Leibniz

3.5.1 Optimal allocation of free time: MRT meets MRS

Alexei wants to get as high an exam grade as possible whilst sacrificing the least possible amount of free time. We have seen diagrammatically that he maximizes his utility by choosing the point where an indifference curve is tangential to the feasible frontier, at which the marginal rate of substitution (MRS) is equal to the marginal rate of transformation (MRT). In this Leibniz we show how to formulate Alexei’s decision mathematically as a constrained optimization problem, and solve it to find the optimal combination of grade and free time.

Alexei’s optimal choice of free time and exam grade is illustrated in Figure 1 below. It combines his feasible set and indifference curves. The optimum is achieved at the point E on the feasible frontier, where the frontier has the same slope as the indifference curve.

Alexei’s utility function is $U(t,\ y)$: utility depends positively on hours of free time t and the exam grade y. He wishes to maximize his utility given the constraint imposed by his feasible set of grades and free time. As in Leibniz 3.4.1, if the production function is $y=f(h)$, where h is hours of study, the equation of the feasible frontier is:

$$y = f(24 - t)$$

Thus Alexei’s problem is to choose t and y to maximize $U(t,\ y)$ subject to the constraint $y=f(24 - t)$.

constrained optimization problem

Problems in which a decision-maker chooses the values of one or more variables to achieve an objective (such as maximizing profit) subject to a constraint that determines the feasible set (such as the demand curve).

This is an example of what is known in mathematics as a problem of constrained optimization. Sometimes in this sort of problem the constraint is written as an inequality: $y \leq f(24 - t)$, which can be interpreted as saying that his choice must lie in the feasible set. But because his utility depends positively on t and y, we know that he will want to choose a point on the frontier. So we can write the constraint as an equation, which makes the problem easier to solve mathematically.

One way to solve Alexei’s problem is to use the constraint to substitute for y in terms of t in the utility function. Then utility is expressed as a function of the single variable t:

$$U(t,\ y) = U(t,\ f\left( 24 - t \right))$$

which may be maximized with respect to t by equating its derivative to zero. This derivative is the total derivative of utility with respect to t, which may be calculated in the usual way via the chain rule:

$$\frac{dU}{dt} = \frac{\partial U}{\partial t} + \frac{\partial U}{\partial y} \frac{dy}{dt}$$

The term $\dfrac{dy}{dt}$ on the right-hand side is calculated by differentiating the production function $y= f(24- t)$:

$$\frac{dy}{dt} = -f'(24- t)$$

by the composite function rule, so:

$$\frac{dU}{dt} = \frac{\partial U}{\partial t} - \frac{\partial U}{\partial y} f'(24 - t)$$

This equation says that as we move along the feasible frontier in the direction of increasing t, the net effect on utility is the result of the direct effect of more free time, which is of course positive, together with the negative indirect effect of a lower exam grade.

At the point that solves Alexei’s maximization problem, $\dfrac{dU}{dt}=0$. So at this point:

$$\frac{\partial U}{\partial t} = \frac{\partial U}{\partial y} f'(24- t)$$

This has an obvious interpretation in terms of the two effects on utility mentioned in the preceding paragraph: at the optimum point, the positive effect of a little more free time and the negative effect of a slightly lower exam grade balance each other out.

Rearranging the last equation, we see that:

$$f^{'}\left( 24 - t \right) = \frac{\frac{\partial U}{\partial t}}{\frac{\partial U}{\partial y}}$$

at the optimum point. The left-hand side is the absolute value of the slope of the feasible frontier, which we called the marginal rate of transformation (MRT) in Leibniz 3.4.1, and as we saw in Leibniz 3.2.1, the right-hand side is the absolute value of the slope of the indifference curve, which we called the marginal rate of substitution (MRS). Thus, at the optimum point, the slopes are equal as in Figure 1. In other words,

$$\text{MRT}=\text{MRS}$$

This is known as a first-order condition for optimization, since it was obtained by equating a first derivative (in this case the total derivative $\frac{dU}{dt}$) to zero. Because much of economics concerns constrained optimization, you will find similar conditions in later Leibniz supplements.

Remember that we want to find the values of t and y that maximize Alexei’s utility. So far we have shown that the values of t and y we are looking for must satisfy the first-order condition. To solve the problem fully and find the optimal values, we need to note that they must also lie on the feasible frontier. So we have a pair of simultaneous equations:

$$f^{'}\left( 24 - t \right) = \frac{\frac{\partial U}{\partial t}}{\frac{\partial U}{\partial y}}$$ $$y = f(24 - t)$$

which must be satisfied by t and y. In the next section, we will derive these equations for particular utility and production functions, and solve them to find the optimal values of t and y.

