2010 3E1 Examination
Q(1)
(a) Classify each of the following equations as elliptic, hyperbolic or parabolic giving
the reasons for your answers:
 ( x, y)  2 ( x, y)
 2 ( x, y )
 2 ( x, y )
 2 ( x, y )
(i)
2
2
 4 x  33 y ,

 8 , (ii)
xy
x 2
y 2
x
xy
(iii)
 2 ( x, y )
 2 ( x, y )
 2 ( x, y )

6

9
 0.
xy
x 2
y 2
[ 3 marks]
(b) The temperature T ( x, t ) in a bar of length L metres is given by
T ( x, t )
 2 T ( x, t )
K
t
x 2
where K is the positive constant of conductivity. If the ends of the bar x  0 and
x  L metres are maintained at zero temperature and if the initial temperature is
T ( x,0)  Cx( L  x) where C is a positive constant show that
T ( x, t ) 
8CL2

3
1
(2m  1) x 
sin[
]e

3
L
m 1 ( 2m  1)

K ( 2 m 1) 2  2t
L2
[ 17 marks]
Q(2)
2n  1


) x) n  1,2,3 is an orthogonal set on 0,1 .
(a) Show that the set sin((
2


[ 3 marks]
(b) The twist angle  ( x, t ) of a torsionally vibrating shaft of unit length is satisfies
 2  2

x 2 t 2
subject to the boundary conditions
0  x  1, t  0
a2
 (0, t )  0,  t
and
 (1, t )
 0,  t
x
and initial conditions
 ( x,0)
 0, 0  x,  1.
t
Use the method of separation of variables to show that the solution is given by
8  (1) n 1
2n  1
2n  1
 ( x, t )  2 
cos[ a(
) t ] sin[(
) x] .
2
2
2
 n1 (2n  1)
[ 17 marks]
 ( x,0)  x, 0  x  1
and
Q(3) A rectangular plate 0  x  20 , 0  y  10 has edges x  0, x  20, y  0
maintained at zero temperature whilst the temperature of the edge y  10
is given by  ( x, y)  20 x  x 2 . If the steady-state temperature satisfies
Laplace’s equation
 2 ( x, y )  2 ( x, y )

0
x 2
y 2
use the method of separation of variables to show that
 ( x, y ) 
3200
3


n 1
sin(
(2n  1)x
(2n  1)y
) sinh(
)
20
20
(2n  1)
(2n  1)3 sinh(
)
2
[ 20 marks]
Q(4)
(a) Consider the homogeneous linear differential equation
d 2 y ( x)
dy ( x)
 p( x)
 q( x) y ( x)  0
2
dx
dx
for some functions p(x) and q(x) defined on any interval I .State and prove the
principle of linear superposition for this system.
[ 4 marks]
(b) Use the method of undetermined coefficients to solve
d 2 y( x) dy ( x)

 2 y ( x)  10 sin( x)
dx
dx 2

dy 
( )  1 .
subject to the initial conditions y ( )  3 and
2
dx 2
[ 16 marks]
Q(5)
(a) The Outdoor Furniture Corporation manufactures two products, benches
and picnic tables, for use in yards and parks. The firm has two main
resources: carpenters(labour force) and a supply of redwood. During the next
production cycle, 1200 hours of labour are available under a union agreement. The
firm also has a stock of 3500 feet of good-quality redwood. Each bench that Outdoor
Furniture produces requires 4 labour hours and 10 feet of redwood: each picnic table
requires 6 labour hours and 35 feet of redwood. Completed benches will yield a
profit of € 9 each, and tables will yield a profit of € 20 each. Management wishes to
decide how many benches and picnic tables it should produce in order to maximise
its profit.
Formulate this as a linear programming problem and solve graphically.
[ 8 marks]
(b) Put the above linear programming problem in canonical form and solve it using the
Simplex algorithm.
[ 12 marks]
Q(6)
(a)State the fundamental theorem of linear programming. Show how this theorem
gives rise to the direct “sledgehammer” method for solving linear programming
models and explain why this direct “sledgehammer” method is not used to solve large
practical linear programming models.
[ 5 marks]
(b)Consider the linear programming model
min imize z  2x1  x2
subject to
x1  x2  5
 x1  x2  1
5x1  x2  40
where x1  0 and x2  0
.
(i) By introducing slack variables x3  0 , x4  0 and x5  0 show that the canonical
form of the above linear programming model is
min imize z  2 x1  x2  0 x3  0 x4  0 x5
subject to
x1  x2  x3
5
 x1  x2  x4
1
=40.
5x1  4 x2
 x5
Show that the all-slack point is not a feasible point.
[ 3 marks]
(ii) Using the two phase Simplex Algorithm, with one artificial variable x6  0
solve the above linear programming model.
[ 12 marks]
Q(7)
(a) State the Newton-Raphson method for solving the unconstrained optimization
problem, namely
minimize z  f ( x1 , x2 , x3 ,, xn )
where the objective function f ( x1 , x2 , x3 ,, xn ) is a nonlinear function of n
variables x1 , x2 , x3 ,, xn .
[ 3 marks]
(b) Taking an initial approximation of x1 , x2 , x3   0,0,0 and using the NewtonRaphson method solve the optimization problem
minimize z  f ( x1 , x2 , x3 )  x12  x 22  x32  2 x1  2 x 2  2 x3  3 .
[ 10 marks]
T
T
(c) Comment on your solution.
[ 3 marks]
(d) Write a note on the main advantage and main disadvantages of the Newton-Raphson
method for solving the unconstrained non-linear optimization problems.
[ 4 marks]
Answers to 2010 Examination
Q(1) Answer
(a) (i) hyperbolic (ii) elliptic and (iii) parabolic
(b) T ( x, t ) 
8CL2

