November 2001 - Course 1 SOA Solutions

24. D
1 1
4 2
100 x y must be maximized subject to x+ y = 150,000
Since y = 150,000 −x
, this reduces to maximizing
( ) ( )
1 1
4 2
S x = 100x 150,000 −x , 0 ≤ x≤150,000
−3 1
4 4
S ( x) x ( x) x ( x)
( 150,000 −x)
− 2x=
0
1 2 −1
2
′ = 25 150,000 − −50 150,000 − = 0
3x
= 150,000
x = 50, 000
(This value of x is a maximum since S ( x) 0
50,000 < x < 150,000 ).
′ > for 0 x 50,000
1 1
4 2
Maximum sales are then ( ) ( )
Alternate solution using Lagrange Multipliers
Solve x+ y− 150,000 = 0
∂ 1 1 ∂
4 2
100x y = λ x+ y−150,000
∂x
∂x
∂ 1 1 ∂
4 2
100x y = λ x+ y−150,000
∂y
∂y
From the last two equations
25x
50x
Eliminating λ
−3 1
4 2
1 4
y
y
−1 2
′ < for
< < and S ( x) 0
100 50,000 150,000 − 50,000 = 472,871
= λ
= λ
−3 1 1 −1
4 2 4 2
25x y = 50x y
25y
= 50x
y = 2x
Using the first equation
x+ 2x− 150,000 = 0
x = 50,000
y = 100,000
( )
( )
1 1
4 2
( ) ( )
The extreme value (which must be a maximum) is 100 50,000 100,000 = 472,871
Course 1 Solutions 14 November 2001