Economics of the Market

1. For \(p=2\), we have \(d(2) = 92\) and \(s(2) = 44\). The market does not clear because \(d(2)>s(2)\). It has a shortage equal to \(d(2) - s(2) = 48\)
2. The market clearing price is obtained by setting \(d(p) = s(p)\) and solving for \(p\). This gives \(p=10\).

1. By \(d(p) = s(p)\), we get \(p=5\). The assumption of the exercise excludes the negative root.
2. The elasticity of demand is given by \[e_{d}(p) = d'(p) \frac{p}{d(p)} = -\frac{250}{p^{2}}\frac{p^{2}}{250} = -1.\] Demand elasticity is constant for all prices.
3. The elasticity of supply is given by \[e_{s}(p) = s'(p) \frac{p}{s(p)} = 10\frac{p}{10 p} = 1.\] Supply elasticity is also constant for all prices.

1. We calculate \(p=3\).
2. The elasticity of demand at \(p=3\) is \[e_{d}(3) = \left. -3 \frac{p}{22 - 3p} \right|_{p=3} = -\frac{9}{13}.\]
3. The elasticity of supply at \(p=3\) is \[e_{s}(3) = \left. \frac{p}{10 + p} \right|_{p=3} = \frac{3}{13}.\]

1. The market clearing price is given by \(d(p) = s(p)\), or equivalently \[\alpha - \beta p = \gamma + \delta p.\] We can solve for prices to get \[p_{m} = \frac{\alpha - \gamma}{\delta + \beta}.\]
2. The elasticity of demand at \(p_{m}\) is \[e_{d}(p_{m}) = -\beta \frac{1}{\frac{\alpha}{p_{m}} - \beta} = -\beta\frac{\alpha-\gamma}{\alpha\delta + \beta\gamma}.\]
3. The elasticity of supply at \(p_{m}\) is \[e_{s}(p_{m}) = \delta \frac{1}{\frac{\gamma}{p_{m}} + \delta} = \delta\frac{\alpha-\gamma}{\alpha\delta + \beta\gamma}.\]

Initial q	Final q	Input scale	Final q / Initial q	Returns to scale
3	9	2	3	Increasing
9	18	1.5	2	Increasing
18	24	1.3333333	1.3333333	Constant
24	30	1.25	1.25	Constant
30	36	1.2	1.2	Constant
36	40	1.1666667	1.1111111	Decreasing
40	42	1.1428571	1.05	Decreasing

1. The production function corresponding to \(\mathcal{Q}^{r}\) is \(f(x) = x^{r}\).

2. For \(r < 1\), the production set is convex.

3. For \(r > 1\), the production set is not convex.

4. Given any \(r>0\), the marginal product of the production function is \(f'(x) = r x^{r - 1}\).
   • If \(r < 1\), then \(f''(x) = r (r - 1) x^{r - 2} \le 0\), which implies that the marginal product of \(f\) is decreasing.
   • If \(r = 1\), then \(f''(x) = r (r - 1) x^{r - 2} = 0\), which implies that the marginal product of \(f\) is constant.
   • If \(r > 1\), then \(f''(x) = r (r - 1) x^{r - 2} \ge 0\), which implies that the marginal product of \(f\) is increasing.

For every \(t>1\), we have

\begin{align*} f(t x_{1}, t x_{2}) &= \left(t x_{1}\right)^{\alpha} \left(t x_{2}\right)^{\beta} \\ &= t^{\alpha + \beta} x_{1}^{\alpha} x_{2}^{\beta} \\ &= t^{\alpha + \beta} f(x_{1}, x_{2}). \end{align*}

• If \(\alpha + \beta > 1\), then \(t^{\alpha + \beta} > t\) and \(f(t x_{1}, t x_{2}) > t f(x_{1}, x_{2})\), which means that \(f\) exhibits increasing returns to scale.
• If \(\alpha + \beta < 1\), then \(t^{\alpha + \beta} < t\) and \(f(t x_{1}, t x_{2}) < t f(x_{1}, x_{2})\), which means that \(f\) exhibits decreasing returns to scale.
• If \(\alpha + \beta = 1\), then \(t^{\alpha + \beta} = t\) and \(f(t x_{1}, t x_{2}) = t f(x_{1}, x_{2})\), which means that \(f\) exhibits constant returns to scale.

The marginal product of the constant elasticity production function with respect to the first input factor is

\begin{align*} \frac{\partial f\left(x_{1}, x_{2}\right)}{\partial x_{1}} &= A \frac{r}{\rho} \left(\alpha x_{1}^{\rho} + \left(1 - \alpha\right) x_{2}^{\rho}\right)^{\frac{r}{\rho}-1} \alpha \rho x_{1}^{\rho-1} \\ &= \frac{f\left(x_{1}, x_{2}\right)}{\alpha x_{1}^{\rho} + \left(1 - \alpha\right) x_{2}^{\rho}} \alpha r x_{1}^{\rho-1}. \end{align*}

Analogously, the marginal product with respect to the second input factor is

\begin{align*} \frac{\partial f\left(x_{1}, x_{2}\right)}{\partial x_{2}} &= \frac{f\left(x_{1}, x_{2}\right)}{\alpha x_{1}^{\rho} + \left(1 - \alpha\right) x_{2}^{\rho}} \left(1 - \alpha\right) r x_{2}^{\rho-1}. \end{align*}

One can then calculate the marginal rate of technical substitution by

\begin{align*} \mathrm{MRTS}(x_{1}, x_{2}) &= \frac{\partial f\left(x_{1}, x_{2}\right) / \partial x_{1}}{\partial f\left(x_{1}, x_{2}\right) / \partial x_{2}} \\ &= \frac{\alpha}{1-\alpha} \left(\frac{x_{1}}{x_{2}}\right)^{\rho-1}, \end{align*}

and, thus,

\begin{align*} \ln \mathrm{MRTS}(x_{1}, x_{2}) &= \ln\frac{\alpha}{1-\alpha} + (\rho-1) \ln\frac{x_{1}}{x_{2}}. \end{align*}

Then, we have

\begin{align*} \frac{\mathrm{d}\, \ln \mathrm{MRTS}\left(x_{1}, x_{2}\right)}{\mathrm{d}\, \ln \left(x_{2}/x_{1}\right) } = 1 - \rho, \end{align*}

which implies that

\begin{align*} e \left(x_{2}, x_{1}\right) = \frac{1}{1 - \rho}. \end{align*}

1. Suppose that \(x_{1} = \min\left\{x_{1}, x_{2}\right\}\). For every \(\rho<0\), we can rewrite the constant elasticity of substitution function as \[ f\left(x_{1}, x_{2}\right) = A x_{1}^{r}\left(\alpha + \left(1 - \alpha\right) \left(\frac{x_{2}}{x_{1}}\right)^{\rho}\right)^{\frac{r}{\rho}}. \] Since \(x_{2}/x_{1} > 1\), we have \(0 \le (x_{2}/x_{1})^{\rho} \le 1\) for all \(\rho<0\). Hence \[ A x_{1}^{r}\alpha^{\frac{r}{\rho}} \le f\left(x_{1}, x_{2}\right) \le A x_{1}^{r}, \] which implies \[ A x_{1}^{r} = \lim_{\rho \to -\infty} A x_{1}^{r}\alpha^{\frac{r}{\rho}} \le \lim_{\rho \to -\infty} f\left(x_{1}, x_{2}\right) \le \lim_{\rho \to -\infty} A x_{1}^{r} = A x_{1}^{r}. \]

   When \(x_{2} = \min\left\{x_{1}, x_{2}\right\}\), a similar argument shows that \[ \lim_{\rho \to -\infty} f\left(x_{1}, x_{2}\right) = A x_{2}^{r}. \]

   We can combine the two cases by writing \[ \lim_{\rho \to -\infty} f\left(x_{1}, x_{2}\right) = A \min\left\{x_{1}, x_{2}\right\}^{r}. \]

2. The result is obtained by substituting \(\rho =1\).

3. We can rewrite \[ f\left(x_{1}, x_{2}\right) = A \exp\left(r \frac{\ln\left(\alpha x_{1}^{\rho} + \left(1 - \alpha\right) x_{2}^{\rho}\right)}{\rho}\right). \] By L'Hospital's

   \begin{align*} \lim_{\rho \to 0} \frac{\ln\left( \alpha x_{1}^{\rho} + \left(1 - \alpha\right) x_{2}^{\rho} \right)}{\rho} &= \lim_{\rho \to 0} \frac{\alpha x_{1}^{\rho} \ln x_{1} + \left(1 - \alpha\right) x_{2}^{\rho} \ln x_{2}}{\alpha x_{1}^{\rho} + \left(1 - \alpha\right) x_{2}^{\rho}} \\ &= \alpha \ln x_{1} + \left(1 - \alpha\right) \ln x_{2} \\ &= \ln x_{1}^{\alpha} + \ln x_{2}^{\left(1 - \alpha\right)}. \end{align*}

   By the continuous mapping theorem, one can then conclude

   \begin{align*} \lim_{\rho \to 0} f\left(x_{1}, x_{2}\right) &= A \exp\left(r \ln x_{1}^{\alpha} + r\ln x_{2}^{\left(1 - \alpha\right)} \right) \\ &= A x_{1}^{\alpha r} x_{2}^{\left(1 - \alpha\right)r}. \end{align*}

Maximize \(\pi(L) = 2\sqrt{L} - L\) to find \(L=1\) and \(q=1\). Then, the optimal profit is \(\pi=1\).

1. The production function exhibits decreasing returns to scale.
2. Maximize \(\pi(K, L) = 2K^{1/4}L^{1/4} - K - L\) to find \(L=K\). Substituting into the production function, we find that the maximum satisfies \(\pi(L) = 2L^{1/2} - 2L\). Not all of these points are maximizing allocations. Maximizing \(\pi(L) = 2L^{1/2} - 2L\) gives \(L=1/4\). Thus, \(K=1/4\) and \(q=1/2\).
3. Substituting \(L=K=1/4\) into the profit function gives \(\pi=1/2\).
4. If we attempt to follow the same process, we find \(\pi(L) = 2L - 2L = 0\). Thus, profits are zero irrespective of the chosen allocation. Every choice of output produced by \(K=L\) constitutes a maximizing allocation. This is because the production function exhibits constant returns to scales.

1. The profits of the firm more than double because the output more than doubles, while the cost doubles.

   \begin{align*} \pi\left(2 x_{1}, 2 x_{2}\right) &= p f(2 x_{1}, 2 x_{2}) - w_{1} 2 x_{1} - w_{2} 2 x_{2} \\ & > 2 \left( p f(x_{1}, x_{2}) - w_{1} x_{1} - w_{2} x_{2} \right) \\ & = 2 \pi\left(x_{1}, x_{2}\right). \end{align*}

2. The profits of the firm less than halve because the output less than halves, while the cost halves.

   \begin{align*} \pi\left(x_{1}, x_{2}\right) &= \pi\left(2 \frac{1}{2} x_{1}, 2 \frac{1}{2} x_{2}\right) \\ &= p f\left(2 \frac{1}{2} x_{1}, 2 \frac{1}{2} x_{2}\right) - w_{1} 2 \frac{1}{2} x_{1} - w_{2} 2 \frac{1}{2} x_{2} \\ & < 2 \left( p f\left(\frac{1}{2} x_{1}, \frac{1}{2} x_{2}\right) - w_{1} \frac{1}{2} x_{1} - w_{2} \frac{1}{2} x_{2} \right) \\ & = 2 \pi\left(\frac{1}{2} x_{1}, \frac{1}{2} x_{2}\right). \end{align*}

   Hence,

   \begin{align*} \pi\left(\frac{1}{2} x_{1}, \frac{1}{2} x_{2}\right) > \frac{1}{2} \pi\left(x_{1}, x_{2}\right). \end{align*}

Since the firm makes positive profits, it is better off than not producing at all, in which case its profits are non-positive. However, the firm is not producing the profit maximizing output because the variational condition of interior solutions, namely \(p f'(x) = w\), is not satisfied. Since \(p\) and \(w\) are fixed, \(f'(x)\) needs to decrease for the variational condition to hold. This implies that the input factor level should increase because the marginal product is decreasing.

1. The maximization problem of the firm is

   \begin{align*} \max_{x_{1}, x_{2}} &\left\{ p x_{1}^{1/2} x_{2}^{1/4} - w_{1} x_{1} - w_{2} x_{2} \right\}, \end{align*}

   with first order conditions

   \begin{align*} \frac{w_{1}}{p} &= \frac{1}{2} x_{1}^{-1/2} x_{2}^{1/4} \\ \underbrace{\frac{w_{2}}{p}}_{Real\ factor\ price} &= \underbrace{\frac{1}{4} x_{1}^{1/2} x_{2}^{-3/4}}_{Marginal\ product} \end{align*}

2. Multiplying the two necessary conditions results in

   \begin{align*} \frac{w_{1} w_{2}}{p^2} &= \frac{1}{8} x_{1}^{1/2-1/2} x_{2}^{1/4-3/4}, \end{align*}

   and solving for \(x_{2}\) gives

   \begin{align*} x_{2} = \frac{p^{4}}{64 w_{1}^{2} w_{2}^{2}}. \end{align*}

   To obtain the factor demand for \(x_{1}\), we substitute the demand for \(x_{2}\) in the first variational condition to get

   \begin{align*} \frac{w_{1}}{p} &= \frac{1}{2} x_{1}^{-1/2} \frac{p}{2 \sqrt{2} w_{1}^{1/2} w_{2}^{1/2}}. \end{align*}

   Then solving for \(x_{1}\) gives

   \begin{align*} x_{1} &= \frac{p^{4}}{32 w_{1}^{3} w_{2}}. \end{align*}
3. Lastly, substituting the prices into the demands gives \(x_{1} = x_{2} = 1\). Substituting the optimal demanded quantities into the production function gives \(q = f(1,1) = 1\). Lastly, the firm's profit is calculated as \(\pi = 4 - 2 - 1 = 1\).

Since all input prices remain constant, we have \(\Delta w_{i} = 0\) for all input factors between the two dates. This implies that the revealed profit maximization inequality reduces to \(\Delta p \Delta q \ge 0\). However, we observe \(\Delta p = 2\) and \(\Delta q = -5\). Thus, by the revealed profit maximization inequality, the firm did not maximize its profit on one of these two (or both) dates.

1. The maximization problem of the firm is

   \begin{align*} \max_{x_{1}, x_{2}} &\left\{ p A x_{1}^{\alpha r} x_{2}^{(1 - \alpha) r} - w_{1} x_{1} - w_{2} x_{2} \right\} \end{align*}

2. The necessary conditions for interior solutions are

   \begin{align*} w_{1} &= p A \alpha r x_{1}^{\alpha r-1} x_{2}^{(1 - \alpha)r} = p \alpha r \frac{f(x_{1}, x_{2})}{x_{1}} \\ w_{2} &= p (1 - \alpha) r x_{1}^{\alpha r} x_{2}^{(1 - \alpha)r - 1} = p (1 - \alpha) r \frac{f(x_{1}, x_{2})}{x_{2}} \end{align*}

3. To shorten the notation, let \(q = f(x_{1}, x_{2})\). With this notation, we can rewrite the first order conditions as

   \begin{align*} x_{1} &= p \alpha r \frac{q}{w_{1}} \\ x_{2} &= p (1 - \alpha) r \frac{q}{w_{2}}. \end{align*}

   Substituting in the production function gives

   \begin{align*} q &= f(x_{1}, x_{2}) \\ &= A \left(p \alpha r \frac{q}{w_{1}}\right)^{\alpha r} \left(p (1 - \alpha) r \frac{q}{w_{2}}\right)^{(1 - \alpha) r} \\ &= A (p r q)^{r} \left(\frac{\alpha}{w_{1}}\right)^{\alpha r} \left(\frac{1 - \alpha}{w_{2}}\right)^{(1 - \alpha) r}, \end{align*}

   which implies

   \begin{align*} q(w_{1}, w_{2}) &= A^{\frac{1}{1-r}} (p r)^{\frac{r}{1-r}} \left(\frac{\alpha}{w_{1}}\right)^{\frac{\alpha r}{1-r}} \left(\frac{1 - \alpha}{w_{2}}\right)^{\frac{(1 - \alpha) r}{1-r}}. \end{align*}

4. We can obtain the input factor demands by replacing the supply function in the first order conditions. This gives

   \begin{align*} x_{1} &= p \alpha r \frac{A^{\frac{1}{1-r}} (p r)^{\frac{r}{1-r}} \left(\frac{\alpha}{w_{1}}\right)^{\frac{\alpha r}{1-r}} \left(\frac{1 - \alpha}{w_{2}}\right)^{\frac{(1 - \alpha) r}{1-r}}}{w_{1}} \\ &= (A p r)^{\frac{1}{1-r}} \left(\frac{\alpha}{w_{1}}\right)^{\frac{1 - (1 - \alpha)r}{1-r}} \left(\frac{1 - \alpha}{w_{2}}\right)^{\frac{(1 - \alpha) r}{1-r}}, \end{align*}

   and analogously

   \begin{align*} x_{2} &= (A p r)^{\frac{1}{1-r}} \left(\frac{\alpha}{w_{1}}\right)^{\frac{\alpha r}{1-r}} \left(\frac{1 - \alpha}{w_{2}}\right)^{\frac{1 - \alpha r }{1-r}}. \end{align*}

5. The easiest way to calculate the profit of the firm is to use the first order conditions of part 2 and the supply function. We then have

   \begin{align*} \pi(p, w_{1}, w_{2}) &= p q - w_{1} p \alpha r \frac{q}{w_{1}} - w_{2} p (1 - \alpha) r \frac{q}{w_{2}} \\ &= p q\left(1 - \alpha r - (1 - \alpha) r \right) \\ &= p (1 - r) q \\ &= (p A)^{\frac{1}{1-r}} \left(1 - r \right) r^{\frac{r}{1-r}} \left(\frac{\alpha}{w_{1}}\right)^{\frac{\alpha r}{1-r}} \left(\frac{1 - \alpha}{w_{2}}\right)^{\frac{(1 - \alpha) r}{1-r}} . \end{align*}
6. If \(r=1\), the production technology has constant returns to scale. Then, the firm has zero profit, and it is indifferent about its level of supply.

1. For a fixed level of cost \(\hat c\), we have \(\hat c = 40 x_{1} + 10 x_{2}\). Solving for \(x_{2}\) gives \[ x_{2} = \frac{1}{10}\hat c - 4 x_{1}, \] so the slope of the isocost with respect to \(x_{1}\) is equal to \(-4\).
2. From the cost minimizing condition, one can calculate \[ \mathrm{MRTS}(x_{1}, x_{2}) = \frac{w_{1}}{w_{2}} = 4. \] The firm substitutes a unit of \(x_{1}\) with \(4\) units of \(x_{2}\) at the minimizing cost level of production.
3. Using the cost minimizing condition, we obtain \[ 4 = \mathrm{MRTS}(x_{1}, x_{2}) = \frac{f_{x_{1}}(x_{1}, x_{2})}{f_{x_{2}}(x_{1}, x_{2})} = \frac{2 x_{1}^{-1/2} x_{2}^{1/2}}{2 x_{1}^{1/2} x_{2}^{-1/2}} = \frac{x_{2}}{x_{1}}. \] Solving for \(x_{2}\) and substituting into the production function gives \[ q = 4 x_{1}^{1/2} x_{2}^{1/2} = 4 x_{1}^{1/2} 2 x_{1}^{1/2} = 8 x_{1}. \] The last equation implies that \(x_{1} = q / 8\) and \(x_{2} = q / 2\). Hence, the cost function is \[ c(q) = w_{1} x_{1} + w_{2} x_{2} = 40 \frac{q}{8} + 10 \frac{q}{2} = 10 q. \]
4. Substituting the given output level results in \(x_{1} = 5\) and \(x_{2} = 20\). Moreover, the minimum cost is \(400\).

The firm does not minimize its cost because the variational condition \(\mathrm{MRTS}(x_{1}, x_{2}) = w_{1} / w_{2}\) is not satisfied. Since \(w_{1}\) and \(w_{2}\) are fixed, \(\mathrm{MRTS}(x_{1}, x_{2})\) needs to decrease for the variational condition to hold. Since the marginal rate of technical substitution is decreasing, the firm can reduce its cost by increasing the input factor level of \(x_{1}\) and decreasing that of \(x_{2}\).

We calculate \(\Delta w_{1}=2\), \(\Delta w_{2}=2\), \(\Delta x_{1} =4\), and \(\Delta x_{2} =-2\). Therefore \[ \Delta w_{1} \Delta x_{1} + \Delta w_{2} \Delta x_{2} = 2 \cdot 4 + 2 \cdot (-2) = 4 > 0. \] Since the revealed cost minimization inequality is not satisfied, the firm does not minimize its cost at one of these two (or both) dates.

1. The production function exhibits decreasing returns to scale.

2. The Lagrangian of the problem is obtained by subtracting from the expenditure of the firm the production constraint multiplied by the Lagrange multiplier \(\lambda\), i.e.,

   \begin{align*} \mathcal{L} = w_{1} x_{1} + w_{2} x_{2} - \lambda \left(2 x_{1}^{\frac{1}{6}} x_{2}^{\frac{1}{3}} - q\right). \end{align*}

3. The first order conditions are obtained by differentiating the Lagrangian with respect to the two input factors ($x₁ and \(x_{2}\)). This gives

   \begin{align*} w_{1} &= \frac{ 2 \lambda x_{1}^{-\frac{5}{6}} x_{2}^{\frac{2}{6}}}{6}, \\ w_{2} &= \frac{ 2 \lambda x_{1}^{\frac{1}{6}} x_{2}^{-\frac{2}{3}}}{3}. \end{align*}

4. From the two first order conditions, we obtain

   \begin{align*} \frac{w_{1}}{w_{2}} &= \frac{1}{2}\frac{x_{2}}{x_{1}}. \end{align*}

   Solving for \(x_{1}\) and substituting to the production function gives the conditional demand of \(x_{2}\), namely

   \begin{align*} x_{2} &= 2^{-\frac{5}{3}}\left(\frac{w_{1}}{w_{2}}\right)^{\frac{1}{3}} q^{2}. \end{align*}

   Instead, the conditional demand of \(x_{1}\) is obtained by solving for \(x_{2}\) and substituting to the production function, i.e.,

   \begin{align*} x_{1} &= 2^{-\frac{8}{3}} \left(\frac{w_{2}}{w_{1}}\right)^{\frac{2}{3}} q^{2}. \end{align*}

5. The cost function is calculated by replacing the conditional factor demands in the expenditure of the firm, namely

   \begin{align*} c(q, w_{1}, w_{2}) &= w_{1} 2^{-\frac{8}{3}} \left(\frac{w_{2}}{w_{1}}\right)^{\frac{2}{3}} q^{2} + w_{2} 2^{-\frac{5}{3}}\left(\frac{w_{1}}{w_{2}}\right)^{\frac{1}{3}} q^{2} \\ &= \frac{3}{2^{8/3}}w_{1}^{\frac{1}{3}} w_{2}^{\frac{2}{3}} q^{2}. \end{align*}

Suppose that a firm maximizes its profits under prices \(p\), \(w_{1}\), \(w_{2}\) when producing \(\hat q\) and demanding \(\hat x_{1}\) and \(\hat x_{2}\). Towards contradiction, suppose that \(\hat x_{1}\) and \(\hat x_{2}\) do not minimize the firm's cost for producing \(\hat q\) under the given prices. Then, there exist \(\tilde x_{1}\) and \(\tilde x_{2}\), different than \(\hat x_{1}\) and \(\hat x_{2}\), that produce \(\hat q\) with less cost. Then

\begin{align*} w_{1} \hat x_{1} + w_{2} \hat x_{2} &> w_{1} \tilde x_{1} + w_{2} \tilde x_{2} \implies \\ \hat \pi = p \hat q - w_{1} \hat x_{1} - w_{2} \hat x_{2} &< p \hat q - w_{1} \tilde x_{1} - w_{2} \tilde x_{2}, \end{align*}

which implies that \(\hat q\), \(\hat x_{1}\), and \(\hat x_{2}\) are not profit maximizing.

The conditional factor demand is obtained by the cost minimization problem, while the factor demand is obtained by the profit maximization problem. The conditional factor demand depends on the output quantity, say \(q\), and factor prices, say \(w_{i}\). Instead, the factor demand depends on the output price, say \(p\), and factor prices.

Let \(x_{i}^{c}\) and \(x_{i}\) correspondingly denote the conditional factor and factor demands. By exercise 1, if a firm is maximizing its profits, then its minimizes it costs. Therefore, if \(q(p, w_{1}, w_{2})\) is the supply function obtained by the profit maximization problem, then the two demands are interrelated by

\begin{align*} x_{i}^{c}\left(q(p, w_{1}, w_{2}), w_{1}, w_{2}\right) = x_{i}\left(p, w_{1}, w_{2}\right). \end{align*}

1. The Lagrangian of the problem is obtained by subtracting from the objective of the firm the production constraint multiplied by the Lagrange multiplier \(\lambda\). Specifically,

   \begin{align*} \mathcal{L} = w_{1} x_{1} + w_{2} x_{2} - \lambda \left(A x_{1}^{\alpha r} x_{2}^{(1 - \alpha) r} - q\right). \end{align*}

2. Differentiating the Lagrangian of the problem gives the first order conditions

   \begin{align*} w_{1} &= \lambda A \alpha r x_{1}^{\alpha r - 1} x_{2}^{(1 - \alpha) r} = \lambda \alpha r \frac{q}{x_{1}}, \\ w_{2} &= \lambda A (1 - \alpha) r x_{1}^{\alpha r} x_{2}^{(1 - \alpha) r - 1} = \lambda (1 - \alpha) r \frac{q}{x_{2}}. \end{align*}

3. Solving the first order conditions for \(x_{1}\) and \(x_{2}\), respectively, and substituting to the production function results in

   \begin{align*} q = A \left(\lambda \alpha r \frac{q}{w_{1}}\right)^{\alpha r} \left(\lambda (1 - \alpha) r \frac{q}{w_{2}}\right)^{(1 - \alpha) r} = A \lambda^{r} r^{r} \left(\frac{\alpha}{w_{1}}\right)^{\alpha r} \left(\frac{1 - \alpha}{w_{2}}\right)^{(1 - \alpha) r} q^{r}, \end{align*}

   from which we obtain

   \begin{align*} \lambda = A^{-\frac{1}{r}} r^{-1} \left(\frac{\alpha}{w_{1}}\right)^{-\alpha} \left(\frac{1 - \alpha}{w_{2}}\right)^{-(1 - \alpha)} q^{\frac{1-r}{r}}, \end{align*}

   Substituting back to the first order conditions, we can eliminate \(\lambda\), and solve for \(x_{1}\), \(x_{2}\) to obtain the conditional demand functions

   \begin{align*} x_{1} &= \lambda \alpha r \frac{q}{w_{1}} = \left(\frac{w_{2}}{w_{1}}\frac{\alpha}{1 - \alpha}\right)^{1-\alpha} \left(\frac{q}{A}\right)^{\frac{1}{r}}, \\ x_{2} &= \lambda (1 - \alpha) r \frac{q}{w_{2}} = \left(\frac{w_{1}}{w_{2}}\frac{1 - \alpha}{\alpha}\right)^{\alpha} \left(\frac{q}{A}\right)^{\frac{1}{r}}. \end{align*}

4. The easiest way to calculate the cost function of the firm is to use the first order conditions of part 2 and the objective function. We then have

   \begin{align*} c(q, w_{1}, w_{2}) &= w_{1} \lambda \alpha r \frac{q}{w_{1}} + w_{2} \lambda (1 - \alpha) r \frac{q}{w_{2}} \\ &= \lambda r q \\ &= \left(\frac{\alpha}{w_{1}}\right)^{-\alpha} \left(\frac{1 - \alpha}{w_{2}}\right)^{-(1 - \alpha)} \left(\frac{q}{A}\right)^{\frac{1}{r}}. \end{align*}

Cost type	Expression
Total cost	\(c(q) = 16 + 4 q^{2}\)
Average cost	\(\bar{c}(q) = \frac{16}{q} + 4 q\)
Average variable cost	\(\bar{\mu}(q) = 4 q\)
Average fixed cost	\(\bar{\sigma}(q) =\frac{16}{q}\)
Marginal cost	\(c'(q) = 8 q\)

The minimum average cost is obtained by the first order condition \[ -\frac{16}{q^{2}} + 4 = 0, \] which implies that \(q = 2\).

The average cost function for the given parameters values is given by

\begin{align*} \bar{c}(q) &= q^{\frac{1 - r}{r}}, \end{align*}

with derivative

\begin{align*} \bar{c}'(q) &= \frac{1 - r}{r}q^{\frac{1 - 2r}{r}}. \end{align*}

If \(r<1\), then \(\bar{c}' >0\) and \(\bar{c}\) is increasing. If \(r>1\), then \(\bar{c}' <0\) and \(\bar{c}\) is decreasing. Finally, if \(r=1\), then \(\bar{c}' =0\) and \(\bar{c}\) is constant.

1. The first order condition of the profit maximization problem is \(20 q = p\). Solving for \(q\) results in the supply function \(s(p) = p / 20\).

2. The average cost is determined by

   \begin{align*} \bar{c}(q) = \frac{1000}{q} + 10 q . \end{align*}

   We can calculate the minimum average cost by solving the equation \(\bar c(q) = c'(q)\), which is equivalent to

   \begin{align*} \frac{1000}{q} + 10 q = 20 q. \end{align*}

   This results to \(q = 10\).

3. The profit function is given by

   \begin{align*} \pi(p) &= p s(p) - c\left(s(p)\right) \\ &= \frac{p^{2}}{20} - 1000 - 10 \frac{p^{2}}{400} \\ &= - 1000 + \frac{p^{2}}{40} . \end{align*}
4. The effect of a marginal price change on profit is \(\pi'(p) = p / 20\).
5. The firm makes non negative profit for prices such that \(\pi(p) \ge 0\), which implies that \(p \ge 200\).

We solve for \(p\) to get

\begin{align*} p(q) &= - 25 + \frac{q}{4}. \end{align*}

Note that \(s_{1}\) is valid for \(p\ge 10\) because the resulting supplied quantities are negative for smaller prices. Similarly, \(s_{2}\) is valid for \(p\ge 15\). Therefore, the market supply is given by

\begin{align*} s(p) &= s_{1}(p) \mathbb{1}_{[10, \infty)}(p) + s_{2}(p) \mathbb{1}_{[15, \infty)}(p) \\ &= \left\{\begin{aligned} &0 & p < 10 \\ &-10 + p & 10 \le p < 15 \\ &-25 + 2p & 15 \le p . \end{aligned}\right. \end{align*}

Thus, the market supply function has two kinks, one at \(p=10\) and one at \(p=15\).

1. The firm \(i\)'s profit maximization condition gives the inverse supply function

   \begin{align*} p = c'(q) = 2 q, \end{align*}

   from which we get the supply \(s_{i}(p) = p / 2\). One can then calculate the profit function of the firm as

   \begin{align*} \pi_{i}(p) = p \frac{p}{2} - \frac{p^{2}}{4} - 1 = \frac{p^{2}}{4} - 1. \end{align*}

2. With \(n\) identical firms, the market supply is

   \begin{align*} s(p) = \sum_{i=1}^{n} s_{i}(p) = \frac{n}{2}p. \end{align*}

3. Equating market demand and supply gives

   \begin{align*} \frac{n}{2}p &= s(p) = d(p) = 52 - p \iff \\ p &= \frac{104}{n + 2}. \end{align*}

4. The last equation implies that the firm i's profit can be rewritten as

   \begin{align*} \pi_{i}(n) = \left(\frac{52}{n + 2}\right)^{2} - 1. \end{align*}

5. As long as entry is free and instantaneous, firms enter the market when they can make positive profits. Therefore, the equilibrium number of firms is exactly such that if one more firm enters the market, profits turn negative. Hence, the equilibrium number of firms is given by the greatest natural number \(n\) for which \(\pi(n)\ge 0\). The non-negative profit condition is satisfied if

   \begin{align*} \frac{52}{n + 2} \ge 1, \end{align*}

   or, equivalently, if \(n\le 50\). Therefore, the equilibrium number of firms is \(n=50\).

