Question 14.1.3: A consumer who has Cobb–Douglas utility function u(x, y) = A......
A consumer who has Cobb–Douglas utility function u(x, y) = Ax^{a}y^{b} faces the budget constraint px + qy = m, where A, a, b, p, q, and m are all positive constants. Find the
only solution candidate to the consumer demand problem
max Ax^{a}y^{b} s.t. px + qy = m (∗)
The Lagrangian is {\mathcal{L}}(x,y)=A x^{a}y^{b}-\lambda(p x+q y-m), so the first-order conditions are
{\mathcal{L}}_{1}^{\prime}(x,y)=a A x^{a-1}y^{b}-\lambda p=0,\ {\mathcal{L}}_{2}^{\prime}(x,y)=b A x^{a}y^{b-1}-\lambda q=0,\enspace{\mathrm{and}}\ p x+q y=mSolving the first two equations for λ yields
\lambda={\frac{a A x^{a-1}y^{b}}{p}}={\frac{b A x^{a}y^{b-1}}{q}}Cancelling the common factor Ax^{a−1}y^{b−1} from the last equality gives ay/p = bx/q. Solving this equation for qy yields qy = (b/a)px, which inserted into the budget constraint gives px + (b/a)px = m. From this equation we find x and then y. The results are the following demand functions:
x=x(p,q,m)={\frac{a}{a+b}}{\frac{m}{p}}~~\mathrm{and}~y=y(p,q,m)={\frac{b}{a+b}}{\frac{m}{q}} (∗∗)
The solution we have found makes good sense. It follows from (∗∗) that for all t > 0 one has x(tp, tq, tm) = x(p, q,m) and y(tp, tq, tm) = y(p, q,m), so the demand functions are homogeneous of degree 0. This is as one should expect because, if (p, q,m) is changed to (tp, tq, tm), then the constraint in (∗) is unchanged, and so the optimal choices of x and y are unchanged—as they should be, according to Example 12.7.4.
Note that in the utility function Ax^{a}y^{b}, the relative sizes of the coefficients a and b indicate the relative importance of x and y in the individual’s preferences. For instance, if a is larger than b, then the consumer values a 1% increase in x more than a 1% increase in y. The product px is the amount spent on the first good, and (∗∗) says that the consumer should spend the fraction a/(a + b) of income on this good and the fraction b/(a + b) on the second good.
Formula (∗∗) can be applied immediately to find the correct answer to thousands of exam problems in mathematical economics courses given each year all over the world! But note that the utility function has to be of the Cobb–Douglas type Ax^{a}y^{b}.^{5}
Another warning is in order here: there is an underlying assumption in problem (∗) that x ≥ 0 and y ≥ 0. Thus, we maximize a continuous function Ax^{a}y^{b} over a closed bounded set S = {(x, y) : px + qy = m, x ≥ 0, y ≥ 0}. According to the extreme value theorem, 13.5.1, a maximum must exist. Since utility is 0 when x = 0 or when y = 0, and positive at the point given by (∗∗), this point indeed solves the problem.Without nonnegativity conditions on x and y, however, the problem might fail to have a maximum. Indeed, consider the problem max x^{2}y s.t. x + y = 1. For real t, the pair (x, y) = (−t, 1 + t) satisfies the constraint, yet x^{2}y = t^{2}(1 + t)→∞ as t→∞, so there is no maximum.
^{5} When u(x, y) = x^{a} + y^{b}, for instance, the solution is not given by (∗∗). To check this, assuming that 0 < a < 1, see: Exercise 9, for the case when b = 1; and Exercise 14.5.4, for the case when a = b.