Question

Maximize:

U = alog(X) + (1 − a)log(Y )

subject to:

I = P_X X + P_Y Y

1. Find X* and Y* using the method of Lagrange multipliers.

2. Can we derive the demand function of X and Y in a general form, e.g:

X = f (P_X , I)andY = f (P_Y , I).

3. Please make some conclusions about the demand derived from the

utility function.

4. Note: a special case, please solve the same problem if U=XY.

Answer 1

Solution

1)

Lagrangian for the Problem :

$L(x,y,\lambda) = a log(x)+(1-a)log(y)+\lambda(I-P_x x-P_y y)$

First-order conditions(derivatives):

$∂L/∂x=a/x-λ^* P_x=0$ (1)

$∂L/∂y=((1-a))/y-λ^* P_y=0$ (2)

$∂L/∂λ=I-P_x x-P_y y=0$ (3)

Using the first two conditions(1 and 2) we have

$a/(1-a) y^*/x^* =P_x/P_y$

So,

$y^*=(1-a)/a(P_x/P_y)x^*$ (4)

Substitute this into the constraint: $I=P_x x+P_y y$

$I=P_x x^* +(P_y ((1−a)/a)(P x ​ /P _ y ​ )x ∗$

$I=P_x x^*+((1-a)/a)P_xx^*$

$I =P_xx^* (1-(1-a)/a)$ (5)

Solve for $x^*$ using (5)

$x^*=(I/P_x )(1+(1-a)/a)$

$x^*=(I/P_x)(1/a)$

$x^*=aI/P_x$ (6)

Substitute this into (4)

$y^*=((1-a)/a)(P_x/P_y)(aI/P_x)$

$y^*=((1-a)/a)(1/P_y)(aI)$

$y^* =(1-a)(I/P_y)$ (7)

2)

The demand function of X and Y can be derived in a general form, e.g: X = f (P_X , I)and Y = f (P_Y , I), reasons been the value of x* in(6) is a function of both P_x and I, and y* is a function of both P_y and I.

So, X and Y can be generalized as

X = f (P_X , I) and Y = f (P_Y , I) respectively.

3)

Derived demands are

$x^*=aI/P_x$

$y^* =(1-a)(I/P_y)$

The derivative of x* with respect to price P_X

$∂x^*/∂P_x=-aI/P_x ^2<0$

The derivative of y* with respect to price P_y

$∂y^*/∂P_y=-(1-a)I/P_y^2<0$

Since the derivative of the two derived demands are less than 0, the demands for goods x and y are decreasing with their respective prices

4)

when U=XY

Lagrangian or the problem:

$L(x,y,\lambda) = xy+ \lambda(I-P_x x-P_y y)$

First-order conditions(derivatives):

$∂L/∂x=y-λ^* P_x=0$ (1)

$∂L/∂y=x-λ^* P_y=0$ (2)

$∂L/∂λ=I-P_x x-P_y y=0$ (3)

Using the first two conditions(1 and 2) we have

$y^*/x^*=P_x/P_y$

So,

$y^*=x^*(P_x/P_y)$ (4)

Substitute this into the constraint: $I=P_x x+P_y y$

$I=P_x x^* + P_y(x^*(P_x/P_y))$

$I=2P_x x^*$ (5)

Solve for x using (5)

$x^*=I/2P_x$

Substitute this into (4)

$y^*=(I/2P_x)(P_x/P_y)$

$y^*=I/2P_y$

So, the derived demand for the two goods(X and Y) are:

$x^*=I/2P_x$

$y^*=I/2P_y$

They are different from the first one, however, they can still be written in general form, X = f (P_X , I)and Y = f (P_Y , I), just like the first one.