6. Since \(n=50\), the equilibrium price is \(p=2\). Substituting the equilibrium number of firms and price in the market supply equation results in \(s(2) = 50\).

7. With the shifted market demand, the market clearing condition implies that

   \begin{align*} p &= \frac{105}{n + 2}, \end{align*}

   hence, profits as a function of \(n\) become

   \begin{align*} \pi_{i}(n) = \left(\frac{52.5}{n + 2}\right)^{2} - 1. \end{align*}

   This implies that \(n\le 50.5\) and, thus, the equilibrium number of firms remains unchanged equal to \(n=50\). One then can calculate the equilibrium price by

   \begin{align*} p &= \frac{105}{52} > 2. \end{align*}

   The supplied market quantity is

   \begin{align*} s &= \frac{50}{2} \frac{105}{52} > 50 \end{align*}

   The equilibrium price increases, the supplied quantity also increases, and individual firm profits turn positive, i.e.

   \begin{align*} \pi_{i}(n) = \left(\frac{52.5}{52}\right)^{2} - 1 > 0. \end{align*}

The monopolist solves \[ \max_{p}\left\{ p d(p) - c\left(d(p)\right) \right\}. \] The first order condition of the problem is \[ d(p) + p d'(p) - c'\left(d(p)\right) d'(p) = 0. \] Substituting the exercise's functional forms and calculating gives \(p_{m} = 26\). Substituting back to demand gives \(q_{m} = d(26) = 48\). The monopolist achieves profit \[ \pi(26) = 2 d(26) - c\left(d(26)\right) = 26 \cdot 48 - 2 \cdot 48 = 1152. \] When demand is affine, the deadweight loss can be calculated using a geometric argument and the formula for the area of a triangle. For this, we need the perfect competition price and quantity. The perfect competition price is given by \(p_{c} = 2\) (why?) and, the perfect competition quantity is \(q_{c} = 96\). One then calculates \[ D\left(q_{c}, q_{m}\right) = \frac{1}{2} \left(q_{c} - q_{m}\right) \left(p_{m} - p_{c}\right) = \frac{1}{2} (96 - 48)(26 - 2) = 576 \]

There are two differences compared to exercise 1. Firstly, demand is hyperbolic instead of affine, and secondly, the cost function is quadratic instead of linear. In this case, the maximization problem of the monopolist is \[ \max_{p} \left\{ p \frac{100}{p} - \frac{100^{2}}{p^{2}} \right\} = \max_{p} \left\{ 100 - \frac{100^{2}}{p^{2}} \right\}. \] The derivative of the objective is \(2 \cdot 10^{4} / p^{3}\), which is positive for all positive prices. This implies that the objective is increasing and, therefore, the maximization problem does not have a solution. The monopolist would like to let prices go to infinity. Taking this limit in the demand equation, we see that the monopolist would like to abstain from producing, i.e., \[ \lim_{p\to\infty} d(p) = \lim_{p\to\infty} \frac{100}{p} = 0. \]

The problem of the monopolist is \[ \max_{s, p} \left\{ p d(s, p) - \frac{s}{2} - 2 d(s, p) \right\} = \max_{s, p} \left\{ 3 \sqrt{s} \left( \frac{1}{p^{2}} - 2 \frac{1}{p^{3}} \right) - \frac{s}{2} \right\}. \] The first order conditions of the problem are

\begin{align*} -2 \frac{1}{p^{3}} + 6 \frac{1}{p^{4}} &= 0, \\ \frac{3}{2} \frac{1}{\sqrt{s}} \left( \frac{1}{p^{2}} - 2 \frac{1}{p^{3}} \right) - \frac{1}{2} &= 0. \end{align*}

The first condition involves only \(p\), and the monopolistic price can be calculated from it. Solving for prices gives \(p=3\). By substituting \(p\) in the second condition, one can then solve for the qualitative attribute to get \(s=1/81\). Using the demand function, the output quantity level that the monopolist produces is \(q = 1/81\). Finally, the profit is \[ \pi = 3 \sqrt{\frac{1}{81}} \left( \frac{1}{9} - 2 \frac{1}{27} \right) - \frac{1}{2}\frac{1}{81} = \frac{1}{162} . \]

No. By the monopolistic pricing condition, one can see that the monopolist would like to set a negative price in such a case. Specifically,

\begin{align*} p = \frac{c'\left(d(p)\right) e_{d}}{e_{d} + 1} < 0. \end{align*}

The monopolistic pricing condition is

\begin{align*} p = \frac{c_{1} e_{d}}{e_{d} + 1}, \end{align*}

where elasticity does not depend on prices by the assumptions of the exercise. We can calculate

\begin{align*} \frac{\mathrm{d}\, p}{\mathrm{d}\, c_{1}} &= \frac{e_{d}}{e_{d} + 1}. \end{align*}

Since \(e_{d} < -1\), then \(\mathrm{d}\, p / \mathrm{d}\, c_{1} > 0\). This implies that whenever a technological improvement reduces marginal cost, the monopolistic price in the market falls.

The difference with exercise 1 is that demand, in this case, is hyperbolic instead of affine. The first order condition of this problem yields \[ \frac{a}{p^{b}} + p \frac{-a b}{p^{b+1}} - c_{1} \frac{-a b}{p^{b+1}} = 0, \] which gives \(p_{m}= c_{1}b / (b-1)\) and \(q_{m} = a (b-1)^{b} / (c_{1}b)^{b}\). The monopolist makes profit \[ \pi_{m} = \frac{c_{1}b}{b-1} \frac{a (b-1)^{b}}{(c_{1}b)^{b}} - c_{1} \frac{a (b-1)^{b}}{(c_{1}b)^{b}} = \frac{c_{1} a (b-1)^{b-1}}{(c_{1}b)^{b}}. \] The competition price is \(p_{c}=c_{1}\) and the competition quantity is \(q_{c} = a / c_{1}^{b}\). The deadweight loss can be calculated by \[ D\left(q_{c}, q_{m}\right) = \int_{q_{m}}^{q_{c}} \left(p(z) - c'(z)\right) \mathrm{d} z, \] where \(p(q) = (a / q)^{1/b}\) is the inverse demand function. For the given parameter values, we get \(q_{c} = 4\), \(q_{m} = 1\), and

\begin{align*} D\left(q_{c}, q_{m}\right) &= \int_{q_{m}}^{q_{c}} \left(a^{1/b} z^{-1/b} - c_{1} \right) \mathrm{d} z \\ &= \int_{1}^{4} \left(4 z^{-1/2} - 2 \right) \mathrm{d} z \\ &= \left. 8 z^{1/2} - 2 z \right|_{1}^{4} \\ &= 16 - 8 - (8 - 2) \\ &= 2 \end{align*}

Suppose that marginal cost is equal to \(c\). The monopolistic pricing condition is

\begin{align*} p = \frac{c e_{d}(p)}{e_{d}(p) + 1}. \end{align*}

Therefore, we can calculate

\begin{align*} \frac{\mathrm{d}\, p}{\mathrm{d}\, c} &= \frac{e_{d}(p)}{e_{d}(p) + 1} + \frac{c}{\left(e_{d}(p) + 1\right)^{2}} \left(e_{d}'(p) \frac{\mathrm{d}\, p}{\mathrm{d}\, c} \left(e_{d}(p) + 1\right) - e_{d}(p) e_{d}'(p) \frac{\mathrm{d}\, p}{\mathrm{d}\, c}\right) \\ &= \frac{e_{d}(p)}{e_{d}(p) + 1} + \frac{c}{\left(e_{d}(p) + 1\right)^{2}} e_{d}'(p) \frac{\mathrm{d}\, p}{\mathrm{d}\, c}, \end{align*}

from which we get

\begin{align*} \frac{\mathrm{d}\, p}{\mathrm{d}\, c} &= \frac{e_{d}(p) \left(e_{d}(p) + 1\right)}{\left(e_{d}(p) + 1\right)^{2} - c e_{d}'(p)}. \end{align*}

