hw7_soln

Thus dividing the two partials by each other we get f

• Notes
• 10

This preview shows page 3 - 7 out of 10 pages.

Thus, dividing the two partials by each other we get (*) f 1 k 1 f 1 l 1 = μ 1 μ 2 3

Subscribe to view the full document.

Similarly, L k 2 = 0 = f 2 k 2 - μ 2 , which is equivalent to f 2 k 2 = μ 1 , L l 2 = 0 = f 2 l 2 - μ 2 , which is equivalent to f 2 l 2 = μ 2 Thus, dividing the two partials by each other we get (**) f 2 k 2 f 2 l 2 = μ 1 μ 2 . Combining (*) and (**) we see that (***) f 1 k 1 f 1 l 1 = f 2 k 2 f 2 l 2 Does (***) look familiar? It looks look an MRS equation (one agent’s MRS equal’s the other’s MRS), EXCEPT u A and u B are replaced by f 1 and f 2 respectively α is replaced with β y 2 is replaced by ¯ u x A 1 is replaced by k 1 , x A 2 is replaced by l 1 , x B 1 is replaced by k 2 , and x B 2 is replaced by k 2 ¯ ω 1 is replaced by 2 K and ¯ ω 2 is replaced by 2 L If you don’t believe me, just compare to the optimization problems. In this problem set the optimization problem is: Y 1 ( y 2 ) = max k 1 ,l 1 ,k 2 ,l 2 f 1 ( k 1 , l 1 ) subject to f 2 ( k 2 , l 2 ) = y 2 k 1 + k 2 = 2 K l 1 + l 2 = 2 L In HW #6, the problem was: U A u ) = max x A 1 ,x A 2 ,x B 1 ,x B 2 u A ( x A 1 , x A 2 ) subject to u B ( x B 1 , x B 2 ) = ¯ u x A 1 + x B 1 = ¯ ω 1 x A 2 + x B 2 = ¯ ω 2 4
Now let’s piggy back o ff of Anne’s solutions: From’s Anne’s solutions, we know that x A 2 = (1 - α ) x A 1 ¯ ω 2 α ¯ ω 1 +(1 - 2 α ) x A 1 So making the replacements, we get (^) l 1 = (1 - β ) k 1 2 L β 2 K +(1 - 2 β ) k 1 At this point, let’s set β = 1 2 . Using this we get l 1 = 2 L 2 K k 1 and since l 1 = 2 L - l 2 and k 1 = 2 K - k 2 so plugging in for l 1 and k 1 in (^) we see that: (^^) l 2 = 2 L 2 K k 2 Now we know that y 2 is being produced so l 2 k 2 = y 2 , so l 2 k 2 = y 2 2 . But if we replace k 2 with the expression in (^^) we get: 2 L 2 K k 2 2 = y 2 2 so k 2 = 2 K 2 L y 2 . Thus, since k 1 = 2 K - k 2 , this implies that (^^^) k 1 = 2 K - 2 K 2 L y 2 Now, remembering that β = 1 2 let’s plug in for k 1 and l 1 in f 1 and get (^^^^) Y 1 ( y 2 ) = (2 K - 2 K 2 L y 2 ) 2 L 2 K (2 K - 2 K 2 L y 2 ) = (2 K - 2 K 2 L y 2 ) 2 L 2 K = (2 K 2 L 2 K - 2 K 2 L 2 L 2 K y 2 ) = 2 KL - y 2 Thus, the PPF is Y 1 ( y 2 ) =2 KL - y 2 . Now before doing the β = 1 2 case, I am going to give the answer to part (c). AFTER PART (c) I will assume β = 1 2 ! (c) For β = 1 2 , MRT = Y 1 ( y 2 ) = -1 Now back to part (b). WARNING/MORAL: THINGS GET MESSY, BUT ENVELOPE THEOREM SAVES THE DAY! We know that l 1 = (1 - β )2 Lk 1 β 2 K +(1 - 2 β ) k 1 . Thus, we get, by the resource constraint that: l 1 = 2 L - l 2 = (1 - β )2 L (2 K - k 2 ) β 2 K +(1 - 2 β )(2 K - k 2 ) . Thus, (^^^^^) l 2 = β 2 Lk 2 (1 - β )2 K +(1 - 2 β ) k 2 Now since we know l 2 k 2 = y 2 , then we know (plugging in the calculations) that l 2 k 2 = β 2 Lk 2 2 (1 - β )2 K +(1 - 2 β ) k 2 = y 2 2 . Hence we have a quadratic equation for k 2 in terms of constants: β 2 Lk 2 2 - (1 - 2 β ) y 2 2 k 2 - (1 - β )2 Ky 2 2 = 0 Using the quadratic formula and eliminating the negative root we get: k 2 = (1 - 2 β ) y 2 2 + y 2 ((1 - 2 β ) 2 y 2 2 +16 β (1 - β ) LK 4 β L , 5

Subscribe to view the full document.

so k 1 = 2 K - (1 - 2 β ) y 2 2 + y 2 ((1 - 2 β ) 2 y 2 2 +16 β (1 - β ) LK 4 β L = 8 β KL - (1 - 2 β ) y 2 2 + y 2 ((1 - 2 β ) 2 y 2 2 +16 β (1 - β ) LK 4 β L Plugging this in to the f 1 production function, we get: Y 1 ( y 2 ) = k 1 ( (1 - β )2
You've reached the end of this preview.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern