Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

What are the dimensions of the rectangular box, open at the top, that has maximum volume when the surface area is \(48 ?\)

Step-by-Step Solution

Answer

The dimensions of the rectangular box are 4 by 4 by 2.

Step 1: Define the Variables

Let's assign variables to the dimensions of the box. Let \( x \) be the length of the base, \( y \) be the width of the base, and \( h \) be the height of the box. Since the box is open at the top, the surface area consists of one bottom (which is \( xy \)) and four sides (which are \( 2xh + 2yh \)).

Step 2: Express the Surface Area Equation

For the open-top box, the total surface area is given by \( xy + 2xh + 2yh = 48 \). This will help us express one of the variables in terms of others.

Step 3: Express the Volume Equation

The volume \( V \) of the box is given by the formula \( V = x y h \). We aim to maximize this volume subject to the surface area constraint.

Step 4: Solve for One Variable from the Surface Area

From \( xy + 2xh + 2yh = 48 \), solve for \( h \) in terms of \( x \) and \( y \): \( h = \frac{48 - xy}{2x + 2y} \).

Step 5: Substitute in the Volume Equation

Substitute the expression for \( h \) in terms of \( x \) and \( y \) into the volume equation: \( V = xy \left(\frac{48 - xy}{2x + 2y}\right) \).

Step 6: Simplify and Differentiate

Simplify the expression for \( V \) and differentiate with respect to \( x \) and \( y \). Set these derivatives equal to zero to find the critical points. Solving these will involve some algebraic manipulation.

Step 7: Solve the Polynomial Equations

For simplification, suppose the base is a square, i.e., \( x = y \). Substitute in the surface area equation: \( x^2 + 4xh = 48 \) and solve for \( h \). Then, replace in the volume \( V = x^2h \) and find \( dV/dx \) to locate critical points.

Step 8: Identify Critical Points and Verify Maximum

Solve the derivative equations to find the dimensions that maximize the volume, given the constraint. Verify the critical points by confirming that second derivative test supports a maximum.

Step 9: Calculate the Dimensions

With \( x = y \), and solving the equations found, we calculate the optimal dimensions: \( x = y = 4 \) and \( h = 2 \).

Key Concepts

Surface Area Constraint

In problems about maximizing or minimizing quantities, constraints define the conditions that the solution must satisfy. Here, the surface area constraint ensures that the box's surface area remains consistent. An open-top rectangular box has its surface area calculated by considering one bottom and four sides.

• The bottom is simply the product of its length and width, represented by the equation: \( A_{bottom} = xy \).
• Each side has an area given by the side's height multiplied by its corresponding base length, meaning the total area for the sides is represented as: \( A_{sides} = 2xh + 2yh \).

Thus, the entire surface area formula becomes \( xy + 2xh + 2yh = 48 \). Understanding and formulating this equation forms the basis for how the constraint steers the solution: no matter how the dimensions are set, this total surface must always equal 48 units. Through this equation, one can express one variable in terms of others, which simplifies the optimization process later.

Volume Maximization

Maximizing the volume of a rectangle-shaped box while adhering to the surface area constraint is a classic optimization problem. The goal here is to express the volume in terms of fewer variables, using the constraint to reduce complexity.The volume of the box is calculated using:\[ V = x \, y \, h \]Where:

• \( x \) and \( y \) stand for the width and the length of the base.
• \( h \) is the height that needs to be determined.

Substituting the expression for \( h \) from the constraint in terms of \( x \) and \( y \) allows you to express the volume equation solely in terms of these two dimensions:\[ V = x \, y \, \left( \frac{48 - xy}{2x + 2y} \right) \]The purpose is to find the configuration of \( x \) and \( y \) that results in the largest possible volume. This involves algebraic steps that require simplifying and sometimes assuming reasonable conditions like \( x = y \) to make calculations more manageable.

Critical Points Identification

The role of critical points is central to optimization problems in calculus. They are where the derivative equals zero or is undefined, often leading to maxima or minima. By finding these points, you can determine when the volume is at its maximum.Once the volume equation is simplified to fewer variables, differentiate it with respect to the respective variables, starting with:

• Use \( \frac{\partial V}{\partial x} \) and \( \frac{\partial V}{\partial y} \) to locate where potential changes in direction occur.
• Setting these partial derivatives equal to zero helps identify critical points, which could be situations where the volume is maximized (or minimized).

For practical scenarios, like when assuming \( x = y \), use these to define a simpler quadratic equation. Solving for \( x \) will reveal potential points. And then, using the second derivative test can confirm whether these points correspond to a maximum volume. Calculating the dimensions under constraints, we derive the box's optimal lengths, ultimately yielding \( x = y = 4 \) and \( h = 2 \). These dimensions ensure maximum volume while respecting the surface area constraint.