Question on consumption theory Question:



What my did is







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

Thank you.
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} &amp; \ (\alpha q + (1-\alpha))x_1 x_2 \\ \text{s.t.} &amp; \ (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} &amp; \ (\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} &amp; \ \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} &gt; 0 &amp; \text{if } p_1(q-1) &gt; c \\ = 0 &amp; \text{if } p_1(q-1) = c  \\ &lt; 0 &amp; \text{if } p_1(q-1) &lt; c \end{cases} \end{eqnarray*}

Therefore, optimal value of $\alpha$ is $1$ when $p_1(q-1) &gt; c$, 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 $p_1(q-1) &lt; c$. In short, optimal value $\alpha^*$ of $\alpha$ satisfy

\begin{eqnarray*} \alpha^*  \in \begin{cases}  \{1\} &amp; \text{if } p_1(q-1) &gt; c \\  [0, 1] &amp; \text{if } p_1(q-1) = c  \\  \{0\} &amp; \text{if } p_1(q-1) &lt; 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) &amp; \text{if } p_1(q-1) &gt; c \\  \left(\dfrac{y}{2(p_1+c\alpha)},\dfrac{y}{2p_2}, \alpha\right) &amp; \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) &amp; \text{if } p_1(q-1) &lt; 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\} &amp; \text{if } p_1 &gt; \dfrac{c}{q-1} \\  \left[\dfrac{y}{2(p_1+c)}, \dfrac{y}{2p_1}\right] &amp; \text{if } p_1= \dfrac{c}{q-1}  \\  \left\{\dfrac{y}{2p_1}\right\} &amp; \text{if } p_1 &lt; \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.



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