Thus, we also need to calculate

\begin{align*} e_{d}'(p) &= \frac{\mathrm{d}\, \left(d'(p)\frac{p}{d(p)}\right)}{\mathrm{d}\, p} \\ &= d''(p)\frac{p}{d(p)} + d'(p)\frac{1}{\left(d(p)\right)^{2}} \left(d(p) - p d'(p) \right) \\ &= d''(p)\frac{p}{d(p)} + \frac{1}{p} d'(p)\frac{p}{d(p)} \left(1 - d'(p)\frac{p}{d(p)} \right) \\ &= d''(p)\frac{p}{d(p)} + \frac{1}{p} e_{d}(p) \left(1 - e_{d}(p) \right) . \end{align*}

Firstly, since demand is affine, we have \(d'' = 0\) and, secondly, by the monopolist's pricing condition, we can rewrite the last expression as

\begin{align*} e_{d}'(p) &= \frac{e_{d}(p) + 1}{c} \left(1 - e_{d}(p) \right) = \frac{1 - \left(e_{d}(p)\right)^{2}}{c} . \end{align*}

Substituting this result in the expression for \(\mathrm{d}\, p / \mathrm{d}\, c\), we obtain

\begin{align*} \frac{\mathrm{d}\, p}{\mathrm{d}\, c} &= \frac{e_{d}(p) \left(e_{d}(p) + 1\right)}{\left(e_{d}(p) + 1\right)^{2} - 1 + \left(e_{d}(p)\right)^{2}} = \frac{1}{2}. \end{align*}

The monopolist solves \[ \max_{p_{1}, p_{2}} \left\{ p_{1} d_{1}(p_{1}) + p_{2} d_{2}(p_{2}) - c_{1}\left(d_{1}(p_{1}) + d_{2}(p_{2})\right) \right\}. \] The optimality conditions for this problem are

\begin{align*} d_{i}(p_{i}) + p_{i} d_{i}'(p_{i}) - c_{1} d_{i}'(p_{i}) &= 0 \qquad (i=1,2). \end{align*}

From them, we deduce that

\begin{align*} p_{i} = \frac{c}{\frac{1}{e_{d,i}} + 1} \qquad (i=1,2). \end{align*}

1. There are two Nash equilibria in this game, namely \(\{Foo, Foo\}\) and \(\{Bar, Bar\}\).
2. The objective of the players is to coordinate. The players would like to choose the same action. Games of this type are called coordination games.

Not necessarily. In general, player \(B\) maximizes her payoff by choosing the best response strategy to player \(A\)'s strategy. If the Nash equilibrium strategy of player B is a dominant strategy, then she should choose it. Dominant strategies are best responses irrespective of the choice of other players. Therefore, if her Nash equilibrium strategy is a dominant strategy, she will still maximize her payoff by choosing it. For example, if player \(A\) chooses \(Deny\) in a prisoner's dilemma game, since the Nash equilibrium strategy of player \(B\), namely \(Confess\), is a dominant strategy, she maximizes her payoff by choosing it.

In other games, in which her Nash equilibrium strategy is not dominant, the best response to the choice of player \(A\) can differ from the Nash equilibrium strategy and, therefore, player \(B\) might switch to a non Nash equilibrium strategy. For example, consider the game obtained by modifying the actions' labels and payoffs of the prisoner's dilemma in the following way.

In this game, if player \(A\) chooses \(Bottom\), the best response of player \(B\) is \(Right\). As is to be expected, \(Left\) is not a dominant strategy.

The Nash equilibrium obtained by backward induction is \[\left\{\left(Bottom, Up', Up \right), \left(Right', Right\right)\right\}\]

1. There are two Nash equilibria given by \(\{Top, Left\}\) and \(\{Bottom, Right\}\).
2. The extensive form of the sequential game is

3. The Nash equilibrium obtained by backward induction is \(\{Top, \left(Left', Right\right)\}\).
4. Nash equilibria in which player \(A\) plays \(Bottom\) cannot be obtained by backward induction in the sequential game. This happens because player \(A\) prefers choosing \(Top\) at date \(1\) as she receives a higher payoff. Even if player \(B\) attempts to convince player \(A\) to choose \(Bottom\) by threatening that she will play \((Right',Right)\), her statement is not credible. Once player \(A\) chooses \(Top\), player \(B\) receives a greater payoff by choosing \(Left'\). In general, dynamic aspects in a game can affect the resulting the credibility of some Nash equilibria. Therefore, when analyzing strategic interactions for which time is a central component, it is important not to neglect it because doing so might lead to less refined equilibrium outcomes.

1. No. There is no set of strategies that constitute best responses to one another.

2. Suppose that player \(A\) randomizes by playing \(Top\) with probability \(p > 0\) and \(Bottom\) with probability \(1-p\). If this mixed strategy is a best response, it must equalize the expected payoff of the strategies of player \(B\) because otherwise, player \(B\) would choose either \(Left\) or \(Right\). The unique best response of player \(A\) to \(Left\) is \(Bottom\), which means that the mixed strategy \((p, 1-p)\) with \(p>0\) cannot be the best response to player \(B\)'s choice of \(Left\). Similarly, we can exclude the case of player \(B\) choosing \(Right\).

   Therefore, for the mixed strategy to be a best response, it should hold \[ \underbrace{p \cdot 2 + (1-p) \cdot 0}_{\text{Player B's expected payoff for $Left$}} = \underbrace{p \cdot 0 + (1-p) \cdot 2}_{\text{Player B's expected payoff for $Right$}}. \] This implies that \(p= 1/2\).

   Suppose that player B randomizes by playing \(Left\) with probability \(q > 0\) and \(Right\) with probability \(1-q\). Using similar arguments as above, the randomization of player B should equalize the payoff of the strategies of player A, namely \[ \underbrace{q \cdot 0 + (1-q) \cdot 2}_{\text{Player A's expected payoff for $Top$}} = \underbrace{q \cdot 2 + (1-q) \cdot 0}_{\text{Player A's expected payoff for $Bottom$}}. \] The last condition implies that \(q= 1/2\).

   Combining the above, the unique Nash equilibrium of the game is given by the mixed strategies \[ \left\{\left(p = \frac{1}{2}, 1-p = \frac{1}{2}\right), \left(q = \frac{1}{2}, 1-q = \frac{1}{2}\right)\right\}. \]

1. We can find the Nash equilibrium by backward induction. At date \(2\), the unique Nash equilibrium of the stage game is obtained when both players choose \(Confess\). Since there are no further dates in which the stage game is repeated, the players cannot coordinate in the Pareto efficient outcome by promising cooperation in the future. At date \(1\), the unique Nash equilibrium of the stage game is obtained once more for both players choosing \(Confess\). Since the only Nash equilibrium of the game at date \(2\) is the pair \(\{Confess, Confess\}\), players cannot coordinate at the Pareto efficient outcome at date \(1\) by threatening to punish non-cooperation in the future. Therefore, the unique Nash equilibrium of the game is obtained by \(\{(Confess, Confess), (Confess, Confess)\}\).
2. If the game is repeated a finite number of times, the unique Nash equilibrium is obtained when both players choose \(Confess\) at every date. This result can be obtained by backward induction. At the final date of the game, the unique Nash equilibrium of the stage game is \(\{Confess, Confess\}\), justified as in the case of the two repetitions. Suppose that we have established that the unique Nash equilibrium of the last \(k\) repetitions of the game is \[ \left\{\underbrace{(Confess, ..., Confess)}_{k-times}, \underbrace{(Confess, ..., Confess)}_{k-times}\right\}. \] With the argument used for date 1 in the case of two repetitions, one concludes that the unique Nash equilibrium of the \((n-k)\) -th date of the game is \(\{Confess, Confess\}\). Continuing in this respect, we conclude that the unique Nash equilibrium of the repeated game is obtained when both players choose \(Confess\) at every date.
3. When a stage game with a unique Nash equilibrium is repeated a finite number of times, the players cannot establish credible promises or threats. However, if the repeated game has an infinite horizon, then coordination strategies that deviate from stage game Nash equilibria are feasible, even if there is only a unique Nash equilibrium in the stage game. An example of this is the infinitely repeated prisoner's dilemma. Promises and threats can be established when a stage game is repeated only a finite number of times, only if the stage game has multiple Nash equilibria, one of which can be used in coordination and one in non-coordination.

