Problem 19 - Calculus Early Transcendentals

Question

A box with a square base is taller than it is wide. In order to send the box through the U.S. mail, the height of the box and the perimeter of the base can sum to no more than 108 in. What is the maximum volume for such a box?

Step-by-Step Solution

Answer

The maximum volume of the box is 11664 cubic inches.

Step 1: Define the Variables

Let the side length of the square base be \( x \) inches. Then the perimeter of the base is \( 4x \) inches. Let the height of the box be \( h \) inches.

Step 2: Establish the Constraint

The constraint given by the problem is that the sum of the height and the perimeter of the base cannot exceed 108 inches. This means:\[h + 4x \leq 108\]

Step 3: Express Height in Terms of Base Length

Rearrange the constraint equation to solve for \( h \):\[h = 108 - 4x\]

Step 4: Write the Volume Equation

The volume \( V \) of the box is given by the product of the area of the base and the height:\[V = x^2 imes h = x^2 (108 - 4x)\]

Step 5: Simplify the Volume Equation

Expand the expression for \( V \):\[V = 108x^2 - 4x^3\]

Step 6: Find the Critical Points

To find the maximum volume, take the derivative of \( V \) with respect to \( x \) and set it equal to zero:\[\frac{dV}{dx} = 216x - 12x^2 = 0\]Factor the equation:\[12x(18 - x) = 0\]Thus, the critical points are \( x = 0 \) and \( x = 18 \).

Step 7: Test the Critical Points

Since \( x = 0 \) does not make sense (the box would have no base), we test \( x = 18 \). Substitute it back into the expression for \( V \):\[V = 108(18)^2 - 4(18)^3 = 11664 - 23328 = 0\]But we find that the maximum point occurs at the boundary constraint \( x = 18 \), where\( h = 108 - 72 = 36 \).

Step 8: Evaluate the Volume Expression with the Boundary

Calculate the volume for the configuration where \( x = 18 \) and \( h = 36 \):\[V = x^2 h = 18^2 imes 36 = 11664\].

Key Concepts

Volume Maximization

Volume maximization is an essential aspect of calculus optimization problems. In this exercise, we are tasked with finding the maximum volume of a box with a square base and a height greater than its width.
The formula for volume, which is a function of the base side length and height, is the key to unlocking this maximum volume. To write it algebraically, the expression for volume is:

• The area of the base, which is \(x^2\).
• The height, noted as \(h\).
• Thus, the volume equation becomes \(V = x^2 h\).

Here, \(x\) is the side length, and \(h\) is the height. Understand that maximizing volume involves finding the best combination of \(x\) and \(h\) within the given constraints. By substituting the expression for \(h\) into the volume equation, we can focus directly on how \(x\) changes the volume.

Constraint Equations

Constraint equations are conditions that must be satisfied in optimization problems. In our task, the constraint ties together the base perimeter and the box’s height. The problem states:

• The perimeter of the base: \(4x\).
• The height of the box \(h\).
• Their sum cannot exceed 108 inches: \(h + 4x \leq 108\).

With this constraint, we balance the box dimensions to not violate the mailing requirements. Rearranging the constraint equation to express \(h\) in terms of \(x\) is crucial to simplifying the problem. Solving for \(h\) gives \(h = 108 - 4x\), which is then used to redefine the volume equation, ensuring that every potential solution adheres to the limitation.

Derivative Applications

Derivatives are powerful tools for identifying critical points in optimization problems, like finding the maximum volume of our box.
To locate these turning points, we differentiate the volume function with respect to \(x\):

• The derivative \(\frac{dV}{dx}\) of the volume function \(V = 108x^2 - 4x^3\).
• This results in: \(\frac{dV}{dx} = 216x - 12x^2\).

Setting \(\frac{dV}{dx} = 0\) helps find the critical points where changes occur. Factoring gives us \(12x(18 - x) = 0\), leading to critical values \(x = 0\) and \(x = 18\). These calculations establish points to test within the allowable dimensions. Derivatives clarify where maximum or minimum values of the volume may occur by disrupting a steady state of the function.

Critical Points

Critical points are crucial in determining where a maximum or minimum value occurs within a function.
This stage involves taking the derivative and examining the resulting values where it equals zero.

• For this box problem, we derived critical points \(x = 0\) and \(x = 18\).
• Testing these points involves substituting them back into our volume formula.

Through analysis, \(x = 0\) was shown as impractical, hinting at zero volume, so \(x = 18\) becomes more viable. Further checks confirm \(x = 18\) aligns with the constraint and represents a realistic maximum. Critical points show pivotal "turns" in a problem where analysis and testing are mandatory for confirming maximum gains.

Mathematical Modeling

Mathematical modeling allows us to represent real-world problems using equations and variables.
Through modeling, a written problem translates into solvable numeric equations.

• In this exercise, a verbal constraint of base perimeter and height fit into \(h + 4x \leq 108\).
• The problem translates into a volume equation \(V = x^2 (108 - 4x)\).
• Then, critical mathematical operations, like taking derivatives, help find optimal solutions.

Modeling bridges word problems with algebraic expressions and solutions, helping us predict outcomes within set boundaries. Good models guide us in predicting practical, efficient solutions, effectively converting constraints into benefits. This strategy allows engineers and scientists to innovate, using accessible and understandable results.