Problem 70 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that of all three-dimensional boxes with a given surface area, the cube has the greatest volume. Hint: The surface area is \(S=2(l w+l h+h w)\) and the volume is \(V=l w h\). Let \(a=l w, b=l h\), and \(c=h w\). Use the previous problem to show that \(\left(V^{2}\right)^{1 / 3} \leq S / 6\). When does equality hold?

Step-by-Step Solution

Answer

A cube has the greatest volume for a given surface area.

Step 1: Understand the Problem

We want to prove that a cube has the greatest volume when constrained by a given surface area. Given the surface area formula \( S = 2(lw + lh + hw) \) and volume formula \( V = lwh \), our task is to show that \((V^2)^{1/3} \leq S/6 \).

Step 2: Calculate Variables for Box

Let \( a = lw \), \( b = lh \), and \( c = hw \). The surface area is then expressed as \( S = 2(a + b + c) \), and the volume is \( V = (abc)^{1/2} \).

Step 3: Apply the AM-GM Inequality

Using the Arithmetic Mean-Geometric Mean (AM-GM) Inequality: \( \frac{a + b + c}{3} \geq (abc)^{1/3} \). Therefore, we have \( \frac{S}{6} = \frac{2(a + b + c)}{6} \geq (abc)^{1/3} \), so \( (abc)^{1/3} \leq \frac{S}{6} \).

Step 4: Express Volume in Terms of Inequality

Since \( V = (abc)^{1/2} \), we can express \( V^2 = abc \). From the inequality \( (abc)^{1/3} \leq \frac{S}{6} \), taking the cube of both sides: \( (abc)^{1/3} \). This implies \( (V^2)^{1/3} \leq \frac{S}{6} \).

Step 5: Determine Condition for Equality

The condition for equality in the AM-GM inequality is when all terms are equal. Thus, \( a = b = c \) which implies \( lw = lh = hw \). Therefore, all edges must be equal, which means \( l = w = h \), forming a cube.

Key Concepts

Surface Area

The surface area of any three-dimensional shape is the sum of the areas of all its surfaces. For a box, this is calculated through the formula:
i \(S = 2(lw + lh + hw)\). Here, the variables \(l\), \(w\), and \(h\) represent the length, width, and height of the box, respectively.
This formula is derived from the fact that a box has three pairs of identical surfaces:

• Two surfaces with area \(lw\).
• Two surfaces with area \(lh\).
• Two surfaces with area \(hw\).

Using this mathematical representation, one can find various configurations of a box, all having the same surface area but possibly different volumes.
Understanding surface area is crucial when dealing with volume maximization, especially in geometric optimization problems like determining which shape has the maximum possible volume for a given surface.

Volume Maximization

The volume of a three-dimensional object refers to the space it occupies and is given by the formula \(V = lwh\) for a box. To maximize the volume of a box with a specified surface area is a common optimization problem in calculus. The formula is simple in appearance, yet it holds vast implications in understanding geometric properties.
For any object, maximizing the volume involves finding the dimensions that allow the most space within the object without changing its surface area. For instance, in the problem statement, the task is to show that a cube (where all sides are equal) provides the greatest volume for a given surface area.
Optimizing volume helps us efficiently use materials and space, influencing fields from packaging to architecture. In this context, achieving the most efficient use involves ensuring each dimension is equal, leading back to the importance of understanding the cubic form when dealing with constraints like surface area.

AM-GM Inequality

The Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality) is a fundamental concept in mathematics. It states that for any non-negative numbers \(a\), \(b\), and \(c\), the inequality \( \frac{a + b + c}{3} \geq (abc)^{1/3} \) holds true.
This inequality relates the average (arithmetic mean) of the numbers to their multiplied result (geometric mean). In the context of the optimization problem, it helps show how different choices of dimensions affect volume.
Through the AM-GM inequality, we illustrate that the volume of a box reaches its maximum when all three numbers (involved in forming that volume) are equal, leading to a cube.

• It is a vital tool for proofs in optimization problems.
• It helps set a framework for determining optimal solutions under constraints.
• The condition for equality in AM-GM (all terms being equal) directly correlates for volumes with maximum values.

Thus, the inequality is pivotal for deriving and understanding results related to volume maximization.