1. The game has three Pareto efficient allocations. One from the strategy profile for which both players do not confess, and both combinations where one player confesses and the other does not. The only Pareto inefficient allocation is the result of the strategy profile in which both players confess since the profile in which both players do not confess constitutes a Pareto improvement.

2. Let \(u_{i}\) denote the payoff of player \(i\in\{A,B\}\). We consider the cases for the strategy profiles \(\{N,N\}\), \(\{N,C\}\), and \(\{C,N\}\), where \(C\) is short for the strategy confess and \(N\) for the strategy not confess.

   For the case \(\{N,N\}\), choose \(\alpha=1/2\) and let \[W_{1}(s_{Α}, s_{Β}) = \frac{1}{2}u_{Α}(s_{Α}, s_{Β}) + \frac{1}{2}u_{Β}(s_{Α}, s_{Β}).\] Then, we have \[\underbrace{W_{1}(N,N)}_{=-1} > \underbrace{W_{1}(C,N) = W_{1}(N,C)}_{=-5} > \underbrace{W_{1}(C,C)}_{-6}.\]

   For the case \(\{N,C\}\), choose \(\alpha=1/20\) (any \(\alpha \le 1/10\) works) and let \[W_{2}(s_{A}, s_{B}) = \frac{1}{20}u_{A}(s_{A}, s_{B}) + \frac{19}{20}u_{B}(s_{A}, s_{B}).\] Then, we have \[\underbrace{W_{2}(N,C)}_{=-1/2} > \underbrace{W_{2}(N,N)}_{=-1} > \underbrace{W_{2}(C,C)}_{-6} > \underbrace{W_{2}(C,N)}_{-9.5}.\]

   The last case \(\{C,N\}\) is symmetric to the case \(\{N,C\}\) and we can choose any \(\alpha\in[9/10, 1)\) to show the claim.

3. Let \(\alpha\) take an arbitrary value in \((0,1)\) and define \[W_{3}(s_{A}, s_{B}) = \alpha u_{A}(s_{A}, s_{B}) + (1 - \alpha)u_{B}(s_{A}, s_{B}).\] We can then calculate \[W_{3}(C, C)=-6 < -1 = W_{3}(N, N).\] Since the choice of \(\alpha\) was arbitrary, the claim follows.

Let \(i\neq j\) for \(i,j=1,2\). Firm \(i\) chooses its strategy by solving \[ \max_{q_{i}} \left\{ p(q_{i} + q_{j}) q_{i} - c_{1} q_{i} \right\}. \] For interior solutions, the firm sets

\begin{align*} p_{0} + 2p_{1}q_{i} + p_{1}q_{j} - c_{1} = 0, \end{align*}

which gives the best response function

\begin{align*} b_{i}(q_{j}) = \frac{c_{1} - p_{0} - p_{1}q_{j}}{2p_{1}} . \end{align*}

Since the problem is symmetric, we can set \(q_{i}=q_{j}\) in the first order condition to easily obtain the equilibrium output

\begin{align*} q_{i} = \frac{c_{1} - p_{0}}{3p_{1}} . \end{align*}

Thus, the total market quantity is

\begin{align*} q = q_{1} + q_{2} = 2\frac{c_{1} - p_{0}}{3p_{1}}, \end{align*}

which implies that the equilibrium price is

\begin{align*} p(q) = \frac{p_{0} + 2 c_{1}}{3}. \end{align*}

Finally, each firm \(i\) makes profit \[ \pi_{i}= (p(q) - c_{1}) q_{i} = - \frac{\left(c_{1} - p_{0}\right)^{2}}{9p_{1}} . \]

Yes. The simultaneous price competition with homogeneous products leads firms to choose the competitive price, which, in turn, results in the production of the efficient output level. In contrast, competition in quantities results in deadweight losses.

1. \(c'(q) = 4\).

2. The perfect competition price is equal to the marginal production cost, i.e., \(p_{c}=4\). By inverting the inverse demand and substituting the competitive price, we obtain

   \begin{align*} q_{c} = \frac{100 - p_{c}}{2} = 48. \end{align*}

3. Each firm solves

   \begin{align*} \max_{q_{i}} \left\{ p\left( q_{i} + q_{j} \right) q_{i} - c(q_{i}) \right\}, \end{align*}

   with first order condition

   \begin{align*} 100 - 4q_{i} - 2 q_{j} -4 = 0. \end{align*}

   Therefore, the best response of each firm is

   \begin{align*} b_{i}(q_{j}) = \frac{48 - q_{j}}{2} \end{align*}

   and the firm's optimal quantity is \(q_{i} = 16\). The market quantity is \(q_{s} = 32\), and the market price is \(p_{s} = 36\).

4. The intersection of the best responses indicates the equilibrium quantities.

5. In this case, the two firms solve the monopolist problem and split the profit. The monopolistic quantity is given by

   \begin{align*} q_{m} = \frac{100 - 4}{4} = 24, \end{align*}

   in which case, the market price is \(p_{m}=52\). With collusion, the market power of the firms increases, the produced output level decreases (\(q_{m} = 24 < 32 =q_{s}\)), while the market price increases (\(p_{m} = 52 > 36 =p_{s}\)).

6. Suppose that firm \(i\) deviates from collusion, while firm \(j\) keeps producing half of the monopolistic quantity. The best deviation of firm \(i\) can be calculated from its best response function, namely

   \begin{align*} q_{b} = b_{i}\left(\frac{q_{m}}{2}\right) = \frac{48 - 12}{2} = 18. \end{align*}

   The best deviation for the firm is to produce at an output level greater than the level of the simultaneous quantity competition (i.e., \(q_{b} = 18 > 16 = q_{s}/2\)). Firm \(i\) takes advantage of the fact that firm \(j\) is not producing at its best response output level and produces more to grab a greater share of the market. The market quantity is

   \begin{align*} q_{w} = q_{b} + \frac{q_{m}}{2} = 18 + 12 = 30. \end{align*}

   Then, the price is \(p_{w} = 40\), and the firms' profits are

   \begin{align*} \pi_{i} &= (p_{w}-c)q_{b} = 36 \cdot 18 = 648 \\ &> 576 = 48 \cdot 12 = (p_{m}-c)\frac{q_{m}}{2} = \frac{\pi_{m}}{2} \end{align*}

   and

   \begin{align*} \pi_{j} &= (p_{w}-c)\frac{q_{m}}{2} = 36 \cdot 12 = 432 < 576 = \frac{\pi_{m}}{2} . \end{align*}
7. No. Each firm has a profitable deviation from producing half of the monopolistic output in this static game. More than a single date is needed for firms to be able to collude successfully.

Suppose that the inverse demand function of the market is given by \(p(q) = p_{0} + p_{1} q = 25 - 4 q\). Let the marginal cost of firm 1 be \(c_{1}=15\) and that of firm 2 to be \(c_{2}=1\). For these values, if there was an equilibrium in this market, according to the best response functions, firm 1 should produce

\begin{align*} q_{1} = \frac{2c_{1} - c_{2} - p_{0}}{3 p_{1}} = -\frac{30 - 1 - 25}{12} < 0. \end{align*}

Therefore, this market does not have an equilibrium. Visually, this results in two best response functions that do not cross each other.