3
1
(2m  1) x 
sin[
]e

3
L
m 1 ( 2m  1)

K ( 2 m 1) 2  2t
L2
.
Q(2) Answer
(a)
See notes
8  (1) n 1
2n  1
2n  1
(b)  ( x, t )  2 
cos[ a(
) t ] sin[(
) x] .
2
2
2
 n1 (2n  1)
Q(3) Answer
(2n  1)x
(2n  1)y
) sinh(
)
 sin(
3200
20
20
.
 ( x, y )  3 
(2n  1)
 n 1
(2n  1)3 sinh(
)
2
Q(4) Answer—I did not do this material this year.
Q(5) Answer
Optimum value is 2862.5 at x1  262.5 and x2  25
Q(6) Answer
(a) See notes
(b) Optimum value is z  1 at x1  2 and x2  3 .
Q(7) Answer---- I did not do this material this year.
.
2009 3E1 Examination
Q (1) The temperature  ( x, t ) of a slender metal bar of length L satisfies the diffusion
equation
 2 ( x, t )
 ( x, t )
K
2
t
x
where K  0 is a constant. The bar is embedded in a perfect insulator so that the
 (0, t )
 ( L, t )
boundary conditions are given by
 0 and
 0 . Initially the
x
x
temperature of the bar is given by  ( x,0)  f ( x) where f (x) is some function of x .
(a) Show that the temperature at a point of the bar distance x from one end of the bar
and at any time t is given by
nx 
 ( x, t )  A0   An cos(
)e
L
n 1

n 2 2t
KL2
where A0 and A1 are constants. Express A0 and A1 in terms of f (x) .
[ 12 marks]
x
(b) If f ( x)  20(1  ) find  ( x, t ) .
L
[ 6 marks]
(c) Sketch the graphs of the temperature distribution  ( x, t ) at time t  0 and at t   .
[ 2 marks]
Q(2)
(a) A steel bar of length L metres is fixed at both ends. It is twisted and vibrates
angularly about the x-axis so that  ( x, t ) , the angular displacement of the bar
satisfies
2
 2  ( x, t )
2   ( x, t )

a
t 2
x 2
where a is a constant. If the initial displacement is f (x) and the initial angular
velocity is g (x ) , use the method of separation of variables to find  ( x, t ) in terms of
f (x ) and g (x ) .
[ 13 marks]
3x
) and g ( x )  0 calculate  ( x, t ) .
(b) If f ( x)  sin(
L
[ 7 marks]
Q(3)
(a) The voltage V ( x, y ) at any point on a rectangular metal plate satisfies Laplace’s
equation
 2V ( x, y )  2V ( x, y )

0.
x 2
y 2
The plate is earthed along the side x  0 , x  a and y  0 so that the boundary
conditions are V (0, y )  0 , V (a, y )  0 and V ( x,0)  0 . A voltage of f (x) is applied
along the fourth side y  b so that the boundary condition is given by
V ( x, b)  f ( x) . By applying the method of separation of variables find V ( x, y ) .
[ 13 marks]
(b) If f ( x)  V0 x(a  x) where V 0 is a constant, show that
V ( x, y ) 
4V0 a

