The outside dimensions of a rectangular box (bottom and four sides, no top), made of aluminum, are 24 by 12 by 4 inches as shown in Figure 1. The wall thickness of the bottom and the sides is x. x in 4 in 12 in 24 in Figure 1 Rectangular aluminum box Given that the specific weight of aluminum is 0.101 kg/in³ and the volume of the aluminum V is calculated from the weight of the box, W by:
The outside dimensions of a rectangular box (bottom and four sides, no top), made of aluminum, are 24 by 12 by 4 inches as shown in Figure 1. The wall thickness of the bottom and the sides is x. x in 4 in 12 in 24 in Figure 1 Rectangular aluminum box Given that the specific weight of aluminum is 0.101 kg/in³ and the volume of the aluminum V is calculated from the weight of the box, W by:
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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Transcribed Image Text:The outside dimensions of a rectangular box (bottom and four sides, no top), made of aluminum,
are 24 by 12 by 4 inches as shown in Figure 1. The wall thickness of the bottom and the sides is
x in.
W
V ==
Y
4 in
12 in
24 in
Figure 1 Rectangular aluminum box
Given that the specific weight of aluminum is 0.101 kg/in³ and the volume of the aluminum V
is calculated from the weight of the box, W by:
Where y is the specific weight. (note: the weight of the box is 15 kg)
a) Derive an expression that relates the weight of the box and the wall thickness, x.
b) Use bisection method to find the root of the function in (a) accurate to within ε = 0.001,
given that x = 0 and x = 0.4.
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