Lecture 7: Corner Solutions

Last week, we looked at situation in which the optimal bundle was characterized by a tangency condition between the indifference curve and the budget line. However, this is not always the case. This week we’ll talk about situations in which the consumer might choose a point along their budget constraint where the MRS is not equal to the price ratio. In particular, we’ll think about three situations:

1. Corner solutions occur when the consumer spends all their money on good 1 or all on good 2.
2. Solutions at kinks occur when the consumer faces a kinked constraint, and buys the bundle at a kink, at which point the price ratio is undefined.
3. Solutions inside the budget constraint occur when the consumer has nonmonotonic preferences, and prefers a point strictly within the budget set to any point along the constraint itself.

To get an intuition for this, try playing with the graphs below. Again, our notion of optimality is that the consumer is choosing a point $X$ that is their best affordable option: that is, $X$ must lie within their budget set (the green area), and there should be no alternative bundle which is both affordable and preferred to $X$.

So, how do we go about solving a problem like this? Let’s start by thinking about how we would find the maximum of a univariate function when that maximum isn’t characterized by the condition $f^\prime(x) = 0$…

Motivation: maximizing a univariate function

The Lagrange method of constrained optimization, with its first-order conditions setting a bunch of stuff equal to zero, is akin to the method in univariate calculus of setting the derivative of a function equal to zero. In many cases that works just fine: in particular, if the function $f(x)$ is continuous, continuously differentiable, and concave (so $f^{\prime \prime}(x) < 0$), and if we are considering the domain of $x$ to be all real numbers, then we do indeed get a nice solution at a point where the derivative is equal to zero:

However, suppose one or more of these conditions doesn’t hold. For example, consider the function \(f(x) = 5 - |x - 5|\) This has a clear maximum at $x = 5$, but the derivative isn’t defined at that point:

The above examples involved unconstrained maxima: that is, we were optimizing the function over all real values of $x$. But suppose we were trying to maximize $f(x)$ over just a closed subdomain of real numbers? Then it’s entirely possible to have a function with an unconstrained maximum that lies outside that range which is characterized by the condition $f^\prime(x) = 0$, but the maximum within that range is not. For example, consider the problem \(\begin{aligned} \max \ & f(x) = 12x - x^2\\ \text{subject to (s.t.) } & 1 \le x \le 5 \end{aligned}\)

If we graph this out, we can see that while the unconstrained maximum occurs when $x = 6$, the highest point of the function within the constraint lies at the right-hand boundary, where $x = 5$:

Furthermore, the derivative at the constrained maximum is not equal to zero: it’s $+2$!

Finally, consider the same domain, but a different function: \(\begin{aligned} \max \ & f(x) = 10 + (4 - x)^2\\ \text{s.t. } & 1 \le x \le 5 \end{aligned}\) This function does have a point within the constraint at which $f^\prime(x) = 0$, but it’s a minimum, not a maximum:

In this case, the highest value on the interval $[1,5]$ occurs at $x = 1$.

Solving for the maximum: use logic, not first-order conditions.

In each of the above scenarios, setting the derivative equal to zero won’t find us the maximum. However, there is one way that the derivative is useful: pointing us toward the solution.

In each of the graphs above, the function is rising when $f^\prime(x) > 0$ and falling when $f^\prime(x) < 0$. So, at any point, we can tell the direction in which higher point lies. This is akin to the “gravitational pull” argument we discussed in the last lecture.

Applying this logic to the three cases above, we can see how this “gravitational pull” draws us to potential solutions:

• For the function $f(x) = 5 - |x - 5|$, we have $f^\prime(x) > 0$ whenever $x < 5$, and $f^\prime(x) < 0$ whenever $x > 5$; so the “gravitational pull” is drawing us to $x = 5$.
• For the function $f(x) = 16x - x^2$, we have $f^\prime(x) > 0$ on the entire region $x \in [0,5]$; so we are always being drawn to the right, and the optimum lies at the right edge of the constraint.
• For the function $f(x) = 10 + (4 - x)^2$, we have $f^\prime(x) < 0$ whenever $x < 4$, and $f^\prime(x) > 0$ whenever $x > 4$. In this case we’re being drawn away from the minimum, so we have two local maxima: one at $x = 1$, and the other at $x = 5$. To figure out which one is the higher value, we need to plug those two values into the function and see which is higher. In this case, it turns out it’s $x = 1$. In short: to solve a problem of this kind, you need to use the logic of the gravitational pull argument, not plug things into a simple formula.

Let’s now see how this works with utility functions and budget constraints in good 1 - good 2 space.

Corner solutions

In the second two examples above, we had a solution at the boundary of a constraint. In this context of consumer choice with two goods, this corresponds to the situation in which the consumer spends all of their money on just one good, buying none of the other.

The gravitational pull argument we developed last time holds that if the $MRS > p_1/p_2$, a consumer can do better by moving to the right along her budget constraint, and vice versa. In some cases, this process leads the consumer to a point along the budget constraint where $MRS = p_1/p_2$, so the indifference curve passing through that point is tangent to the constraint. But what happens if $MRS$ never gets as low as the price ratio? That is, if even the last dollar spent on good 1 gives the consumer more utility than if she’d spent that dollar on good 2?

