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. PART A: This student has made multiple errors. Find at least three independent errors the student has made.PART B: Write down the correct response the student should have given, showing all your work as usual. STUDENT RESPONSE: Suppose the length of the box is x and its height is h. Then the sideshave area xh. Meanwhile, the width is 2x (twice the length), making the base of the box have area2x · x = 2x2. Thus, the surface area of the box isS = 4xh + 2x?Since we know the surface area will be 800 (the amount of material we have to work with), we canuse this to solve for h inrms of x:800 – 2x2-800 = 4xh + 2x2 → h =4xNow, the volume we wish to maximize is800 – 2x2-Volume = x · 2x · h = 2x²4xyielding our objective functionf (x)400x – x* & Domain = (0, 800]-Finally, we wish to find the absolute maximum of f(x) (the maximum volume) by examining thesign of f"(x) = -6x. Since f"(x) goes from positive to negative at x = 0, the absolute maximummust be at x = 0.

(A) To determine the errors in the students response. (1) The surface area of the open-top cardboard…

Answered: We wish to construct the world's…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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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.

PART A: This student has made multiple errors. Find at least three independent errors the student has made.

PART B: Write down the correct response the student should have given, showing all your work as usual.

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 · x = 2x2. Thus, the surface area of the box is
S = 4xh + 2x?
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
rms of x:
800 – 2x2
-
800 = 4xh + 2x2 → h =
4x
Now, the volume we wish to maximize is
800 – 2x2
-
Volume = x · 2x · h = 2x²
4x
yielding our objective function
f (x)
400x – x* & 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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