This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Economics 3010 Professor Daniel Benjamin Fall 2009 Cornell University Problem Set 0 solutions Prepared by Ahmed Jaber 1. Solve max x & ; y & c ln ( x )+ d ln ( y ) subject to p x x + p y y = m , where c , d , p x , p y , and m are all strictly positive constants. (Hint: There are two ways to solve this constrained maximization problem. If you have taken multivariable calculus, you can use the Lagrangian method. If not, you can use the constraint to substitute out x in terms of y, and then solve a single-variable unconstrained optimization problem.) Note that we do not need to worry about corner solutions in this problem since lim z ! ln( z ) = &1 , and so the agent would never choose a bundle of the form (0, e y ) or ( e x ,0) where e x and e y . Substitution Using the constraint p x x + p y y = m , we get x = m p y y p x since p x > . Substituting for x into the objective function, the problem can be rewritten as: max y c ln & m p y y p x + d ln ( y ) . Taking the &rst order condition with respect to y and setting it to to get the maximum, we get c py px m & pyy px + d 1 y = 0 which simpli&es to c p y m p y y + d 1 y = 0 . Rearranging, we get c p y y = d ( m & p y y ) ) ( c + d ) p y y = dm ) y = d c + d m p y . By substituting for y into the formula for x , x = m p y y p x , we get x = c c + d m p x . Remark 1 Rather than substituting for y into the expression x , we could have remarked that the problem is symmetric in x and y and so directly concluded that x = c c + d m p x (by replacing d by c and p y by p x in the expression we found for y ). Remark 2 We could just as well have substituted out y in terms of x and proceeded in a similar way to get the same &nal answers....
View Full Document