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.

Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter9: Systems Of Linear Equations
Section: Chapter Questions
Problem 10MPS
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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 c and its height is h. Then the sides
have area ch. 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 + 2x2
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 r:
800 = 4xh +2x² → h =
800 2x2
4x
Now, the volume we wish to maximize is
Volume = x · 2x · h = 2x²
800-2x2
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) = -6z. Since f"(x) goes from positive to negative at r = 0, the absolute maximum
must be at I= 0.
19
tv
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 c and its height is h. Then the sides have area ch. 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 + 2x2 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 r: 800 = 4xh +2x² → h = 800 2x2 4x Now, the volume we wish to maximize is Volume = x · 2x · h = 2x² 800-2x2 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) = -6z. Since f"(x) goes from positive to negative at r = 0, the absolute maximum must be at I= 0. 19 tv
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