Doubt from Exercise 1

[Abhinandan_Ghosh]

This is Question No. 12 from Part B: Utility Maximization and Demand

A consumer’s utility function for sugar (x) and bread (y) is given by

u(x,y) = x^{\alpha} y^{1-\alpha} , 0<{\alpha}<1

A rationing scheme is in place, so that the consumer needs both money and ration coupons to
purchase sugar and bread. The consumer’s income is M = Rs. 1000, and the per-unit prices of
sugar and bread are p_{x} = Rs. 10 and p_{y} = Rs. 5, respectively. He also has 1000 ration coupons, and must give up 5 coupons per unit of sugar and 10 coupons per unit of bread.
(a) Set up the consumer’s utility maximization problem and draw the constraint set clearly.
(b) Can either x = 0 or y = 0 be a solution to the utility maximization problem? Explain clearly.
(c) Suppose 0 < {\alpha} ≤ \frac{1}{3} Obtain a solution to the utility maximization problem, and draw an appropriate diagram to illustrate your solution.
(d) Suppose \frac{1}{3} < {\alpha} < \frac{2}{3}. Obtain a solution to the utility maximization problem, and draw an appropriate diagram to illustrate your solution.
(e) Suppose \frac{2}{3} ≤ {\alpha} < 1. Obtain a solution to the utility maximization problem, and draw an appropriate diagram to illustrate your solution.

[Archishman]

u(x,y) = x^{\alpha} y^{1-\alpha}, \quad \text{where } 0 < \alpha < 1

The budget constraint is given by:

p_x x + p_y y \leq M

Substituting the parameters M = 1000, p_x = 10, and p_y = 5:

10x + 5y \leq 1000

\boxed{2x + y \leq 200}

Coupon Constraint

5x + 10y \leq 1000

\boxed{x + 2y \leq 200}

Optimization Problem:

\max_{x, y} \quad x^{\alpha} y^{1-\alpha} \quad \text{s.t.} \quad 2x + y \leq 200, \quad x + 2y \leq 200, \quad x, y \geq 0

2x + y = 200 \implies \text{Intercepts: } (100,0), \, (0,200) \quad \text{Slope: } -2

x + 2y = 200 \implies \text{Intercepts: } (200,0), \, (0,100) \quad \text{Slope: } -\frac{1}{2}

\begin{cases} 2x + y = 200 \\ 2x + 4y = 400 \end{cases} \implies 3y = 200 \implies y = \frac{200}{3}

2x + \frac{200}{3} = 200 \implies 2x = \frac{400}{3} \implies x = \frac{200}{3}

\therefore \boxed{\left(x, y\right) = \left( \frac{200}{3}, \frac{200}{3} \right)}

Now B part

u(x,y) = x^{\alpha}y^{1-\alpha}

x = 0 \implies u = 0, \quad y = 0 \implies u = 0

\exists \, (x,y) \in \text{Feasible Set} \text{ s.t. } x > 0, y > 0 \implies u(x,y) > 0

\therefore \boxed{x^* \neq 0 \quad \text{and} \quad y^* \neq 0}

so they can’t be optimal

c part

For small values of α\alphaα, the consumer places relatively more importance on bread (y) than sugar (x). Since bread requires more coupons per unit than sugar, it is natural to expect the coupon constraint to bind

x + 2y = 100
Solves using Lagrangian

\mathcal{L} = x^{\alpha} y^{1-\alpha} + \lambda (200 - x - 2y)

first-order conditions :

(x, y) = (200\alpha, \, 100(1-\alpha))

To verify :

2x + y \leq 200

Substituting our candidate values:

2(200\alpha) + 100(1-\alpha) \leq 200

400\alpha + 100 - 100\alpha \leq 200

300\alpha \leq 100 \implies \alpha \leq \frac{1}{3}

Hence, for the parameter range 0 < \alpha \leq \frac{1}{3}, the money constraint is not violated, and the global optimal bundle is:

\boxed{(x^*, y^*) = (200\alpha, \, 100(1-\alpha))}

Now part d

Here, neither constraint alone can determine the optimum. Hence, both constraints must bind simultaneously at a corner vertex of the budget set:

2x + y &= 200
x + 2y &= 200

Solving this linear system of equations gives:

x = y = \frac{200}{3}

Thus, for this specific parameter range, the optimal bundle is stuck at the kink:

\boxed{(x^*, y^*) = \left( \frac{200}{3}, \frac{200}{3} \right)}

Now part e

For larger values of \alpha, the consumer places relatively more importance on sugar (x). Since sugar is more expensive in terms of money (p_x = 10 > p_y = 5), we expect the money constraint to bind.

money constraint :

2x + y = 200

using the Lagrangian:

\mathcal{L} = x^{\alpha} y^{1-\alpha} + \lambda (200 - 2x - y)

first-orders :

(x, y) = (100\alpha, \, 200(1-\alpha))

Substituting :

x + 2y \leq 200

(100\alpha) + 2(200(1-\alpha)) \leq 200

100\alpha + 400 - 400\alpha \leq 200

200 \leq 300\alpha \implies \alpha \geq \frac{2}{3}

Hence, for the range \frac{2}{3} \leq \alpha < 1, the candidate bundle is feasible and constitutes the optimal bundle:

\boxed{(x^*, y^*) = (100\alpha, \, 200(1-\alpha))}