A. $Expected Value = .5*20,000 + .5*10,000 = $15,000

1. 
There is a 50-50 chance that an individual with utility u=√($) and with wealth of $20,000 will contract a debilitating disease and suffer a loss of $10,000.

a.  $expected value = .5*20,000 + .5*10,000 = $15,000.

b.  the expected utility = .5*√20,000 + .5*√10,000 = .5*141.42 + .5*100 = 120.71 utils

c.  the $certainty equivalent = 120.712 = $14,571

d.  Set Wgood=Wbad=$CE=$14,571. So 14571=20000-Prem è Prem = $5429. And then, 14571=20000-10000-5429+Ben è Ben = $10,000.
Expected profit = 5429-(.5*10000)=$429

e.  $expected value = .25*20,000 + .75*10,000 = $12,500.
expected utility = .25*√20,000 + .75*√10,000 = .25*141.42 + .75*100 = 110.36 utils $certainty equivalent = 110.362 = $12,179
Wgood=Wbad=$CE=$12,179. So 12179=20000-Prem è Prem = $7821. And then, 12179=20000-10000-7821+Ben è Ben = $10,000
Expected profit = 7821-(.75*10000)=$321

2.  Diminishing returns to scale implies that if x=f(L,K) and y=f(sL, sK) where s>1, then y<sx. All inputs must be scaled by the same factor, s, in the concept of returns to scale. With diminishing returns to a factor of production, look at how the marginal product of a factor varies, holding the other factors constant. The concept of diminishing returns to a factor of production implies that eventually the marginal product of the factor declines. A production function CAN exhibit diminishing returns to scale but NOT have diminishing returns to a factor. For example, consider the production function: x=L + √K. Let L=1 and K=4èx=3. Now let s=2, so that L=2 and K=8 and x=4.83 which is less than 2*3=6, so you doubled inputs and less than doubled output, so we have decreasing returns to scale. But we DO NOT have diminishing returns to the factor L. Its marginal product is constant and equal to 1, in this case.

3.  Q=KL - .8K2 - .2L2

a.  Suppose K=10. TPL = 10L – 80 -.2L2 and APL = Q/L = 10 – 80/L -.2L and also note that MPL = dQ/dL = 10 - .4L. TPL reaches a max when MPL = 0. So setting MPL = 0 è
10 -.4L = 0 è L = 25. APL is at its max when APL=MPL, so set the two equal to each other and solve for L è 10 – 80/L - .2L = 10 - .4L è L=20. And note that when L=20, Qtotal=40. And APL = 0 at L=10 and L=40. Also note that TPL=0 at L=10 and L=40.

b.  From above, MPL=0 at L=25.

c.  If K=20, then: TP=20L-320-.2L2; AP=20 – 320/L -.2L and reaches a max at L=40 and Qtotal=160; MP = 20 - .4L and MP=0 at L = 50. (You can draw the pictures for this, right?)

4.  Q=S1/2J1/2, Ws=$3 and Wj=$12, and S=900 è Q=√900√J = 30√J

a.  For Q=150, need J=25. For Q=300, need J=100. For Q=450, need J=225.

b.  Note that Q=30√J è J*(Q)=Q2/900 è vc(Q) = $12* Q2/900 è mc(Q) = (2*12)/900*Q, so marginal cost of the 150th page = $4 and of the 300th page = $8 and of the 450th page = $12