Problem 31 - Calculus for Biology and Medicine

Question

Find the minimum surface area of a rectangular closed (top, bottom, and four sides) box with volume \(64 \mathrm{~m}^{3}\).

Step-by-Step Solution

Answer

The minimum surface area is 96 m².

Step 1: Understand the problem

We need to find the minimum surface area of a closed rectangular box with a fixed volume of 64 cubic meters. The box has length (l), width (w), and height (h), and it is closed on all sides.

Step 2: Write down the volume and surface area formulas

For a rectangular box with dimensions length (l), width (w), and height (h), the volume is given by:\[ V = l imes w imes h \]Since the volume is 64, we have:\[ l imes w imes h = 64 \]The surface area (A) of the box is given by:\[ A = 2(lw + lh + wh) \]

Step 3: Use the volume constraint to express one variable in terms of the others

From the volume equation, we solve for height (h):\[ h = \frac{64}{l \cdot w} \]

Step 4: Substitute h into the surface area formula

Substitute the expression for h into the surface area equation:\[ A = 2(lw + l\left(\frac{64}{lw}\right) + w\left(\frac{64}{lw}\right)) \]Simplify the equation:\[ A = 2(lw + \frac{64}{w} + \frac{64}{l}) \]

Step 5: Find critical points to minimize the surface area

To find the critical points, first set the partial derivatives of the surface area equation with respect to l and w to zero. Calculate:\[ \frac{\partial A}{\partial l} = 2w - \frac{128}{l^2} = 0 \]Solving for \( l \), we find:\[ w = \frac{64}{l^2} \]Similarly for \( w \),\[ \frac{\partial A}{\partial w} = 2l - \frac{128}{w^2} = 0 \]Solving for \( w \), we find:\[ l = \frac{64}{w^2} \]

Step 6: Substitute critical points back into the volume equation

From the expressions \( w = \frac{64}{l^2} \) and \( l = \frac{64}{w^2} \), assume \( l = w \), due to symmetry. Then:\[ l^3 = 64 \]Taking the cube root gives:\[ l = w = 4 \] m. Consequently, find:\[ h = \frac{64}{l^2} = 4 \] m.

Step 7: Calculate the minimum surface area

Substitute \( l = 4 \), \( w = 4 \), \( h = 4 \) into the surface area formula:\[ A = 2(4 \times 4 + 4 \times 4 + 4 \times 4) = 2(3 \times 16) = 96 \] m².

Key Concepts

Surface Area Minimization

Surface area minimization is a fundamental problem in calculus optimization where the goal is to reduce the surface area of a given shape subject to certain conditions or constraints. In this particular problem, we're dealing with a rectangular box where the volume is fixed at 64 cubic meters. The objective is to find the box dimensions that will minimize its surface area while keeping the volume constant. To draw a clear picture:

• The box has a fixed volume, meaning the amount of space inside the box cannot change.
• The surface area is the sum of the areas of all the six faces (top, bottom, and four sides) of the box.

To minimize the surface area, calculus optimization typically involves finding where the derivative of the surface area function equals zero. This means these points are either a minimum or maximum value. In our scenario, the goal is to identify the critical points that represent a minimized surface area. This is achieved through mathematical techniques like substituting the volume constraint into the surface area equation, simplifying it, and then using calculus to find the values that minimize the equation.

Rectangular Box Volume Constraint

The volume constraint is an essential part of solving optimization problems involving geometric shapes. In this exercise, the constraint relates to the volume of a rectangular box, which must remain constant at 64 cubic meters. Here, the volume formula for the box is given by:\[ V = l \times w \times h \]Where:

• \( l \) is the length,
• \( w \) is the width, and
• \( h \) is the height.

The idea is to express one dimension in terms of the other two dimensions using this volume constraint. This often involves isolating one variable so that it can be substituted into another equation—in this case, the surface area equation. By rearranging the volume formula to express the height as:\[ h = \frac{64}{l \times w} \]This adjustment allows the complex problem to become simpler. The volume constraint is critical because it ensures that while minimizing the surface area, we do not alter the necessary volume of the box.

Partial Derivatives

Partial derivatives are a critical tool in multivariable calculus, especially when dealing with optimization problems involving functions of multiple variables, like our rectangular box's surface area. Understanding how changes in one variable affect the outcome while keeping others fixed is essential for calculus optimization.Given the surface area formula:\[ A = 2(lw + \frac{64}{w} + \frac{64}{l}) \]We need to find the partial derivatives with respect to both length \( l \) and width \( w \):

• \( \frac{\partial A}{\partial l} = 2w - \frac{128}{l^2} \)
• \( \frac{\partial A}{\partial w} = 2l - \frac{128}{w^2} \)

Setting these derivatives to zero allows us to find the critical points where the surface area is minimized. These critical points indicate where minor changes in \( l \) or \( w \) produce little to no change in surface area, signifying possible minimum values.Partial derivatives give a breakdown of how each variable specifically contributes to changes in surface area, identifying optimal sizes for the box dimensions.