3
2

1   1 
n 1
n3

n
 ny 
sinh 

 a  sin  nx  .


 nb   a 
sinh 

 a 
[ 7 marks]
Q(4) Consider the following initial boundary value problem
 2 u ( x, t )  2 u ( x , t )

, x  0, t  0
x 2
t 2
subject to the boundary conditions
u (0, t )  0 ,
Lim u ( x, t )  0 t  0
x
and the initial conditions
u ( x,0)
 0 x  0.
t
Use the Laplace transform method to show that the solution is given by
u ( x,0)  xe x and
u( x, t )  (t  x) sinh( t  x) H (t  x)  e  x t sinh( t )  xe x cosh( t ) .
[ 20 marks]
Q(5)
(a) In a given factory there are two machines M 1 and M 2 used in making two products
1
1
hours and M 2 for
hours
P1 and P2 respectively. One unit of P1 occupies M 1 for
12
30
1
respectively. The corresponding figures for one unit of P2 are for M 1
hours and
60
1
for M 2
hours respectively. The net profit per unit of P1 produced is € 50 and for
15
per unit of P2 produced is € 30(independent of whether the machines are used to full
capacity or not). What production plan gives the most profit? Formulate this linear
programming problem and solve it graphically.
Hint: Let x1  number of units of P1 produced per hour and let x2  number of units
of P2 produced per hour.
[ 9 marks]
(b) Put the above linear programming problem in canonical form and solve it using the
Simplex algorithm.
[ 11 marks]
Q(6) Consider the following linear programming problem in canonical form
Minimize f  4 x1  9 x2
subject to
 x1  x2  x3
 40
x1
 x4
 50
x1  2 x2
 x5  80 .
Use the 2 stage Simplex algorithm with one artificial x 6 to solve the above linear
programming problem.
[ 20 marks]
Q(7)
(a) Find the stationary points of the function


2
f ( x, y)  x 2  9  y 2
and investigate their nature.
[ 9 marks]
(b) The geometric buckling of the neutron flux in a cylindrical nuclear reactor is given
by
a  
B2      
R H 
where R and H are respectively, the effective radius and height of the core and the
constant a has the approximate value of 2.40. Use the method of Lagrange
multipliers to show that the smallest volume for a specified fixed buckling is given
148
approximately by 3 and find the corresponding values of H and R .
B
[ 11 marks]
2
2
Answers to 2009 Exam
Q(1) Answer
nx 
(a)  ( x, t )  A0   An cos(
)e
L
n 1

n 2 2t
KL2
.
1
nx 
)e
(b)  ( x, t )  10  2  2 cos(
L
 n1,3,5 n n

80
n 2 2t
KL2
( 2 m 1) 2  2t
1
(2m  1)x  KL2
 ( x, t )  10  2 
cos(
)e
L
 m1 (2m  1) 2
Q(2) Answer

an t
an t
n x
)  Dn sin(
)] sin(
)
(a)  ( x, t )   [C n cos(
L
L
L
n 1
where
2 L
n x
2 L
n x
C n   f ( x) sin(
)dx , Dn 
g ( x) sin(
)dx .

L 0
L
an 0
L
80

(b) C n  0, n  3 , C3  1 and Dn  0, n .
 ( x, t )  cos(
3a t
3x
) sin(
).
L
L
Q(3)Answer
(a)

V ( x, y )   C n sinh(
n 1
where
a
Cn 
ny
nx
) sin(
)
a
a
n x
)dx
a
n b
a sinh(
)
a
2 f ( x) sin(
0
(b)
V ( x, y ) 
4V0 a

3
2

1   1 
n 1
n3

n
 ny 
sinh 

 a  sin  nx 


 nb   a 
sinh 

 a 
Q(4) Answer –Did not do this material this year.
Q(5) Answer Maximum value is 800 and occurs at x1  10 and x2  10 .
Q(6) Answer Optimum is z  1010 at x1  50 and x2  90 .
Q(7) Answer
(a) Stationary points at (0,0) --saddle point
at (3,0) --maxima
at (3,0) --maxima.
(b) Vmin 
3
148
and H 
and R 
3
B
B
3 a
.
2B