Problem 32 - Calculus for Biology and Medicine

Question

Find the minimum surface area of a rectangular open (bottom and four sides, no top) box with volume \(256 \mathrm{~m}^{3}\).

Step-by-Step Solution

Answer

The minimum surface area is achieved when \(l = w = \sqrt[3]{128}\) and \(h = \frac{256}{128}\).

Step 1: Define the Variables

To find the dimensions of the box that minimize the surface area, let's define the variables. Let the length be \( l \), the width be \( w \), and the height be \( h \) of the box. The volume is given by \( V = lwh = 256 \mathrm{~m}^{3} \).

Step 2: Express Surface Area in Terms of Dimensions

The surface area of the open box (without the top) is given by the formula \( A = lw + 2lh + 2wh \). Our aim is to express the surface area \( A \) in terms of two variables instead of three.

Step 3: Substitute Volume into Surface Area Equation

From the volume equation \( lwh = 256 \), express \( h \) in terms of \( l \) and \( w \): \( h = \frac{256}{lw} \). Substitute this into the surface area equation: \[A = lw + 2l\left(\frac{256}{lw}\right) + 2w\left(\frac{256}{lw}\right) \]Simplify to get \[A = lw + \frac{512}{w} + \frac{512}{l} \]

Step 4: Set Up the Optimization Problem

We now have a function of two variables: \( A(l, w) = lw + \frac{512}{w} + \frac{512}{l} \). To find the minimum surface area, we need to take partial derivatives with respect to \( l \) and \( w \), and solve them to find the critical points.

Step 5: Calculate Partial Derivatives

Calculate the partial derivatives:\( \frac{\partial A}{\partial l} = w - \frac{512}{l^2} \)\( \frac{\partial A}{\partial w} = l - \frac{512}{w^2} \).Set each partial derivative to zero and solve for \( l \) and \( w \): \( w = \frac{512}{l^2} \) and \( l = \frac{512}{w^2} \).

Step 6: Solve the System of Equations

Substitute \( w = \frac{512}{l^2} \) into \( l = \frac{512}{w^2} \):\( l = \frac{512}{\left(\frac{512}{l^2}\right)^2} = \frac{512l^4}{512^2} \). Simplifying gives \( l = w \).

Step 7: Find the Dimensions

With \( l = w \), substitute into the volume equation \( (l^2)(h) = 256 \):\( h = \frac{256}{l^2} \). Substitute \( h \) back into the surface area equation to find consistent values:If \( l = w = h = \sqrt[3]{256/2} \).

Step 8: Verify Solution

Verify by checking if dimensions give correct volume and minimal computation. Substitute back \( l = w = \sqrt[3]{128} \) into volume and surface area equations to check they fit. Confirm that surface area is minimized and \( v = 256 \mathrm{~m}^{3} \) holds.

Key Concepts

Surface Area Minimization

Finding the minimum surface area of a geometric shape, like our rectangular box without a top, involves using calculus techniques to reduce its surface coverage while holding its volume constant. In our case, we are tasked with minimizing the surface area of a box with given constraints.

To start, it's essential to understand how the shape's surface area is calculated. For the open box we're considering, the surface area is found using the formula:

• The box has a base area of \( lw \), where \( l \) is the length and \( w \) is the width.
• Additionally, the box has four side panels, but since it's open-topped, only two heights are considered: \( 2lh + 2wh \).

We need to express this surface area in terms of only two variables rather than three, which helps streamline our work using calculus. By substituting one variable (here, height \( h \)) using the volume condition, we could rewrite the surface area in terms of \( l \) and \( w \) only.

This minimization technique is valuable in practical scenarios to ensure material efficiency, addressing industrial needs where eliminating waste is crucial.

Partial Derivatives

Partial derivatives are powerful tools in calculus, helping to find the rate at which a function changes along each independent variable. Particularly in this exercise, partial derivatives help pinpoint the minimum of our surface area.By taking the partial derivatives of the surface area equation \( A(l, w) = lw + \frac{512}{w} + \frac{512}{l} \), we gain insight into how changes in length and width affect the total surface area.For instance:

• The partial derivative with respect to \( l \), \( \frac{\partial A}{\partial l} = w - \frac{512}{l^2} \), gauges how surface area varies when only the length changes.
• Similarly, \( \frac{\partial A}{\partial w} = l - \frac{512}{w^2} \) focuses on changes when adjusting only the width.

By setting these derivatives to zero, we're finding points where the slope of the surface area "curve" in relation to each variable is flat — indicating potential minima, maxima, or saddle points. Solving these equations allows us to identify the dimension values where the surface area can be minimized.

Rectangular Box Optimization

Optimization in the context of an open rectangular box focuses on making the best or most effective use of the box's dimensions to fulfill the volume requirement while minimizing its surface area. This problem has practical applications in design, manufacturing, and logistics.The optimization process involves:

• Establishing a clear **objective function**, in this case, minimizing the surface area \( A \).
• Ensuring **constraints** — like the box's volume \( lwh = 256 \) — are maintained through calculations.
• Utilizing mathematical techniques to discern the smallest achievable surface area while also meeting the volume requirement.

By substituting derived expressions into the volume equation, and using logical reasoning with mathematical tools, we've determined that the optimal dimensions occur when the length equals the width. Solving and verifying provides a solution that adheres to all constraints and delivers the minimum surface area possible given the specified volume.

This exercise not only hones calculus skills but also deepens understanding of optimizing real-world scenarios where resource efficiency is fundamental.