*Monday, February 7, 2022*

## Overview

Today we put all of our tools together thus far to solve the consumer’s constrained optimization problem. To solve the problem the “traditional way” would be to use Lagrangian multipliers and calculus and solve for the first order conditions. [Take the full Lagrangian equation and solve for all partial derivatives, set equal to 0.] We instead will find the solution by looking graphically, and use an algebraic rule that should make a lot of intuitive sense. That rule, equivalently expressed in two ways:

\[ \begin{align*}\frac{MU_x}{MU_y} &= \frac{p_x}{p_y}\\ \frac{MU_x}{p_x} &= \frac{MU_y}{p_y}\\ \end{align*} \] The first rule states that the slope of the indifference curve (left) equals the slope of the budget constraint (right). This is stating that the marginal benefit of consuming \(x\) (left) equals the marginal cost (right) of consuming \(y\), and that the individual is trading off \(y\) for \(x\) at the same rate as the market is.

The second (version of the) rule, solved for \(x\) and \(y\) terms on either side, shows that the marginal utility per dollar spent is equal across both (all) goods.

Both forms are often called the **equimarginal rule**, that at the optimum choice, the marginal benefit equals the marginal cost (which is the marginal benefit of the next best foregone opportunity).

We will do some practice to make sure you can solve this type of model, and then you are fully prepared for Problem Set 1.

## Readings

- Ch. 4.4-4.5 in Goolsbee, Levitt, and Syverson, 2019
- Math Review Guide

## Practice

Today we will be working on practice problems. Answers will be posted later on that page.

## Slides

Below, you can find the slides in two formats. Clicking the image will bring you to the html version of the slides in a new tab. Note while in going through the slides, you can type `h` to see a special list of viewing options, and type `o` for an outline view of all the slides.

The lower button will allow you to download a PDF version of the slides. I suggest printing the slides beforehand and using them to take additional notes in class (*not everything* is in the slides)!

## Assignments

### Problem Set 1 Due Wednesday February 16

Problem set 1 (on 1.1-1.4) is due by class time Wednesday February 16.

# Math Appendix

## Solving the Constrained Optimization Problem with Calculus

### Example

A consumer has a utility function of \[u(x, y) = xy\] and faces prices of \(p_x = 5\), \(p_y = 10\), and has an income of \(m = 50\). Find the consumer’s optimum consumption bundle.

The problem can be expressed as

\[\max_{x,y} xy\] \[s. t. 5x+10y=50\]

There are a few strategies you can use to solve this problem using calculus.

**Substitution Method**: take the budget constraint and solve it for one good in terms of the other good and income. Let’s solve for \(x\):

\[\begin{align*} 5x+10y&=50 && \text{The budget constraint}\\ 5x&=50-10y && \text{Subtracting } 10y \text{ from both sides}\\ x&=10-2y && \text{Dividing both sides by 5}\\ \end{align*}\]

Now we can plug this in for \(x\) in the utility function to get an *unconstrained* maximization problem:

\[\begin{align*} \max_{y} (10-2y)y && \text{Plugging into the original utility function}\\ \max_{y} 10y-2y^2 && \text{Distributing the }y\\ \end{align*}\]

Then we take the derivative with respect to \(y\) and set it equal to zero to find the maximum value of \(y\) in the function.

\[\begin{align*} u&=10y-2y^2 && \text{The new utility function}\\ \frac{d u}{d y} = 10-4y&=0 && \text{Taking the first derivative} \\ 10&=4y && \text{Adding } 4y \text{ to both sides}\\ 2.5&=y && \text{Dividing both sides by } 4\\ \end{align*}\]

Now that we know \(y\) is 2.5. To find \(x\), plug 2.5 in for \(y\) in the first equation we found from the budget constraint:

\[\begin{align*} x&=10-2y && \text{The equation}\\ x&=10-2(2.5) && \text{Plugging in 2.5}\\ x&=10-5 && \text{Distributing the -2}\\ x&=5 && \text{Simplifying}\\ \end{align*}\]

So we’ve found the **optimal bundle is 5 of \(x\) and 2.5 of \(y\), or (\(x^*,y^*)=(5,2.5)\).**

**Lagrangian Method**: Recall the Lagrangian adds the objective function and the Lagrange multiplier \((\lambda)\) times the constraint set equal to 0:

\[\max_{x, y, \lambda} \Lambda(x, y, \lambda)=xy+\lambda(5x+10y-50)\]

Solving for the First Order Conditions (setting all partial derivatives to 0):

\[\begin{align*} \frac{\partial \Lambda}{\partial x} &= y+5\lambda &= 0 \\ \frac{\partial \Lambda}{\partial y} &= x+10\lambda &= 0 \\ \frac{\partial \Lambda}{\partial \lambda} &= 5x+10y-50 &= 0 \\ \end{align*}\]

Taking the first two equations, and rearranging each to equal \(\lambda\):

\[\begin{align*} \frac{y}{5}&=\lambda \\ \frac{x}{10}&=\lambda \\ \end{align*}\]

Setting them equal to one another, and solving for \(x\):

\[\begin{align*} \frac{y}{5}&=\frac{x}{10} \\ 2y&=x \\ \end{align*}\]

So the consumer will buy twice the amount of \(y\) as \(x\) (note this should be intuitive as \(y\) costs twice as much as \(x\)!). To find the exact amounts of \(x\) and \(y\) she will buy, plug what we just found into the budget constraint:

\[\begin{align*} 5(2y)+10y&=50 \\ 20y&=50 \\ y&=2.5 \\ \end{align*}\]

Since we now know \(y^*\), we can find \(x^*\):

\[\begin{align*} x&=2y\\ x&=2(2.5)=5 \\ \end{align*}\]

So the **optimum consumption bundle (\(x^*, y^*\)) again is (5,2.5).**

You may be asking: *so what in the world is \(\lambda\)?* Besides being instrumental in solving constrained optimization problems, the Lagrangian multiplier \(\lambda\) does have some economic interpretation. First, let’s return to the FOCs, knowing \(x\) and \(y\), to solve for \(\lambda\):

\[\begin{align*} y+5\lambda &=0\\ 2.5+5\lambda &=0\\ \lambda &=0.5 \\ \end{align*}\]

Mathematically, it can be proved that \(\lambda=\displaystyle \frac{\partial u(x,y)}{\partial m}\): the rate of instantaneous change in your objective function with respect to your constraint (i.e. your income). That is, if you were to change your constraint (i.e. your income) by one unit, \(\lambda\) tells us how much your objective function would increase (in units of the objective function).

Economically, it is known as the “shadow price,” or sometimes the “marginal utility of wealth.” It tells us how much your utility would increase if you were able to increase your constraint (income) by one unit (e.g. $1). Alternatively, we can think of it as the marginal benefit of relaxing the constraint (having $1 more to spend), or the marginal cost of strengthening the constraint (having $1 less).

In our example, \(\lambda=0.5\), so if we had one +(-)$1 in income to spend, utility would increase(decrease) by 0.5.

Recall that since utility is *ordinal* and not *cardinal*, the number of “utils” your utility were to change by is meaningless. \(\lambda\) does have meaningful quantitative interpretations when we use constrained optimization to talk about production, where units are dollars and output, rather than dollars and utility.

Let’s lastly confirm that this is the optimum using the definition

\[\begin{align*} |MRS| &= |\frac{p_x}{p_y}| && \text{At the optimum, slopes of I.C. and B.C. are equal}\\ \frac{MU_{x}}{MU_{y}}&=\frac{p_x}{p_y} && \text{Definition of each slope}\\ \frac{y}{x} &= \frac{p_x}{p_y} && \text{MU's for this specific utility function}\\ \frac{(2.5)}{(5)} &= \frac{5}{10} && \text{At optimum, } y=2.5, x=5\\ \frac{1}{2} &= \frac{1}{2} && \checkmark \end{align*}\]

## General Case for \(n\) Goods & Marshallian Demand Functions

In the case of \(n\) number of goods (from good \(x_1\) to \(x_n)\), the optimal consumption portfolio solves the utility maximization problem:

\(\max_{x_1,\cdots,n_n}U(x_1,\cdots, x_n)\) \(s.t. \sum^{n}_{i=1} x_ip_i \leq m\)

That is, as usual, choosing the consumption bundle that maximizes utility, subject to the budget constraint.

Using the Lagrangian method, the Lagrangian is:

\[\mathbb{L} = U(x_i, \cdots, x_n)+\lambda(M - \sum_{i=1}^{n} x_ip_i)\]

The first order conditions are a series of \(n+1\) equations, with \(n+1\) unknowns:

\[\begin{align*} \frac{\partial u}{\partial x_1} &= \lambda p_i\\ \vdots\\ \frac{\partial u}{\partial x_n} &= \lambda p_n\\ \sum^{n}_{i=1} x_ip_i &= m\\ \end{align*}\]

Solving for each individual \(x_i^*\) gives us a **Marshallian Demand System**:

\[\begin{align*} x_1^* &= x_1^M(p_1,\cdots,p_n,m)\\ \vdots\\ x_n^* &= x_n^M(p_1,\cdots,p_n,m)\\ \lambda^* &= \lambda^M(p_1,\cdots,p_n,m)\\ \end{align*}\]

Each equation here is a demand function for that respective good, as a function of all market prices \((p_1, \cdots, p_n)\) and income \(m\).