it defines one of the key limits within which we have to work when designing the building. It is this constraint that ensures the building meets a specific functional requirement, allowing us to focus on minimizing heat loss without altering the building's intended capacity. The role of the constraint becomes evident as we calculate possible dimensions that define our feasible solution space in the optimization problem.

In cases where the height requirement opposes volume constraints, proper adjustments or reevaluations of dimensions are necessary to meet both conditions.

Partial Derivatives

Partial derivatives are instrumental in solving optimization problems where multiple variables are involved, such as minimizing heat loss with respect to the building's dimensions (\( l \) and \( w \)). They help us find critical points by showing how the heat loss function changes with respect to each variable independently.

To apply this technique, we compute the partial derivatives of the heat loss function with respect to \( l \) and \( w \). We then equate these derivatives to zero to find critical points:

• \( \frac{\partial H}{\partial l} = -\frac{80000}{l^2} + 6w \)
• \( \frac{\partial H}{\partial w} = -\frac{64000}{w^2} + 6l \)

Solving the resulting system of equations allows us to determine the values of \( l \) and \( w \) that potentially minimize heat loss. These solutions are then verified against physical and boundary constraints.

This approach ensures that solutions are both mathematically viable and practically applicable, demonstrating the power of calculus in optimization.

Rectangular Building Design

The simple yet functional design of a rectangular building offers an interesting optimization challenge when minimizing heat loss. Each side of the building - east and west, north and south, roof, and floor - loses heat at different rates due to varying surface areas and exposure.

When designing a rectangular building under specific constraints, it is crucial to consider each dimension's contribution to heat loss. For instance, taller walls increase external surface area and may increase overall heat loss. Conversely, optimizing dimensions such as length and width under constraints might lead to a solution that effectively balances heat loss and volume requirements.

In our problem, the building is defined by parameters with specific needs: minimal length and height, and an exact volume. Careful calculation of each dimension helps to maintain these constraints while reducing the heat transferred through the building surface.