This result is the analog of the upper bound for the elasticity of monopolistic markets discussed in exercise 4. In particular, the exact result of the monopolistic case is obtained for \(n=1\). To simplify notation, let \(q = \sum_{j=1}^{n} q_{j}\). Each firm solves

\begin{align*} \max_{q_{i}} \left\{ p\left( q \right) q_{i} - c(q_{i}) \right\}, \end{align*}

with the first order condition of firm \(i\) being

\begin{align*} p'\left(q\right) q_{i} + p\left(q\right) - c'(q_{i}) = 0. \end{align*}

By the inverse function theorem, we can rewrite the above condition as

\begin{align*} 0 &= \frac{1}{d'\left(p(q)\right)} q_{i} + p\left(q\right) - c'(q_{i}), \end{align*}

where \(d\) is the demand function, and, then, express it in terms of the demand elasticity as

\begin{align*} \frac{c'(q_{i})}{p\left(q\right)} &= \frac{1}{d'\left(p(q)\right)} \frac{q}{p(q)} \frac{q_{i}}{q} + 1 = \frac{1}{e_{d}\left(p(q)\right)} \frac{q_{i}}{q} + 1. \end{align*}

Since all firms are identical, the equilibrium, if it exists, is symmetric. This means that \(q_{i} = q / n\). Substituting the symmetry condition in the first order condition gives

\begin{align*} p\left(q\right) &= \frac{c'(q_{i})}{\frac{1}{e_{d}\left(p(q)\right)} \frac{1}{n} + 1}. \end{align*}

Since the marginal cost is positive, positive prices require that

\begin{align*} \frac{1}{e_{d}\left(p(q)\right)} \frac{1}{n} + 1 > 0, \end{align*}

which is equivalent to \(e_{d}\left(p(q)\right) < - 1 / n\).

1. Firm 1 solves

   \begin{align*} \max_{q_{1}} \left\{ 48 q_{1} - q_{1}^2 \right\}, \end{align*}

   which results in \(q_{1} = 24\). Firm 2 solves

   \begin{align*} \max_{q_{2}} \left\{ (60 - q_{1}) q_{2} - q_{2}^2 \right\}, \end{align*}

   which gives

   \begin{align*} q_{2} = 30 - \frac{q_{1}}{2} = 30 - 12 = 18. \end{align*}

   Then, we can calculate the profits

   \begin{align*} \pi_{1} &= 48 \cdot 24 - 24^{2} = 576, \\ \pi_{2} &= (60 - 24) 18 - 18^{2} = 324, \\ \pi &= \pi_{1} + \pi_{2} = 900. \end{align*}

2. In this case, \(q_{1}=0\) and the quantity that maximizes profit in the second market is

   \begin{align*} q_{2} = 30 - \frac{q_{1}}{2} = 30. \end{align*}

   The firm's profit is

   \begin{align*} \pi = \pi_{2} = (60 - 0) 30 - 30^{2} = 900. \end{align*}

3. When firm \(1\) produces \(q_{1}\), the damage caused by the externality to firm \(2\) is equal to \(q_{1}q_{2}\). With the transfer, firm 1 solves

   \begin{align*} \max_{q_{1}} \left\{ 48 q_{1} - q_{1}^2 - q_{1}q_{2}\right\}, \end{align*}

   which essentially moves the externality to the first market. This results in the necessary condition

   \begin{align*} q_{1} = 24 - \frac{q_{2}}{2}. \end{align*}

   Firm \(2\) solves

   \begin{align*} \max_{q_{2}} \left\{ 60 q_{2} - q_{2}^2 \right\}, \end{align*}

   which gives \(q_{2} = 30\). Substituting into the best response of the first firm, we get \(q_{1} = 9\). Then, the profits are given by

   \begin{align*} \pi_{1} &= (48 - 30) \cdot 9 - 9^{2} = 81, \\ \pi_{2} &= 60 \cdot 30 - 30^{2} = 900, \\ \pi &= \pi_{1} + \pi_{2} = 981. \end{align*}

4. The merged firm maximizes

   \begin{align*} \max_{q_{1}, q_{2}} \left\{ 48 q_{1} - q_{1}^2 + 60 q_{2} - q_{2}^{2} - q_{1}q_{2}\right\}. \end{align*}

   The first order conditions of the problem are

   \begin{align*} 48 - 2 q_{1} - q_{2} &= 0, \\ 60 - 2 q_{2} - q_{1} &= 0. \end{align*}

   Solving the above system gives \(q_{1}=12\) and \(q_{2}=24\). The profit of the merged firm is

   \begin{align*} \pi = 48 \cdot 12 - 12^2 + 60 \cdot 24 - 24^{2} - 12 \cdot 24 = 1008. \end{align*}

5. The merged case is the Pareto efficient structure because the externality is internalized.

   Case	Structure	\(\pi_{1}\)	\(\pi_{2}\)	\(\pi\)
   1	Externality in the second market	576	324	900
   2	First market shuts down	0	900	900
   3	Externality in the first market	81	900	981
   4	Internalized externality			1008

1. The good \(G\) is non-rivalrous, as both players consume the exact same good \(G\), and non-excludable as the good is not priced in the market.
2. The budget of individual \(i\) is \(p x_{i} + g_{i} \le B\).

3. Individual \(i\)'s optimization problem is

   \begin{align*} \max_{x_{i}, g_{i}} \left\{ u(x_{i}, G) + \lambda(B - p x_{i} - g_{i}) \right\}, \end{align*}

   where \(\lambda\) is the Lagrange multiplier. The first order conditions are

   \begin{align*} \alpha \frac{u(x_{i}, G)}{x_{i}} &= \lambda p, \\ \beta \frac{u(x_{i}, G)}{g_{i}} &= \lambda, \end{align*}

   which imply

   \begin{align*} \frac{\alpha}{\beta}G &= px_{i}. \end{align*}

   Combining the last condition with the budget constant, we get

   \begin{align*} \frac{\alpha}{\beta}G &= B - g_{i}, \end{align*}

   from which we conclude that the best response is

   \begin{align*} g_{i} &= \frac{\beta}{\alpha + \beta}B - \frac{\alpha}{\alpha + \beta}g_{j}. \end{align*}

4. Due to symmetry, the Nash equilibrium can be easily calculated by setting \(g_{i}=g_{j}\), which results in

   \begin{align*} g_{n} &= \frac{\beta}{2\alpha + \beta}B. \end{align*}

5. The Pareto efficient solutions are obtained by solving the joint maximization problem

   \begin{align*} \max_{x_{1}, x_{2}, G} \left\{ u(x_{1}, G) + u(x_{2}, G) + \mu(2B - p (x_{1} + x_{2}) - G) \right\}, \end{align*}

   where \(\mu\) is the Lagrange multiplier of this problem. The first order conditions are

   \begin{align*} \alpha \frac{u(x_{1}, G)}{x_{1}} &= \mu p, \\ \alpha \frac{u(x_{2}, G)}{x_{2}} &= \mu p, \\ \beta \frac{u(x_{1}, G) +u(x_{2}, G)}{G} &= \mu. \end{align*}

   The first two conditions and the budget constraint imply

   \begin{align*} \alpha \left( u(x_{1}, G) + u(x_{2}, G) \right) &= \mu p \left( x_{1} + x_{2} \right) = \mu \left( 2B - G \right). \end{align*}

   Combining with the third first order condition gives

   \begin{align*} G_{p} &= \frac{\beta}{\alpha + \beta}2B. \end{align*}

   For \(\alpha \neq 0\), \(G_{p} \neq 2 g_{n}\), from which we conclude that the Nash equilibrium is not Pareto efficient.

6. In the Nash equilibrium, fewer than the Pareto efficient resources are allocated in the production of the public good. Specifically, for all \(\alpha>0\), we have \(G_{p} > 2 g_{n}\).

1. Player 1 prefers consuming the public good if and only if

   \begin{align*} \pi_{1}(1) \ge \pi_{1}(0) \iff 2 (B_{1} - x_{1}) \ge B_{1} \iff x_{1} \le \frac{B_{1}}{2}. \end{align*}

2. For player 2, an analogous approach gives

   \begin{align*} \pi_{2}(1) \ge \pi_{2}(0) \iff 3 (B_{2} - x_{2}) \ge 2 B_{2} \iff x_{2} \le \frac{B_{2}}{3}. \end{align*}

3. The total production cost of the public good, say \(x\), has to be covered by the sum of the individual contributions, i.e., \(x_{1} + x_{2} \ge x\). Both players prefer covering the cost if

   \begin{align*} x &\le x_{1} + x_{2} \\ &\le \frac{B_{1}}{2} + \frac{B_{2}}{3} \\ &\le \frac{3B_{1} + 2B_{2}}{6} \\ &= \frac{300 + 150}{6} = 75 =: \hat x \end{align*}
4. Suppose that \(x_{2}=50\) and \(x_{1}=20\). We then have \(x_{1} + x_{2} = 70 < 75 = \hat x\), but player 2 prefers not to contribute to the public good production because \(x_{2}>25\).