Here goes...
1. Consider a model in which an individual lives two periods: this period (time one) and next period (time two). This period his budget constraint requires that his consumption, c1; plus his saving, s; equals his income, y1:
c1 + s = y1:
Next period his budget constraint requires that his consumption, c2; equals his income, y2; plus the value of his saving from the initial period:
c2 = y2 + (1 + r)s
where r is a known interest rate.
The individual's problem is to maximize is
U(c1) + �βU(c2)
by choices of c1; c2; and s subject to his budget constraints. What are his optimal choices ofconsumption (in both periods) and saving when
U(c) = -1/2c^2
y1 = 1000
y2 = 200
r = 0:03;
and � β = 0.97
NOTE: For this utility function, MU(c)= 1/c^3
To find the individual's optimal choices of consumption and saving, we need to solve the maximization problem subject to the budget constraints.
1. Start by rewriting the problem in terms of the utility function and the budget constraints:
Maximize U(c1) + βU(c2)
Subject to: c1 + s = y1
c2 = y2 + (1 + r)s
2. Substitute the given utility function U(c) = -1/2c^2 and simplify the objective function:
Maximize -1/2c1^2 + β(-1/2c2^2)
3. Rewrite the budget constraints in terms of c1 and c2:
c1 + s = 1000
c2 = 200 + (1 + 0.03)s
4. Now, we have an optimization problem with two variables (c1, c2) and one constraint.
To solve this problem, we can use an optimization technique called the Lagrange multiplier method. Let's set up the Lagrangian function:
L(c1, c2, s, λ) = -1/2c1^2 + β(-1/2c2^2) + λ(c1 + s - 1000)
5. We need to find the critical points of the Lagrangian function by taking partial derivatives with respect to c1, c2, s, and λ, and setting them equal to zero:
∂L/∂c1 = -c1 + λ = 0
∂L/∂c2 = -βc2 + λ = 0
∂L/∂s = λ = 0
∂L/∂λ = c1 + s - 1000 = 0
6. Solve the system of equations to find the values of c1, c2, s, and λ:
From the first equation: c1 = λ
From the second equation: c2 = λ/β
From the third equation: λ = 0
From the fourth equation: c1 + s = 1000
Substituting λ = 0 into the third equation, we get s = 1000.
Therefore, the optimal choices of consumption and saving are:
c1 = λ = 0
c2 = λ/β = 0/0.97 = 0
s = 1000
So, the individual should consume 0 in both periods and save 1000.