Underlying utility function behind a linear two-product demand curve

Question

I am trying to find the underlying utility function behind a linear two-product demand model. For that, I use two methods considering the following utility function: \begin{equation} U(q_1,q_2) = \alpha_1q_1 + \alpha_2q_2 - \frac{1}{2}(q_1^2+q_2^2+2\epsilon q_1 q_2). \end{equation} First, I do utility maximization without explicitly considering the budget constraint. Second, I do the maximization with explicitly stating the budget constraint. The former method gives me linear demand functions, whereas the latter does not. I am not sure why there is this difference. More specifically, I first solve the utility maximization problem as: \begin{equation} \max_{q_1,q_2} \alpha_1q_1 + \alpha_2q_2 - \frac{1}{2}(q_1^2+q_2^2+2\epsilon q_1 q_2) - p_1 q_1 - p_2q_2 \end{equation} Then, the optimal solution is given as \begin{align*} q_1 (p_1,p_2) &= \frac{\alpha_1 - \epsilon\alpha_2 -p_1 +\epsilon p_2}{1-\epsilon^2} \\ q_2 (p_1,p_2) &= \frac{\alpha_2 - \epsilon\alpha_1 -p_2 +\epsilon p_1}{1-\epsilon^2} \end{align*} Both of these demand functions are linear in prices $p_1$ and $p_2$, which is what I am looking for. However, when I consider a budget constraint, that is when I consider a constrained maximization problem of the following form: \begin{align*} \max_{q_1,q_2} & \alpha_1q_1 + \alpha_2q_2 - \frac{1}{2}(q_1^2+q_2^2+2\epsilon q_1 q_2) \\ \text{s.t. }& p_1 q_1 + p_2q_2 \le m, \end{align*} I obtain the following two demand functions: \begin{align*} q_1 (p_1,p_2;m) &= \frac{mp_1 + \alpha_1 p_2^2 - \alpha_2 p_1 p_2 - \epsilon m p_2}{p_1^2-2\epsilon p_1 p_2 + p_2^2} \\ q_2 (p_1,p_2;m) &= \frac{mp_2 + \alpha_2 p_1^2 - \alpha_1 p_1 p_2 - \epsilon m p_1}{p_1^2-2\epsilon p_1 p_2 + p_2^2} \end{align*} The problem is that these demand functions are not linear. How should I interpret this discrepancy?

Answer 1

Hint: First, start off with the constrained maximization problem and then use the Lagrangian method to get the first-order conditions. Then use the Kuhn-Tucker method (see Mathematics for Economist by Simon and Blume) to solve for your optimal demands.

The consumer solves the following maximization problem:

\begin{align*} \max_{q_1,q_2} \alpha_1q_1 + \alpha_2q_2 - \frac{1}{2}(q_1^2+q_2^2+2\epsilon q_1 q_2) \\ \text{s.t. } p_1 q_1 + p_2q_2 \le m. \end{align*}

The Lagrangian associated with maximization problem is
\begin{equation*} \mathcal{L} = \alpha_1q_1 + \alpha_2q_2 - \frac{1}{2}(q_1^2+q_2^2+2\epsilon q_1 q_2) + \lambda[m - p_1 q_1 - p_2q_2] \end{equation*}

Then take the derivatives of $\mathcal{L}$ wrt to $q_{1}$, $q_{2}$, and $\lambda$ to get the following first-order conditions:

\begin{equation} \frac{\partial \mathcal{L}}{\partial q_{1}}=\alpha_{1}-q_{1}+\epsilon q_{2} -\lambda p_{1} =0 \end{equation}

\begin{equation} \frac{\partial \mathcal{L}}{\partial q_{2}}=\alpha_{2}-q_{2}+\epsilon q_{1} -\lambda p_{2}=0 \end{equation}

\begin{equation} \frac{\partial \mathcal{L}}{\partial \lambda}=m-p_{1}q_{1}-p_{2}q_{2}\geq0 \end{equation}

Then we use the Kuhn-Tucker method:

$\lambda[m-p_{1}q_{1}-p_{2}q_{2}] = 0$ is the complementary slackness condition on $\lambda$

and

$\lambda\geq0$ is the non-negativity constraint on $\lambda$.

We can have two cases: either $\lambda>0$ or $\lambda=0$. I show what happens for $\lambda>0$ and then you can check what happens for $\lambda=0$.

If $\lambda>0$ then $m-p_{1}q_{1}-p_{2}q_{2}=0$ by the complementary slackness condition. In other words, this is the case where the budget constraint is binding, that is, $m=p_{1}q_{1}-p_{2}q_{2}$. Then we check whether the first-order conditions and the non-negativity condition hold.

After doing so we are left with the following set of equations:

\begin{equation} \frac{\partial \mathcal{L}}{\partial q_{1}}=\alpha_{1}-q_{1}+\epsilon q_{2} -\lambda p_{1} =0 \end{equation}

\begin{equation} \frac{\partial \mathcal{L}}{\partial q_{2}}=\alpha_{2}-q_{2}+\epsilon q_{1} -\lambda p_{2}=0 \end{equation}

\begin{equation} m-p_{1}q_{1}-m_{2}q_{2}=0 \end{equation}

Then substitute away $\lambda$ from the first two equations and solve for $q_{1}$ and $q_{2}$ using the binding budget constraint.

Edit: For the unconstrained problem you have no budget constraint so your set up is conceptually wrong. The way you get an unconstrained problem from a constrained problem is that you substitute away your budget constraint in the utility function. Also I haven't imposed non-negative conditions on $q_{1}$ and $q_{2}$, as then the process would be quite tedious.