Question

Assume a consumer has the utility function U (x1 , x2 ) = ln x1 + ln x2 and faces prices p1 = 1 and p2 = 3 . [He,She] has income m = 200 and [his,her] spending on the two goods cannot exceed her income.

Write down the non-linear programming problem. Use the Lagrange method to solve for the utility maximizing choices of x1 , x2 , and the marginal utility of income λ at the optimum.

Answer 1

Consumer Utility (U)= p1 *x1 +   p2*x2

Where x1 and x2 represents the quantity of goods

Consumer income constraint of 200

   200= p1 *x1 + p2*x2

Consumer has 200 to spend to by the products x1 and x2

Goal is to get maximum utility subject to income constraint of 200

Maximize Z = Objective function + L Constraint

                   Z= (p1 *x1 +p2*x2) + L (200-(p1 *x1 +p2*x2)

                      = (1 *x1 +3*x2) + L (200-(1 *x1 +3*x2) {given p1=1 and p2 =3}

                      =(x1+3x2) +L(200-(x1+3x2)

Take three partial derivatives setting each one equal to zero

Z_{x 1}= 3x2-L=0 ------------(1)

Z_{x 2}= x1-L=0 -------------- (2)

Z_L=200-x1-3x2=0----------(3)

Solve (1) and (2) for L

3x2=L

X1=L

Implies 3x2=x1

Putting this equation in (3) we get

200-3x2-3x2=0

200-6x2=0

200=6x2

X2=200/6

So x1=3* 200/6

         =600/6

          =100

Therefore, utility U (x1, x2) = (100,200/6)

L here represents λ(lambda)