Problem 28 - Calculus with Concepts in Calculus

Question

A 12 -foot wire is cut into 12 pieces, which are soldered together to form a rectangular frame whose base is twice as long as it is wide (as in Figure 4.57). The frame is then covered with paper. a. How should the wire be cut if the volume of the frame is to be maximized? b. How should the wire be cut if the total surface area of the paper is to be maximized?

Step-by-Step Solution

Answer

a. Cut the wire into lengths of 2, 1, 1 feet for maximum volume. b. Cut the wire into lengths of \( \frac{9}{8}, \frac{15}{8}, \frac{9}{16} \) feet for maximum surface area.

Step 1: Define the Variables

Identify the variables to use in this problem. Let \( l \) be the length, \( w \) be the width, and \( h \) be the height of the rectangular frame. Since the base is twice as long as it is wide, we have \( l = 2w \).

Step 2: Write the Perimeter Equation

The total length of the wire is 12 feet, and since the wire forms the edges of the rectangle, the perimeter equation combining the edges can be written as \( 4w + 4h + 2l = 12 \). Substituting \( l = 2w \), we get \( 4w + 4h + 2(2w) = 12 \) simplifying to \( 8w + 4h = 12 \).

Step 3: Solve for Height \( h \) in Terms of Width \( w \)

Rearrange the equation from Step 2 to solve for \( h \): \[ 4h = 12 - 8w \] \[ h = 3 - 2w \]

Step 4: Derive Volume in Terms of \( w \)

The volume \( V \) of the rectangular frame is given by \( V = lwh = (2w)w(3-2w) \), which simplifies to: \[ V = 2w^2(3-2w) = 6w^2 - 4w^3 \]

Step 5: Maximize the Volume Function

Differentiate \( V \) with respect to \( w \) and set equal to zero to find critical points. \( V = 6w^2 - 4w^3 \) Differentiate to get \( V' = 12w - 12w^2 = 12w(1-w) \). Setting \( V' = 0 \) gives \( 12w(1-w) = 0 \), so \( w = 0 \) or \( w = 1 \). Since \( w = 0 \) does not make sense in context, use \( w = 1 \), then \( l = 2 \) and \( h = 1 \).

Step 6: Surface Area Derivation in Terms of \( w \)

The total surface area \( S \) is calculated as \( S = 2lw + 2lh + 2wh \), so \( S = 2(2w)w + 2(2w)(3-2w) + 2w(3-2w) \) which simplifies to \[ S = 4w^2 + 12w - 8w^2 + 6w - 4w^2 \] Further reduction gives \( S = 18w - 16w^2 \).

Step 7: Maximize the Surface Area Function

Differentiate \( S \) with respect to \( w \) and set equal to zero to find critical points. \( S = 18w - 16w^2 \) Differentiate to get \( S' = 18 - 32w \). Setting \( S' = 0 \) yields \( 18 = 32w \), so \( w = \frac{18}{32} = \frac{9}{16} \). Calculating \( l = 2(\frac{9}{16}) = \frac{18}{16} = \frac{9}{8} \) and substituting back in \( h = 3 - 2(\frac{9}{16}) = \frac{30}{16} = \frac{15}{8} \).

Key Concepts

rectangular prism volume

The volume of a rectangular prism is one of the fundamental concepts in geometry and calculus. In our exercise, the rectangular frame is modeled by a prism with a base and a height. The base of the frame is twice as long as it is wide, leading to the relationship \( l = 2w \), where \( l \) is the length and \( w \) is the width. To maximize volume, we first express the volume formula in terms of a single variable. Given that \( V = lwh \), substituting \( l = 2w \) and solving for \( h \) using the perimeter equation allows us to write the volume as a function of \( w \). This yields \( V = 6w^2 - 4w^3 \). Finding the maximum volume involves deriving its critical points and confirming any potential maxima or minima.

perimeter equations

Perimeter measures the total distance around a shape. In the case of a rectangular prism, the perimeter is derived from the sum of all its edges. The exercise introduces a constraint: a wire that is 12 feet long forms the edges, allowing us to set up a perimeter equation. The perimeter equation is crafted based on the relation \( 4w + 4h + 2l = 12 \). By substituting \( l = 2w \), the equation simplifies to \( 8w + 4h = 12 \). Simplifying perimeter equations is crucial in optimization problems to express key variables like \( h \) in terms of one other variable like \( w \), facilitating the process of finding critical points efficiently.

critical points

Critical points occur where the derivative of a function reaches zero or becomes undefined, often revealing maximum or minimum values. These points are instrumental in solving optimization problems. To find the critical points of the volume function \( V = 6w^2 - 4w^3 \), we differentiate with respect to \( w \) to get \( V' = 12w - 12w^2 \). By setting \( V' = 0 \), we find \( w = 0 \) or \( w = 1 \). The value \( w = 0 \) is not practical for this context as it negates the existence of a frame. Thus, \( w = 1 \) proves usable, giving us maximal dimensions. Validating critical points helps confirm whether they meet the problem's constraints and practical needs.

surface area maximization

Maximizing surface area often involves calculating derivatives just as with volume but applied to a different formula. The surface area \( S \) for our rectangular prism requires careful expression: \( S = 2lw + 2lh + 2wh \). After substituting dimensions and simplifying, the expression for \( S \) becomes \( S = 18w - 16w^2 \). Differentiating the surface area function \( S \) with respect to \( w \) gives \( S' = 18 - 32w \). Solving \( S' = 0 \) helps determine critical points necessary for optimizing the surface area. Once computed, these critical points, when practical, provide the dimensions \( l \), \( w \), and \( h \) under the given constraints to maximize surface area.