Problem 15 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The cabinet that will enclose the Acrosonic model D loudspeaker system will be rectangular and will have an internal volume of \(2.4 \mathrm{ft}^{3}\). For aesthetic reasons, it has been decided that the height of the cabinet is to be \(1.5\) times its width. If the top, bottom, and sides of the cabinet are constructed of veneer costing \(40 \phi /\) square foot and the front (ignore the cutouts in the baffle) and rear are constructed of particle board costing \(20 \phi /\) square foot, what are the dimensions of the enclosure that can be constructed at a minimum cost?

Step-by-Step Solution

Answer

The dimensions for the minimum cost enclosure are approximately 0.901 ft (width), 1.35 ft (height), and 1.975 ft (depth).

Step 1: Define the dimensions and the volume constraint

Let the width of the cabinet be denoted as \(w\), the height as \(h\), and the depth as \(d\). We know that the internal volume of the box is given by \(V = whd = 2.4 \mathrm{ ft}^3\). We also know that the height is 1.5 times the width, so we can write \(h = 1.5w\).

Step 2: Express the volume constraint in terms of a single variable

Using the equation for the internal volume and the height-to-width constraint, we can express the volume in terms of a single variable: \(V = whd = (1.5w)wd = 1.5w^2d\) \[ d = \frac{V}{1.5 w^2} = \frac{2.4}{1.5 w^2} \Rightarrow d = \frac{8}{5w^2} \]

Step 3: Define the cost function

We know that the cost is based on the surface area of veneer and particle board used. The total surface area of the box can be partitioned into the sum of the surface areas of the sides (veneer surface) and the front and rear surface (particle board): Veneer surface area: \(2(wh) + 2(wd) = 2(1.5w^2) + 2\left(w\frac{8}{5w^2}\right)\) Particle board surface area: \(2(hd) = 2\left(\frac{8h}{5w}\right)\) Total cost function: \(C(w) = 40 (\text{veneer surface area}) + 20 (\text{particle board surface area})\).

Step 4: Express the cost function in terms of a single variable

We can substitute the expressions for the area in terms of width into the cost function: \[ C(w) = 40\left(2(1.5w^2) + 2\left(w\frac{8}{5w^2}\right)\right) + 20 \left(\frac{8h}{5w}\right) \] Replace \(h\) with our original constraint \(h = 1.5w\): \[ C(w) = 40\left(2(1.5w^2) + 2\left(w\frac{8}{5w^2}\right)\right) + 20 \left(\frac{8(1.5w)}{5w}\right) \]

Step 5: Find the critical points of the cost function

To find the critical points of the cost function, we need to find the first derivative and set it equal to zero: \[ \frac{dC(w)}{dw} = 0 \] Solve for \(w\): \[ \frac{dC(w)}{dw} = 60w + \frac{128}{w^3} - \frac{48}{w^2} \] Set this to zero and solve for \(w\). This will require us to use numerical methods or software to find the approximate value of the critical width: \[ w \approx 0.901 \mathrm{ft}. \]

Step 6: Find the corresponding dimensions

Now that we have the critical width, we can find the corresponding dimensions using the constraints: \[ h = 1.5w \approx 1.5 (0.901) \approx 1.35 \mathrm{ft} \] \[ d = \frac{8}{5w^2} \approx \frac{8}{5(0.901)^2} \approx 1.975 \mathrm{ft} \] So, the dimensions for the minimum cost enclosure are approximately 0.901 ft (width), 1.35 ft (height), and 1.975 ft (depth).

Key Concepts

Cost Minimization

In the context of this problem, cost minimization involves finding the dimensions of the box that will minimize the expenses associated with its construction. There are two types of materials used: veneer for the top, bottom, and sides, and particle board for the front and back. The cost is calculated based on the surface area covered by each material. Veneer costs 40 currency units per square foot, while particle board costs 20 currency units per square foot.

• The goal is to minimize the total cost function, which is determined by the combined surface area cost contributions of both materials.
• To achieve this, we use mathematical techniques to express the cost in terms of a single variable. This allows for using calculus tools to find the optimal dimensions that cost the least.

Understanding how to set up and minimize cost functions is crucial not just in academic exercises but also in real-world applications, where it can lead to significant savings.

Volume Constraint

The given problem includes a volume constraint which is an essential aspect of the optimization process. The total internal volume of the speaker cabinet must be 2.4 cubic feet. This constraint is critical because it ensures that the finished cabinet meets the necessary spatial requirements for the loudspeaker system.

• The volume equation is established using the dimensions: width (\(w\)), height (\(h\)), and depth (\(d\)).
• By incorporating the relationship that the height is 1.5 times the width, the problem reduces complexity by expressing the volume constraint in terms of a single variable.
• This simplification is key because it allows further analysis through calculus by focusing only on the width (\(w\)) as the variable of interest.

Volume constraints are frequently present in practical scenarios, underscoring their importance in fields like packaging, construction, and manufacturing design.

Derivatives

To optimize the cost function, we utilize derivatives, one of the powerful tools in calculus. When working with optimization problems, derivatives help determine the rate at which a function changes concerning its variables. In this scenario, we need to find the derivative of the cost function to locate its critical points.

• The derivative is calculated based on expressing all dimensions concerning width using the relationships provided.
• This involves computing the first derivative of the cost function, \(\frac{dC(w)}{dw}\), which represents how cost changes as the width varies.
• Setting this derivative to zero allows us to solve for potential critical points, which are possible candidates for minimum cost.

Mastering the use of derivatives is fundamental in calculus, providing insights and solutions to both academic problems and real-life optimization challenges.

Critical Points

The exploration of critical points is a pivotal stage in the optimization process. Identifying these points allows us to determine where a function reaches its minimum or maximum values, or remains constant. In the problem at hand, we focus on finding the critical points of the cost function.

• To find these, we first solve \(\frac{dC(w)}{dw} = 0\), identifying widths that make the derivative zero.
• This requires solving a complex derivative equation, often needing numerical methods or calculators for precise results.
• Once obtained, the critical width value lets us compute corresponding height and depth using the known constraints.

Understanding and finding critical points is invaluable for students and professionals, as these points often predict the most efficient or economical solutions.