<p>Consumer's Budget Constraint is</p>

<p>$$p_1x_1 + p_2x_2 + c\alpha x_1 = y$$</p>

<p>Holding $\alpha$ fixed, consumer's utility maximization problem is</p>

<p>\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*}</p>

<p>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*}</p>

<p>Corresponding utility as a function of $\alpha$ and other parameters is</p>

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

<p>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*}</p>

<p>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*}</p>

<p>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*}</p>

<p>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</p>

<p>\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*}</p>

<p>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*}</p>

<p>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*}</p>

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

<p><a href="https://i.sstatic.net/XB9XD.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/XB9XD.png" alt="enter image description here"></a></p>

<p>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*}</p>

<p>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*}</p>

<p>Maximizing profit with respect to $p_1$, we get the optimal price as:</p>

<p>\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*}</p>

<p>Now you can solve the remaining two parts in a similar way.</p>

CC BY-SA 4.0
2018-06-07T17:03:13.043Z
Thank you very much Amit. Really nice answer. If you have time is it possible to have a look at this one https://economics.stackexchange.com/questions/22211/subgame-perfect-nash-equilibrium-two-stage-game
Stefanos Makridis
CC BY-SA 4.0
2018-06-07T16:36:04.920Z
2018-06-07T16:36:04.920Z
Amit
<p><strong>Question:</strong></p>

<p><a href="https://i.sstatic.net/rwbii.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/rwbii.jpg" alt="enter image description here"></a></p>

<p><strong>What my did is</strong></p>

<p><a href="https://i.sstatic.net/M0oZ5.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/M0oZ5.jpg" alt="enter image description here"></a>
<a href="https://i.sstatic.net/th6fe.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/th6fe.jpg" alt="enter image description here"></a>
<a href="https://i.sstatic.net/3K3V8.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3K3V8.jpg" alt="enter image description here"></a>
<a href="https://i.sstatic.net/9nj7W.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/9nj7W.jpg" alt="enter image description here"></a>
<a href="https://i.sstatic.net/tR1lQ.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/tR1lQ.jpg" alt="enter image description here"></a></p>

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

<p>Thank you. </p>

<microeconomics><consumer-theory><self-study>
Question on consumption theory
b11bb
# Question on consumption theory

(Asked by: username_0 on 2018-06-06)

**Question:**

[![enter image description here](https://i.sstatic.net/rwbii.jpg)](https://i.sstatic.net/rwbii.jpg)

**What my did is**

[![enter image description here](https://i.sstatic.net/M0oZ5.jpg)](https://i.sstatic.net/M0oZ5.jpg)
[![enter image description here](https://i.sstatic.net/th6fe.jpg)](https://i.sstatic.net/th6fe.jpg)
[![enter image description here](https://i.sstatic.net/3K3V8.jpg)](https://i.sstatic.net/3K3V8.jpg)
[![enter image description here](https://i.sstatic.net/9nj7W.jpg)](https://i.sstatic.net/9nj7W.jpg)
[![enter image description here](https://i.sstatic.net/tR1lQ.jpg)](https://i.sstatic.net/tR1lQ.jpg)

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

Thank you.

**Tags:** microeconomics, consumer-theory, self-study


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## 1 answers

### Answer by username_1 on 2018-06-07. Score: 2 | Accepted answer

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.

[![enter image description here](https://i.sstatic.net/XB9XD.png)](https://i.sstatic.net/XB9XD.png)

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.

> **Comments:**
> **username_2**: Thank you very much username_1. Really nice answer. If you have time is it possible to have a look at this one https://economics.stackexchange.com/questions/22211/subgame-perfect-nash-equilibrium-two-stage-game  

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