We wish to construct the world's largest open-top cardboard box. We have exactly 800 square feet of cardboard to use, and the box we make must (due to Amazon regulations) be twice as wide as it is long. Of all the dimensions we could choose from, which box dimensions will result in the box with the largest volume? List the dimensions for the bottom of the box, i.e., no need to find the final height of the box.

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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a) the student has made multiple errors. Find at least three errors the student has made B) write down the correct response the student should have given, showing all work
We wish to construct the world's largest open-top cardboard box. We have exactly 800 square feet of cardboard to use, and the box we make must (due to Amazon regulations) be twice as wide as it is long. Of all the dimensions we could choose from, which box dimensions will result in the box with the largest volume? List the dimensions for the bottom of the box, i.e., no need to find the final height of the box.

**Student Response:**
Suppose the length of the box is \( x \) and its height is \( h \). Then the sides have area \( xh \). Meanwhile, the width is \( 2x \) (twice the length), making the base of the box have area \( 2x \cdot x = 2x^2 \). Thus, the surface area of the box is

\[
S = 4xh + 2x^2
\]

Since we know the surface area will be 800 (the amount of material we have to work with), we can use this to solve for \( h \) in terms of \( x \):

\[
800 = 4xh + 2x^2 \implies h = \frac{800 - 2x^2}{4x}
\]

Now, the volume we wish to maximize is

\[
\text{Volume} = x \cdot 2x \cdot h = 2x^2 \left( \frac{800 - 2x^2}{4x} \right)
\]

yielding our objective function

\[
f(x) = 400x - x^3 \quad \text{& Domain = (0, 800]}
\]

Finally, we wish to find the absolute maximum of \( f(x) \) (the maximum volume) by examining the sign of \( f''(x) = -6x \). Since \( f''(x) \) goes from positive to negative at \( x = 0 \), the absolute maximum must be at \( x = 0 \).
Transcribed Image Text:We wish to construct the world's largest open-top cardboard box. We have exactly 800 square feet of cardboard to use, and the box we make must (due to Amazon regulations) be twice as wide as it is long. Of all the dimensions we could choose from, which box dimensions will result in the box with the largest volume? List the dimensions for the bottom of the box, i.e., no need to find the final height of the box. **Student Response:** Suppose the length of the box is \( x \) and its height is \( h \). Then the sides have area \( xh \). Meanwhile, the width is \( 2x \) (twice the length), making the base of the box have area \( 2x \cdot x = 2x^2 \). Thus, the surface area of the box is \[ S = 4xh + 2x^2 \] Since we know the surface area will be 800 (the amount of material we have to work with), we can use this to solve for \( h \) in terms of \( x \): \[ 800 = 4xh + 2x^2 \implies h = \frac{800 - 2x^2}{4x} \] Now, the volume we wish to maximize is \[ \text{Volume} = x \cdot 2x \cdot h = 2x^2 \left( \frac{800 - 2x^2}{4x} \right) \] yielding our objective function \[ f(x) = 400x - x^3 \quad \text{& Domain = (0, 800]} \] Finally, we wish to find the absolute maximum of \( f(x) \) (the maximum volume) by examining the sign of \( f''(x) = -6x \). Since \( f''(x) \) goes from positive to negative at \( x = 0 \), the absolute maximum must be at \( x = 0 \).
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