FQM Solutions to Exercises 6
1
a = 2
2
1. We have
1
b= 1
4
2
c = b a = 1
2
The cosines of the three angles are given by
ab
1 + 2 + 8
1
=
= ;
ab
9 18
2
2+24
ac
=
= 0;
ac
9 9
21+8
bc
1
= =
bc
18 9
2
Thus the triangle has a right-angle, and two angles

FQM Solutions to Exercises 7
1. The gradient vector for the function h(x, y, z) = x2 + y 2 + z 2 is given by
)
(
2x
1
1 1 1
h = 2y
At
, ,
, h = 1
2 2
2
2z
2
so this is the normal vector to the surface at that point. The equation of the tangent plane
i

FQM Solutions to Exercises 9
1. (a) f (x, y) = x2 + 4xy + 2y 2 + 6x + 6y + 1 has stationary points when
fx = 2x + 4y + 6 = 0
3
x=0 y=
2
fy = 4x + 4y + 6 = 0
The quadratic part of the function is the quadratic form
(
T
2
x Ax = x + 4xy + 2y
2
where
A=
1

FQM Solutions to Exercises 5
1. The auxiliary equation of yt 4yt1 + 4yt2 = 5
with two equal roots, = 2.
is 2 4 + 4 = ( 2)2 = 0,
The general solution of the associated homogeneous equation is yt = (C + Dt)(2)t .
A particular (constant) solution is y = 5/(1

FQM Solutions to Exercises 1
1.
Y R = ( 500
10000
1.05 0.95
1000 ) 1.05 1.05 = ( 12395
1.37 1.42
12395 )
so it is riskless since the guaranteed return is 12395 in each state.
Z = ( 1000
2000 1000 )
cost(Z) = 1000 2000 + 1000 = 0
ZR = ( 1000
1.05
2000 1000

FQM Solutions to Exercises 4
1. In matrix form, the equations are
(
ft
rt
)
(
=
.4
.4
.3
1.2
)(
ft1
rt1
)
= Axt
Looking for the eigenvalues of A we get
0.3
= 2 1.6 + 0.60 = ( 0.6)( 1) = 0
1.2
0.4
|A I| =
0.4
The eigenvalues of A are 0.6 and 1.
To dia

FQM Solutions to Exercises 10
1. (a) u(x, y) = x3 y 3 2xy + 1 The partial derivatives are
ux (x, y) = 3x2 2y
3x2 = 2y
27y 4 8y = y(27y 3 8) = 0
2
2
uy (x, y) = 3y 2x
2x = 3y
Solve the second equation for x, substitute into the first, and obtain the las