University of London The London School of Economics and Political Science
Further Quantitative Methods
MATHEMATIC MA207

Spring 2016
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
is
x
University of London The London School of Economics and Political Science
Further Quantitative Methods
MATHEMATIC MA207

Spring 2016
FQM Solutions to Exercises 10
1. (a) u(x, y) = x3 y 3 2xy + 1
ux (x, y) = 3x2 2y
The partial derivatives are
3x2 = 2y
uy (x, y) = 3y 2x
2
2x = 3y
2
27y 4 8y = y(27y 3 8) = 0
Solve the second equation for x, substitute into the rst, and obtain the last equ
University of London The London School of Economics and Political Science
Further Quantitative Methods
MATHEMATIC MA207

Spring 2016
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 2
University of London The London School of Economics and Political Science
Further Quantitative Methods
MATHEMATIC MA207

Spring 2016
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
=
= ;
a b
9 18
2
2+24
ac
=
= 0;
a c
9 9
21+8
bc
1
= =
b c
18 9
2
Thus the triangle has a rightangle, and two angl
University of London The London School of Economics and Political Science
Further Quantitative Methods
MATHEMATIC MA207

Spring 2016
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
cost(Z) = 1000 2000 + 1000 = 0
1000 )
1.05 0.95
1000 ) 1.05 1.05
University of London The London School of Economics and Political Science
Further Quantitative Methods
MATHEMATIC MA207

Spring 2016
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
University of London The London School of Economics and Political Science
Further Quantitative Methods
MATHEMATIC MA207

Spring 2016
FQM Solutions to Exercises 4
1. In matrix form, the equations are
ft
rt
.4
.4
=
.3
1.2
ft1
rt1
= Axt
We will give two methods which can be used to solve this system. The solutions will be linear
combinations of (1 )t and (2 )t where 1 and 2 are the eigenv