Read more: Sections 8.1 to 8.3 for maximization, and Section 14.2 for the distinction between total and partial derivative, of Malcolm Pemberton and Nicholas Rau. 2015. Mathematics for economists: An introductory textbook, 4th ed. Manchester: Manchester University Press

Optimal allocation of free time: an example

We now illustrate the principles of the previous section with specific production and utility functions.

budget constraint

An equation that represents all combinations of goods and services that one could acquire that exactly exhaust one’s budgetary resources.

The constrained optimization problem has two parts: the objective function, which describes the utility Alexei wants to maximize, and the constraint, which in this case is Alexei’s production function for exam grades.

We assume as in Leibniz 3.2.1 that Alexei has a Cobb-Douglas utility function:

$$U (t,\ y)= t^a y^b$$

where a and b are positive constants. (We use a and b rather than $\alpha$ and $\beta$ because $\alpha$ will be used in the production function.)

As in Leibniz 3.1.1, we assume that the relationship describing how Alexei turns hours of study h into an exam grade y is

$$y = Ah^{\alpha}$$

where A and $\alpha$ are positive constants and $\alpha \lt 1$. Also, as before, we may write this production function in terms of hours of free time t, since $h = 24 -t$. Doing this, and assuming for simplicity that $A=1$, we see that:

$$y = {(24 - t)}^{\alpha}$$

This is the equation of the feasible frontier.

Alexei’s problem is to choose t and y to maximize $t^a y^b$, subject to the constraint:

$$y = \left(24- t \right)^\alpha$$

The previous section gives us two ways of solving this problem: we may either use the substitution method or apply the formula. We will demonstrate both, and confirm that they give us the same answer.

Applying the formula

By ‘the formula’ we mean the first-order condition $\text{MRT} = \text{MRS}$. We know that the solution to the problem must satisfy this condition, so we calculate the MRT from the feasible frontier, and the MRS from the utility function, and equate the two.

As we saw in Leibniz 3.4.1, the MRT is the absolute value of the slope of the feasible frontier. Using the equation above for the frontier:

$$\text{MRT} = \left|\frac{dy}{dt}\right| = \alpha \left( 24 - t \right)^{\alpha -1} = \frac{\alpha y}{ 24 - t}$$

Also, we showed in Leibniz 3.2.1 that the MRS is the ratio of the marginal utilities. The marginal utilities are found by differentiating the utility function:

$$\frac{\partial U}{\partial t} = at^{a - 1}y^{b}$$ $$\frac{\partial U}{\partial y} = bt^{a}y^{b - 1}$$

So the MRS is given by:

$$\text{MRS} = \frac{\partial U}{\partial t} \left/ \frac{\partial U}{\partial y} \right. = \frac{a y}{b t}$$

Equating MRT to MRS and multiplying through by $\frac{t}{y}$,

$$\frac{\alpha t}{24 - t} = \frac{a}{b}$$

Solving this equation for t, we see that $t = \frac{24}{(1+c)}$, where $c=\frac{\alpha b}{a}$. Substituting this into the production function we obtain the full solution to Alexei’s problem:

$$t = \frac{24}{1+c}, \; \; y = \left(\frac{24c }{1+c} \right)^\alpha, \, \text{where}\ c=\frac{\alpha b}{a}$$

These are the values of t and y that give Alexei the highest utility he can achieve within the feasible set.

Using the substitution method

The method consists of substituting the constraint into the objective function to make it a function of just one variable and maximizing that function. Substituting the constraint $y = \left(24- t \right)^\alpha$ into the utility function gives us an expression for utility as a function of t alone:

$$U = t^a \left(24 - t \right)^{\alpha b}$$

To maximize U, we calculate $dU/dt$ and equate it to zero. Using the product rule,

$$\frac{dU}{dt} = a t^{a-1}(24 - t )^{\alpha b} - \alpha b t^a (24- t )^{\alpha b-1}$$

As in the general analysis of the previous section, this expresses the net effect on utility of an increase in t as the result of a positive direct effect and the negative effect of a lower exam grade.

Setting $dU/dt$ equal to zero and dividing the resulting equation by $t^{a-1}(24 - t )^{\alpha b}$, we see that $a = \frac{\alpha b t}{24 – t}$. Rearranging,

$$\frac{\alpha t}{24 - t} = \frac{a}{b}$$

This is the same equation for t that we obtained using $\text{MRT} = \text{MRS}$, and the rest of the solution is as before.

Read more: Sections 8.1 to 8.3 of Malcolm Pemberton and Nicholas Rau. 2015. Mathematics for economists: An introductory textbook, 4th ed. Manchester: Manchester University Press.