Question

Suppose all families have utility from food (F) and from money spent on other items (M) equal to U(F, M) = 10ln(F) + M. Suppose also that the unit price of food is PF = 1 and money spent on other items is measured in dollars so that PM = 1.

Consider families with a steady daily income of $25. We will label this type of family as “middle income”.

a) Using the Lagrangian method, find the optimal daily bundle of a “middle income” family.

b) What share of income does a “middle income” family spend on food?

c) In a diagram, illustrate the budget constraint of a “middle income” family. Then add an indifference curve to illustrate the

family’s optimal bundle.

Now consider families with a steady daily income of $9. We will label this type of family as “extremely poor”.

d) Find the daily optimal bundle of an “extremely poor” family.

e) What share of income does an “extremely poor” family spend on food?

f) In a second diagram, illustrate the budget constraint and the optimal bundle of an “extremely poor” family.

Answer 1

Consider the given problem, here the utility fun f a family is given by U = 10 ln(F) + M, F = Food & M = other item. the price of Food is p_F = 1 & price of other item is also 1 p_M = 1, I = 25 (middle in ) the Lagrangian Function is given below. alpha = 10 ln(F) + M + lambda (I - P_F F - P_M M) FOC for maximization are given below partialdifferential alpha/partialdifferential F = partialdifferential alpha/partialdifferential M = 0 partialdifferential alpha/partialdifferential F = 0 10/F + lambda(-P_F) = 0 10/F = lambda P_F equation (1) Similarly by partialdifferential alpha/partialdifferential M = 0 1 + lambda(-P_M) = 0 lambda(P_M) = 1 equation (2) by (1) Divisionsign (2) we will get a condition given below 10/F/1 = lambda P_F/lambda P_M 10/F = 1 F^* = 10 putting the optimum value of on the budget. we will get optimum value of M.