Consumer Choice Problem: Shifted Log Utility

[amitgoyal]

Suppose a consumer has preferences represented by the utility function

u(x,y)=\ln(x+2)+2\ln(y+3)

where x,y\geq 0. Let the prices p_x >0, p_y>0 and income M\geq 0 be given.
Solve the consumer’s utility-maximisation problem:

\begin{align} \max_{(x,y)\in\mathbb{R}^2_+} & \ln(x+2)+2\ln(y+3) \\ \text{s.t.} & \quad p_xx+p_yy\leq M\end{align}

[Archishman]

X= M-4px+3py/3px

Y= 2M+4px-3py/3py

[Tejasvi_Bhaskar_007]

Problem Statement
We want to maximize u(x,y) = \ln(x + 2) + 2 \ln(y + 3) subject to p_x x + p_y y \le M and x,y \ge 0.

The utility function is strictly increasing in both x and y, and the budget constraint is: p_x x + p_y y = M.

Also, the utility function is strictly quasi-concave and differentiable for all x, y \ge 0. This ensures that if an interior stationary point exists where the tangency condition holds, it represents a global maximum.

First-Order Conditions (Interior Solution)
Now, we find the marginal utilities:

MU_x = \frac{1}{x + 2}

MU_y = \frac{2}{y + 3}

At interior optimum, the Marginal Rate of Substitution equals the price ratio:

MRS = \frac{MU_x}{MU_y} = \frac{p_x}{p_y}

\frac{y + 3}{2(x + 2)} = \frac{p_x}{p_y}

Now, we cross-multiply and isolate p_y y:

p_y (y + 3) = 2p_x (x + 2)

p_y y + 3p_y = 2p_x x + 4p_x

p_y y = 2p_x x + 4p_x - 3p_y

Now, we substitute above expression for p_y y into the binding budget constraint:

p_x x + (2p_x x + 4p_x - 3p_y) = M

3p_x x + 4p_x - 3p_y = M

3p_x x = M - 4p_x + 3p_y

Finally, we isolate x and get the Marshallian demand for x:

x^* = \frac{M - 4p_x + 3p_y}{3p_x}

Substituting x^* back into the budget constraint to get y we get:

p_y y = M - p_x \left( \frac{M - 4p_x + 3p_y}{3p_x} \right)

p_y y = \frac{3M - (M - 4p_x + 3p_y)}{3}

p_y y = \frac{2M + 4p_x - 3p_y}{3}

y^* = \frac{2M + 4p_x - 3p_y}{3p_y}

The interior solution (x^*, y^*) is only valid if the non-negativity constraints hold.
For x \ge 0, we need M \ge 4p_x - 3p_y.
For y \ge 0, we need M \ge \frac{3p_y - 4p_x}{2}.

(Note for Amit Sir – Please explain in class how to proceed graphically if M does not satisfy the above condition)

[amitgoyal]

Solving the utility maximisation problem yields the following demand function:

\begin{align}(x^d,y^d)(p_x,p_y,M)=\begin{cases} \left(0,\dfrac{M}{p_y}\right) & \text{if } M\leq 4p_x-3p_y \\ \left(\dfrac{M}{p_x},0\right) & \text{if } M\leq \dfrac{3p_y-4p_x}{2} \\ \left(\dfrac{M-4p_x+3p_y}{3p_x},\dfrac{2M+4p_x-3p_y}{3p_y}\right) & \text{otherwise}\end{cases}\end{align}