UC Berkeley ME128: Minimizing Chain Energy & Maximizing Box Volume, Assignments of Mechanical Engineering

This document from the university of california, berkeley's mechanical engineering department, me128 course, spring 2006, includes two problems. The first problem deals with minimizing the potential energy of a chain suspended between two hooks using lagrangian mechanics. The second problem involves maximizing the volume of a cardboard box subject to a given amount of cardboard. Students are asked to write the lagrangian function, first order necessary conditions, and second order necessary conditions for the chain problem, and to find the dimensions of the cardboard box that maximize its volume.

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University of California at Berkeley
College of Engineering
Mechanical Engineering Department
ME128, Spring 2006 Liwei Lin
Problem Set #3
Due March 13 (Monday)
Problem 1 (Lagrangian Function)
A chain is suspended from two twin hooks that are 8 feet apart on a horizontal line. The
chain itself consists of 10 links. Each link is one foot in length. We wish to determine
the equilibrium shape of the chain. The equilibrium shape is the shape that has the
minimum potential energy. We let each link i span a horizontal distance of xi and a y
distance of yi measured with respect to the start of each link (the value could be
negative). Assuming unit weight, the potential energy (objective function) is
characterized by:
=
+=++++++++++++ n
iinn yinyyyyyyyyyyy
12
1
2
1
13213
2
1
212
2
1
11
2
1)()()()( LL
The chain is subject to two constraints. The total y displacement is zero and the total x
displacement is 8. Therefore, the problem is formulated as:
Minimize
=
+
n
ii
yin
12
1)(
Subject To
=
=
=
=
n
ii
n
ii
y
y
1
2
1
81
0
n=10
1. Write the Lagrangian function for the minimization problem.
2. Write the first order necessary conditions for the problem.
3. Write the second order necessary conditions for the problem.
Problem 2 (Optimization Practice)
A cardboard box for packing is to be manufactured. The top, bottom and front faces must
be of double weight, i.e., two pieces of cardboard, as shown in the figure below. Find the
dimensions of such a box that maximizes the volume for a given amount of cardboard,
equal to 72 sq. ft.
a) The objective is to maximize the volume V=xyz. The constraint is expressed as 4xy +
3xz + 2yz – 72 = 0.
b) Find x, y, z.
c) (optional) Verify the second order conditions.
z
y
front
pf2

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University of California at Berkeley College of Engineering Mechanical Engineering Department ME128, Spring 2006 Liwei Lin Problem Set # Due March 13 (Monday)

Problem 1 (Lagrangian Function) A chain is suspended from two twin hooks that are 8 feet apart on a horizontal line. The chain itself consists of 10 links. Each link is one foot in length. We wish to determine the equilibrium shape of the chain. The equilibrium shape is the shape that has the minimum potential energy. We let each link i span a horizontal distance of xi and a y distance of yi measured with respect to the start of each link (the value could be negative). Assuming unit weight, the potential energy (objective function) is characterized by:

=

                    • + − + = − +

n

i

y y y y y y y y y yn yn n i yi 1

2

1 2

1 2 3 1 2 3 1

1 2 2 1 2

1 2 1 1

(^1) ( ) ( ) L ( L ) ( )

The chain is subject to two constraints. The total y displacement is zero and the total x displacement is 8. Therefore, the problem is formulated as:

Minimize ∑

=

n

i

n i yi 1

2

(^1 )

Subject To

=

= n

i

i

n

i

i

y

y

1

2

1

1 8

n=

  1. Write the Lagrangian function for the minimization problem.
  2. Write the first order necessary conditions for the problem.
  3. Write the second order necessary conditions for the problem.

Problem 2 (Optimization Practice) A cardboard box for packing is to be manufactured. The top, bottom and front faces must be of double weight, i.e., two pieces of cardboard, as shown in the figure below. Find the dimensions of such a box that maximizes the volume for a given amount of cardboard, equal to 72 sq. ft. a) The objective is to maximize the volume V=xyz. The constraint is expressed as 4xy + 3xz + 2yz – 72 = 0. b) Find x, y, z. c) (optional) Verify the second order conditions.

z y

front

x