MY-ASSIGNMENTEXPERT™可以为您提供 campuspress ECON121 Microeconomics微观经济学的代写代考和辅导服务！
ECON121 课程简介
Course Description: A presentation and study of national income aggregates and accounting; equilibrium analysis of output, employment and the price level; general equilibrium analysis; and an introduction to economic dynamics.
Course Prerequesites: ECON 1010 and 1020 or their equivalents; QA and MATH 2200/2350
Primary Text: Macroeconomics by N. Gregory Mankiw ( $10^{\text {th }}$ edition)
Classroom Polls: We’ll be using REEF Polling to take classroom polls. It is free to download the app to your phone, tablet or laptop.
Prerequisites
Course Description: Microeconomic theory is the study of models economists use to describe how agents (consumers, firms, governments, etc.) make decisions and how these decisions affect market outcomes and welfare. We begin by analyzing how consumers and firms make optimal decisions given the budgetary and physical/technological constraints they face. We then study how these decisions by individuals translate into competitive market equilibria, and look at the Conditions under which the “invisible hand” of the market optimizes welfare.
The second half of the class discusses deviations from the competitive ideal. We look at important sources of market failures, including including externalities, market power (monopolies), and asymmetric information. An important tool for economists in analyzing such situations where strategic interactions between people influences outcomes is game theory. The last section of the course will be a basic introduction to game theory and its applications in economics, including imperfectly competitive markets (e.g., oligopolies), public goods, and prinicpal-agent problems.
ECON121 Microeconomics HELP（EXAM HELP， ONLINE TUTOR）
Cost minimizing problem:
$$
\begin{gathered}
\min _{x_1, x_2} p_1 x_1+p_2 x_2 \
\text { sub. to } x_1^\alpha x_2^{1-\alpha} \geq y
\end{gathered}
$$
F.O.C.s are therefore given by:
$$
\begin{aligned}
p_1-\lambda \alpha x_1^{\alpha-1} x_2^{1-\alpha} & =0 \
p_2-\lambda(1-\alpha) x_1^\alpha x_2^{-\alpha} & =0 \
y-x_1^\alpha x_2^{1-\alpha} & =0
\end{aligned}
$$
To solve this problem, we can use the Lagrangian method. The Lagrangian for this problem is:
$$
L=p_1 x_1+p_2 x_2-\lambda\left(x_1^\alpha x_2^{1-\alpha}-y\right)
$$
Taking the partial derivative of $L$ with respect to each of the decision variables and the Lagrange multiplier, we get:
$$
\begin{aligned}
\frac{\partial L}{\partial x_1} & =p_1-\lambda \alpha x_1^{\alpha-1} x_2^{1-\alpha}=0 \
\frac{\partial L}{\partial x_2} & =p_2-\lambda(1-\alpha) x_1^\alpha x_2^{-\alpha}=0 \
\frac{\partial L}{\partial \lambda} & =x_1^\alpha x_2^{1-\alpha}-y=0
\end{aligned}
$$
Solving the first equation for $\$ \backslash a m b d a \$$ and substituting into the second equation gives:
$$
\lambda=\frac{p_1}{\alpha x_1^{\alpha-1} x_2^{1-\alpha}}=\frac{p_2}{(1-\alpha) x_1^\alpha x_2^{-\alpha}}
$$
Equating these two expressions for $\$ \backslash$ lambda $\$$ and solving for $\$ x_{-} 2 \$$ gives:
$$
x_2=\left(\frac{(1-\alpha) p_1}{\alpha p_2}\right)^{\frac{1}{\alpha 1}} x_1
$$
Substituting this expression for $\$ x_{-} 2 \$$ into the third equation gives:
$$
y=x_1^\alpha\left(\frac{(1-\alpha) p_1}{\alpha p_2}\right)^{1-\alpha} x_1^{1-\alpha}=\left(\frac{(1-\alpha) p_1}{\alpha p_2}\right)^{1-\alpha} x_1
$$
Solving for $\$ x_{-} 1 \$$ and substituting back into the expression for $\$ x_{-} 2 \$$ gives the optimal values for $\$ x_{-} 1 \$$ and $\$ x_{-} 2 \$$ :
$$
\begin{aligned}
& x_1=y^{\frac{1}{\alpha}}\left(\frac{\alpha p_2}{(1-\alpha) p_1}\right)^{\frac{1}{\alpha}} \
& x_2=y^{\frac{1}{\alpha}}\left(\frac{(1-\alpha) p_1}{\alpha p_2}\right)^{\frac{1}{\alpha-1}}
\end{aligned}
$$
Therefore, the optimal values for $\$ x_{-} 1 \$$ and $\$ x_{-} 2 \$$ can be found by plugging in the values of $\$ p_{-} \$ \$, \$ p_{-} 2 \$, \$ \backslash a l p h a \$$, and $\$ y \$$ into the above expressions.
Solving the F.O.C.s give us the solution for input demands:
$$
\begin{aligned}
& x_1^=y\left(\frac{(1-\alpha) p_1}{\alpha p_2}\right)^{\alpha-1} \ & x_2^=y\left(\frac{(1-\alpha) p_1}{\alpha p_2}\right)^\alpha
\end{aligned}
$$
Yes, that’s correct. To get these input demands, we can solve the first two F.O.C.s for $\$ x_{-} 1 \$$ and $\$ x_{-} 2 \$$, respectively, and then substitute one into the constraint to solve for the other.
From the first F.O.C., we have:
$$
x_1^*=\frac{\lambda \alpha}{p_1} x_2^{\alpha-1}
$$
Substituting this expression for $\$ x_{-} 1^{\wedge *} \$$ into the second F.O.C., we get:
$$
p_2=\lambda(1-\alpha)\left(\frac{\lambda \alpha}{p_1}\right)^\alpha x_2^{-\alpha}
$$
Solving for $\$ \backslash$ lambda $\$$ and simplifying, we get:
$$
\lambda=\left(\frac{p_1 p_2^\alpha}{\alpha^\alpha(1-\alpha)^{1-\alpha}}\right)^{\frac{1}{\alpha+1}} x_2^{\frac{\alpha}{\alpha+1}}
$$
Substituting this expression for $\$$ Vambda $\$$ into the first F.O.C., we get:
$$
x_1^=\left(\frac{\alpha p_2}{\lambda(1-\alpha)}\right)^{\frac{1}{\alpha+1}} x_2 $$ Substituting this expression for $\$ x_{-} 1^{\wedge} \$ \$$ into the constraint, we get: $$ y=\left(\frac{\alpha p_2}{\lambda(1-\alpha)}\right)^{\frac{\alpha}{\alpha-1}} x_2^{1-\alpha} $$ Substituting the expression for $\$ \backslash a m b d a \$$ and simplifying, we get: $$ x_2^=y\left(\frac{(1-\alpha) p_1}{\alpha p_2}\right)^\alpha
$$
Substituting this expression for $\$ x _2^{\wedge} \$$ back into the expression for $\$ x_{-} 1 \wedge \$$, we get:
$$
x_1^*=y\left(\frac{(1-\alpha) p_1}{\alpha p_2}\right)^{\alpha-1}
$$
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