Problem 32 - Calculus with Concepts in Calculus

Question

A rectangular box without top is to have a volume of 32 cubic meters. Find the dimensions of such a box having the smallest possible surface area.

Step-by-Step Solution

Answer

The box dimensions with the smallest surface area are 4m x 4m x 2m.

Step 1: Define Variables and Equations

Let the dimensions of the rectangular box be denoted as follows: length = \( l \), width = \( w \), and height = \( h \). Since the volume \( V \) is given by \( V = l \times w \times h \), we substitute \( V = 32 \) to get \( l \times w \times h = 32 \).

Step 2: Express the Surface Area

The surface area \( S \) of the box (without the top) is given by \( S = lw + 2lh + 2wh \). Our goal is to minimize \( S \).

Step 3: Use the Volume to Express One Dimension

Solve for one of the variables in terms of the others using the volume constraint. For simplicity, express \( h \) in terms of \( l \) and \( w \): \( h = \frac{32}{lw} \).

Step 4: Substitute and Simplify the Surface Area Equation

Substitute \( h = \frac{32}{lw} \) into the surface area equation: \( S = lw + 2l\left(\frac{32}{lw}\right) + 2w\left(\frac{32}{lw}\right) = lw + \frac{64}{w} + \frac{64}{l} \).

Step 5: Minimize the Surface Area Using Calculus

To find the minimum surface area, take partial derivatives \( \frac{\partial S}{\partial l} \) and \( \frac{\partial S}{\partial w} \), set them to zero, and solve the resulting equations for \( l \) and \( w \). This step involves using the derivatives: \( \frac{\partial S}{\partial l} = 1 - \frac{64}{l^2} = 0 \) and \( \frac{\partial S}{\partial w} = 1 - \frac{64}{w^2} = 0 \). Solve \( l = w = 4 \).

Step 6: Find the Corresponding Height

Use the value of \( l = 4 \) and \( w = 4 \) to find \( h \) using the volume equation: \( h = \frac{32}{4\times4} = 2 \).

Step 7: Conclusion on Dimensions

The dimensions of the box with the smallest possible surface area are \( l = 4 \), \( w = 4 \), and \( h = 2 \).

Key Concepts

Rectangular Box

Imagine a rectangular box as a 3D shape made up of six sides, or faces, with all angles right angles. When dealing with boxes, it’s common to refer to three dimensions: length, width, and height. The volume of a rectangular box, at its simplest, is calculated by multiplying these three dimensions:

• Volume = length × width × height

For instance, in the given problem, the volume is specified as 32 cubic meters. We're trying to find the best dimensions for a box without a top, in terms of its surface area.
Rectangular boxes are useful because they simplify complex shapes into something easy to understand and calculate. When one dimension is changed, such as height, the whole situation adapts but remains straightforward. That makes rectangular boxes a great starting point for optimization problems in calculus.

Volume and Surface Area

In any practical problem involving a box, two primary attributes are typically optimized: volume and surface area. Volume is the amount of space a box can contain, while surface area is the total area covering the box.

• Surface Area = front area + back area + left area + right area + bottom area + (top area, if present)

For boxes without a top, like in the problem, the formula for surface area changes slightly. It includes the bottom, front, back, and sides but omits the top. Given:

• Total Surface Area = bottom area + 2 × front area + 2 × side area

In this box problem, we optimize surface area because we want to use the least material for construction while maintaining the required volume. Finding the minimal surface area encloses the volume efficiently, showcasing an essential principle in calculus for minimization.

Partial Derivatives

Partial derivatives are a calculus tool used to study how a function behaves as just one of several input variables is modified. They are especially handy when we aim to optimize functions of multiple variables.
In terms of a surface area function, which can change with length and width in a box scenario, partial derivatives help us find points where these changes result in maximum or minimum values. This process begins by calculating the derivative of the surface area function with respect to each variable independently.

• Set each derivative to zero to find critical points.

For our problem, taking the partial derivatives of the surface area concerning length and width and setting them to zero helps determine the optimal dimensions. This means that these derivatives being zero indicates no increase or decrease in surface area with small changes in dimensions, pointing us towards either a minimum or maximum. In optimization problems, this will hopefully lead to the dimensions that minimize our required surface area.