Solutions 5 6

Some solutions for Exercise 5-6, 2006 Mathematics for Economics and Finance Prof: Norman Schürho¤ TAs: Zhihua Chen (Cissy), Natalia Guseva Exercise Session 5 4. (a) Putting (L) = f (L) wL; then we have 0 (L) = f 0 (L) w, 00 (L) = 00 f (L) < 0, therefore, (L) is strictly concave and the …rst order condition 0 (L) = 0 characterizes a maxima. (b) Write the FOC in the form F (L; w; ) = f 0 (L) w = 0; and we observe 0 FL = f 00 (L) < 0; Fw0 = 1 < 0; F 0 = f 0 (L) > 0:By the IFT, the derivatives of the solution L = L(w; ) are given by @L = @w

Fw0 = FL0

( ) <0 ( )

@L = @

F0 = FL0

(+) >0 ( )

Hence, the demand for labor decreases with the wage rate , and increase with the productivity parameter :

Exercise Session 6 3. Step 1#: A stationary point of f can be found by FOC: 8 @f > < @x1 = 2x1 x2 + 2x3 = 0 @f x1 + 2x2 + x3 = 0 @x2 = > : @f = 2x + x + x = 0 1 2 3 @x3

Solve the above linear system of equations, we have only one stationary point, x = (0; 0; 0): Step 2#: Check the Hessian matrix’s de…niteness: 2 3 2 1 2 H(x) = 4 1 2 1 5 ; 2 1 6

and the leading principal minors of the Hessian matrix are D1 = 2; D2 = 3; D3 = 4: H(x) is positive de…nite, and thus f (x) is strictly convex. The point x = (0; 0; 0) is a local (and global) minimum.

1

4. Assume Cobb-Douglas utility function is concave (you should be able to show it by yourself) and it is increasing at all (x1 ; x2 ) >> 0; hd(1): We need to max u(x1 ; x2 ), s.t. p1 x1 + p2 x2 = w It turns out to be easier to use increasing x1 ;x2

transformation max ln x1 + (1

) ln x2

x1 ;x2

s:t p1 x1 + p2 x2 = w: Since we have x2 = max ln x1 + (1

1 p2 (w

) ln

x1

p1 x1 ); we can rewrite

1 (w p2

p1 x1 )

Using FOC ; take the …rst derivative w.r.t x1 ; x1

p1 (1 w p1 x1

)=0

Solve it, we have x1

=

x2

=

w p1 (1

)w p2

5. (a). For = 1; we have u(x) = 1 x1 + 2 x2 :Thus the indi¤erence curves are linear. (b). Since every monotonic transformations of a utility function represents the same preference, we shall consider u ~(x) = ln u(x) =

1

ln(

1 x1

+

2 x2 )

By L’Hopital’s Rule,

!0

ln x1 + 1 x1 +

u(x)g 2 )~

= x1 1 x2 2 ; we have obtained a Cobb-Douglas util-

lim u ~(x) = lim !0

ity.

Since exp f(

1

+

(c). Suppose x1

1 x1

! 1

< 0; since x1 1 x1

ln x2

2 x2

=

1

ln x1 + 1+

2

ln x2

0; x2 1 x1

+

1 x1

+

2 x2 ]

x1

(

1

0 we have 2 x2

)(

1)

2

1

1 x1

+

:

2

x2 ; we want to show x1 = lim [

Let

2 x2

2 x2 )

1

On the other hand , since x1 1 x1

+

x2 , we have

2 x2

1 x1

Hence (

1 x1

+

2 x2 )

+

1

2 x1

(

1

=( +

1

2)

1

+

2 )x1

x1

Therefore ( Letting

!

1)

1

x1

(

1 x1

+

1; we obtain lim [ ! 1

2 x2 )

1 x1

3

+

1

( 2 x2 ]

1 1

+

2)

1

x1

= x1 by Sandwich Theorem.