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Question:

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What my did is

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I proceed this question until that point. But, I will be happy if help me to solve this question.

Thank you.

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Consumer's Budget Constraint is

$$p_1x_1 + p_2x_2 + c\alpha x_1 = y$$

Holding $\alpha$ fixed, consumer's utility maximization problem is

\begin{eqnarray*} \max_{x_1, x_2} & \ (\alpha q + (1-\alpha))x_1 x_2 \\ \text{s.t.} & \ (p_1 + c\alpha)x_1 + p_2x_2 = y \end{eqnarray*}

Solving it we get the optimal bundle of goods as: \begin{eqnarray*} (x_1, x_2) = \left(\dfrac{y}{2(p_1+c\alpha)},\dfrac{y}{2p_2}\right) \end{eqnarray*}

Corresponding utility as a function of $\alpha$ and other parameters is

\begin{eqnarray*} (\alpha q + (1-\alpha))\left(\dfrac{y^2}{4p_2(p_1+c\alpha)}\right) \end{eqnarray*}

Now we find optimal value of $\alpha$ by solving \begin{eqnarray*} \max_{0 \leq \alpha \leq 1} & \ (\alpha q + (1-\alpha))\left(\dfrac{y^2}{4p_2(p_1+c\alpha)}\right) \end{eqnarray*}

Equivalently, we can find the optimal value of $\alpha$ by maximizing $\log$ of the objective \begin{eqnarray*} \max_{0 \leq \alpha \leq 1} & \ \ln(\alpha q + (1-\alpha)) - \ln(p_1+c\alpha) + 2\ln y - \ln 4 - \ln p_2 \end{eqnarray*}

Differentiating the objective with respect to $\alpha$, we get \begin{eqnarray*} \frac{p_1(q-1) - c}{(\alpha(q-1) + 1)(p_1+c\alpha)} \begin{cases} > 0 & \text{if } p_1(q-1) > c \\ = 0 & \text{if } p_1(q-1) = c \\ < 0 & \text{if } p_1(q-1) < c \end{cases} \end{eqnarray*}

Therefore, optimal value of $\alpha$ is $1$ when $p_1(q-1) > c$, all values of $\alpha$ in the interval $[0, 1]$ are optimal when $p_1(q-1) = c$ and optimal value of $\alpha$ is $0$ when $p_1(q-1) < c$. In short, optimal value $\alpha^*$ of $\alpha$ satisfy

\begin{eqnarray*} \alpha^* \in \begin{cases} \{1\} & \text{if } p_1(q-1) > c \\ [0, 1] & \text{if } p_1(q-1) = c \\ \{0\} & \text{if } p_1(q-1) < c \end{cases} \end{eqnarray*}

Consequently, final demand $(x_1^*, x_2^*)$ for $(x_1, x_2)$ satisfy \begin{eqnarray*} (x_1^*, x_2^*, \alpha^*) = \begin{cases} \left(\dfrac{y}{2(p_1+c)},\dfrac{y}{2p_2}, 1\right) & \text{if } p_1(q-1) > c \\ \left(\dfrac{y}{2(p_1+c\alpha)},\dfrac{y}{2p_2}, \alpha\right) & \text{if } p_1(q-1) = c \text{ and } 0 \leq \alpha \leq 1 \\ \left(\dfrac{y}{2p_1},\dfrac{y}{2p_2}, 0\right) & \text{if } p_1(q-1) < c \end{cases} \end{eqnarray*}

Alternatively, we can write demand for $x_1$ as \begin{eqnarray*} x_1^* \in \begin{cases} \left\{\dfrac{y}{2(p_1+c)}\right\} & \text{if } p_1 > \dfrac{c}{q-1} \\ \left[\dfrac{y}{2(p_1+c)}, \dfrac{y}{2p_1}\right] & \text{if } p_1= \dfrac{c}{q-1} \\ \left\{\dfrac{y}{2p_1}\right\} & \text{if } p_1 < \dfrac{c}{q-1} \end{cases} \end{eqnarray*}

Here is the figure demonstrating the relationship between demand for $x_1$ and its price $p_1$, holding $q$, $y$ and $c$ fixed.

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Now we consider Uno's revenue as a function of price $p_1$ : \begin{eqnarray*} r(p_1) = \begin{cases} \dfrac{y}{2} & \text{if } p_1 \leq \dfrac{c}{q-1} \\ \dfrac{yp_1}{2(p_1+c)} & \text{if } \dfrac{c}{q-1} < p_1 \leq \overline{p}_1 \end{cases} \end{eqnarray*}

Consequently, profit as a function of price $p_1$ is \begin{eqnarray*} \pi(p_1) = \begin{cases} \dfrac{y}{2} - \dfrac{ky}{2p_1} & \text{if } p_1 \leq \dfrac{c}{q-1} \\ \dfrac{y(p_1-k)}{2(p_1+c)} & \text{if } \dfrac{c}{q-1} < p_1 \leq \overline{p}_1 \end{cases} \end{eqnarray*}

Maximizing profit with respect to $p_1$, we get the optimal price as:

\begin{eqnarray*} p_1^* = \begin{cases} \dfrac{c}{q-1} & \text{if } 1 - \dfrac{k(q-1)}{c} \geq \dfrac{\overline{p}_1-k}{\overline{p}_1+c} \\ \overline{p}_1 & \text{if } 1 - \dfrac{k(q-1)}{c} < \dfrac{\overline{p}_1-k}{\overline{p}_1+c} \end{cases}\end{eqnarray*}

Now you can solve the remaining two parts in a similar way.

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