Active Learning Exercise #1 Assume the utility function is C

Active Learning Exercise #1 Assume the utility function is π‘ˆ 𝐢, 𝐿 = 𝐢 !/! + 2𝐿!/! Optimization problem is π‘šπ‘Žπ‘₯!,! 𝐢 !/! + 2𝐿!/! 𝑠𝑒𝑏𝑗𝑒𝑐𝑑 π‘‘π‘œ 𝐢 ≀ π‘Š 100 βˆ’ 𝐿 π‘œπ‘Ÿ 𝐢 + π‘ŠπΏ ≀ 100π‘Š, 0 ≀ 𝐿 ≀ 100 Find the optimal values for C and L. SOLUTION Derive MRS πœ•π‘ˆ
𝐿!!/!
𝐢
𝑀𝑅𝑆 = βˆ’ πœ•πΏ = βˆ’
=2
1
πœ•π‘ˆ
𝐿
(2)𝐢 !!/!
πœ•πΆ
!/!
Substitute π‘Š 100 βˆ’ 𝐿 for C so that all income is used (budget constraint is satisfied) 𝐢
𝑀𝑅𝑆 = βˆ’2
𝐿
!
!
π‘Š 100 βˆ’ 𝐿
= βˆ’2
𝐿
!/!
Equate with slope of budget line: π‘Š 100 βˆ’ 𝐿
𝑀𝑅𝑆 = βˆ’2
𝐿
π‘Š 100 βˆ’ 𝐿
2
𝐿
4π‘Š 100 βˆ’ 𝐿
!/!
=π‘Šβ†’4
!/!
= βˆ’π‘Š = βˆ’
π‘Š 100 βˆ’ 𝐿
𝐿
= π‘Š ! 𝐿 β†’ 4 100 βˆ’ 𝐿
𝑃!
𝑃!
= π‘Š ! = π‘ŠπΏ β†’ πΏβˆ— =
400
π‘Š+4
Confirm that the optimum respects the condition: 0 ≀ 𝐿 ≀ 100. It does as long as W is non-­β€negative. Alternative Solution Use the FOC by taking the derivative of the objective function and set it equal to zero. The optimization problem is π‘šπ‘Žπ‘₯!,! 𝐢 !/! + 2𝐿!/! 𝑠𝑒𝑏𝑗𝑒𝑐𝑑 π‘‘π‘œ 𝐢 ≀ π‘Š 100 βˆ’ 𝐿 , 0 ≀ 𝐿 ≀ 100 Substitute: π‘šπ‘Žπ‘₯! (π‘Š 100 βˆ’ 𝐿 )!/! + 2𝐿!/! 𝑠𝑒𝑏𝑗𝑒𝑐𝑑 π‘‘π‘œ 0 ≀ 𝐿 ≀ 100 Derive the FOC: πœ•βˆ™
1
=βˆ’
πœ•πΏ
2
π‘Š 100 βˆ’ 𝐿
!
!
!π‘Š
!
+ 𝐿!! = 0 β†’ πΏβˆ— =
400
π‘Š+4
Active Learning Exercise #2 Assume the utility function is π‘ˆ 𝐢, 𝐿 = 𝐢 + 2𝐿 Optimization problem is π‘šπ‘Žπ‘₯!,! 𝐢 + 2𝐿 𝑠𝑒𝑏𝑗𝑒𝑐𝑑 π‘‘π‘œ 𝐢 ≀ π‘Š 100 βˆ’ 𝐿 π‘œπ‘Ÿ 𝐢 + π‘ŠπΏ ≀ 100π‘Š, 0 ≀ 𝐿 ≀ 100 Find the optimal values for C and L. SOLUTION πœ•π‘ˆ
2
𝑀𝑅𝑆 = βˆ’ πœ•πΏ = βˆ’ = βˆ’2 πœ•π‘ˆ
1
πœ•πΆ
Given MRS = -­β€2, consumers are left as well off when they trade-­β€off one more unit of leisure in exchange for two units of consumption and that is true regardless of the values for C and L. Indifference curves are then parallel lines with a constant slope of -­β€2. The slope of the budget line is βˆ’
𝑃!
π‘Š
= βˆ’ = βˆ’π‘Š 𝑃!
1
There is a corner solution except when W = 2. The problem can also be solved as follows. After substituting π‘Š 100 βˆ’ 𝐿 for C, the problem is π‘šπ‘Žπ‘₯! π‘Š 100 βˆ’ 𝐿 + 2𝐿 𝑠𝑒𝑏𝑗𝑒𝑐𝑑 π‘‘π‘œ 0 ≀ 𝐿 ≀ 100 π‘šπ‘Žπ‘₯! π‘Š100 + (2 βˆ’ π‘Š)𝐿 𝑠𝑒𝑏𝑗𝑒𝑐𝑑 π‘‘π‘œ 0 ≀ 𝐿 ≀ 100 Optimal solution: If W < 2 then L =100. If W > 2 then L = 0. If W = 2 then any value of L is optimal. Intuition: MU of leisure and MU of consumption are constant. The price of leisure is W and the price of consumption is 1. By increasing leisure by one unit, utility is raised by 2 due to more leisure but is reduced by W due to less consumption. If 2 > W then utility goes up so that the consumer should take on more leisure. Given that the net gain in utility is 2-­β€W regardless of how much leisure is consumed, a consumer will want to take as much leisure as she can. Thus, the optimal bundle is a corner solution in which L = 100, C = 0. The reason is that MU is constant. If instead MU of leisure is decreasing in the amount of leisure then, perhaps at some point, the net effect on taking more leisure will be negative. In that case, there will be some amount of leisure between 0 and 100 at which MRS = slope of the budget line which will be the optimal level of leisure.