In order for this to occur, it must be the case that the consumer’s indifference curves cross the axes. Two examples we’ve seen of this kind of utility function are the quasilinear utility function, whose indifference curves are parallel transforms of each other (shown in the left diagram below), and the perfect substitutes utility function, whose indifference curves are just straight lines (shown in the right diagram below).

In the left graph, the consumer maximizes her utility by buying only good 1 (drag the bundle all the way to the right), and in the right-hand graph she maximizes her utility by buying only good 2 (drag the bundle all the way to the left). Let’s examine these two cases mathematically to see what’s going on. In each case, we’ll assume the consumer has $m = 100$ and faces prices $p_1 = 1$ and $p_2 = 2$, so their budget constraint is given by the equation \(x_1 + 2x_2 = 100\) and the price ratio is $p_1/p_2 = 1/2$.

Example 1: Quasilinear utility

Suppose the consumer’s preferences could be represented by the quasilinear utility function \(u(x_1,x_2) = 100 \ln x_1 + x_2\) which has the associated MRS \(MRS = \frac{MU_1}{MU_2} = \frac{\frac{100}{x_1}}{1} = \frac{100}{x_1}\) This is infinite if the consumer buys only good 2 ($x_1 = 0$), and 1 if she buys only good 1 (resulting in $x_1 = 100$).

Since her MRS is always greater than the price ratio, there is always an area of overlap between the preferred region and the budget set, and the consumer always wants to move to the right along the budget line — up to and including the point where they’re spending all their money on good 1! If we plot the utility (top right graph), we can see that her utility is increasing at every point along the budget line. And if we plot MRS and price ratio along the budget line (bottom right graph), the two curves never touch, and that MRS is always greater than price ratio:

For cases like this, it’s worth noting that you could try to solve using the tangency condition and budget constraint. As a reminder, this method finds the point along the budget constraint where the MRS is equal to the price ratio. But if we set the MRS equal to the price ratio, we get \(\begin{aligned} MRS &= {p_1 \over p_2}\\ {100 \over x_1} &= {1 \over 2}\\ x_1 &= 200 \end{aligned}\) In other words, the tangency condition is a vertical line at $x_1 = 200$. This intersects the equation of the budget line, $x_1 + 2x_2 = 100$, at the point $(200,-50)$, but while this makes mathematical sense, it doesn’t make economic sense:

This illustrates an important point: a lot of times, the solution you get by plugging something into a mathematical formula may not be the actual solution to an economic problem. It’s important to check your work after you solve a problem to make sure that the solution you found is actually feasible! A key way to do this is to check if any values are negative that cannot be. In this case, the mathematical “solution” yields a negative value for good 2; since you can’t go into a store and buy a negative quantity of a good, that can’t be the economic solution.

Example 2: Perfect Substitutes

Let’s consider the same budget constraint as above \(x_1 + 2x_2 = 100\) but this time use the utility function \(u(x_1,x_2) = x_1 + 4x_2\) Note that the price ratio is 1/2, meaning that good 2 is twice as expensive as good 1. But with this utility function, each unit of good 1 brings you 1 util, while each unit of good 2 brings you 4 utils; so your MRS is \(MRS = {MU_1 \over MU_2} = {1 \over 4}\) That is, you enjoy each unit of good 2 four times as much as each unit of good 1, no matter how many of each good you have!

Logically, it follows that if you like each unit of good 2 four times as much as each unit of good 1, but it only costs twice as much, you’re not going to spend any of your money on good 1. And indeed that’s the case: the optimal bundle involves you spending all of your money on good 2:

In this case, the MRS is always less than the price ratio, so the consumer is always drawn to the left.

Note that you can’t even use the tangency condition here! Why not? The tangency condition finds the point along the budget line where the MRS equals the price ratio. But here, the MRS is always 1/4, and the price ratio is always 1/2. There is no point at which $1/4 = 1/2$, so the tangency condition is of no use to us.

Conditions for an interior (non-corner) solution

Notice that in the first example above, the MRS was greater than the price ratio even when the consumer was spending all their money on good 1: the utility function was always increasing along the budget line. By contrast, in the second example, the MRS was less than the price ratio even when the consumer was spending all their money on good 2, so the utility function was always decreasing along the budget line.

From these two examples we can derive one necessary condition for an interior solution to an optimization problem: specifically, that \(\begin{aligned} MRS > p_1/p_2 & \text{ when }x_1 = 0\\ MRS < p_1/p_2 & \text{ when }x_2 = 0\end{aligned}\) This means that at the left corner of the budget constraint, the consumer is being pulled to the right; and at the right corner of the budget constraint, the consumer is being pulled to the left.

We might notice that there are some utility functions that guarantee an interior solution. For example, a Cobb-Douglas utility function of the form $u(x_1,x_2) = a \ln x_1 + b \ln x_2$ has an MRS of \(MRS = \frac{ax_2}{bx_1}\) This is infinite when $x_1 = 0$, which must be greater than any finite price ratio; and it’s zero when $x_2 = 0$, which must be less than any finite price ratio. Therefore a Cobb-Douglas utility function will never yield a corner solution.

On the other hand, there are some utility functions which guarantee a corner solution. For example, a concave utility function of the form \(u(x_1,x_2) = ax_1^2 + bx_2^2\) has an MRS of \(MRS = \frac{ax_1}{bx_2}\) This is zero when $x_1 = 0$, and infinite when $x_2 = 0$.

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