#OK to post homework
#George Spahn, 2-6-2022, Assignment 5

# 5 

deriv1 := diff(q[1]*(1-q[1]-q[2])^(d[1]),q[1]):
deriv2 := diff(q[2]*(1-q[2]-q[1])^(d[2]),q[2]):
solve({deriv1 = 0, deriv2 = 0}, {q[1],q[2]}):

# the produced output is 
# q[1] = d[2]/(d[1]*d[2] + d[1] + d[2])
# q[2] = d[1]/(d[1]*d[2] + d[1] + d[2])

# when d1 and d2 are both equal to 1,
# q1 and q2 are both equal to 1/3, as we computed before


#6 

# Let's define nashut to be the combined utility of both players
# in the nash equilibrium described above.

ut := q[1]*(1-q[1]-q[2])^(d[1]) + q[2]*(1-q[2]-q[1])^(d[2]):
nashut := subs(q[1] = d[2]/(d[1]*d[2] + d[1] + d[2]),
               q[2] = d[1]/(d[1]*d[2] + d[1] + d[2]),
			ut):

# Let ut1 be the best possible utility of player 1 if q2 = 0.

u1 := q[1]*(1-q[1])^(d[1]):
s1 := solve(diff(u1,q[1]) = 0, q[1]):
ut1 := subs(q[1]=s1,u1):

# Similarly let ut2 be the best possible utility of player 2 if q1 = 0.

u2 := q[2]*(1-q[2])^(d[2]):
s2 := solve(diff(u2,q[2]) = 0, q[2]):
ut2 := subs(q[2]=s2,u2):

# Let m be the max of ut1 and ut2.

m := max(ut1,ut2):

# We now claim that m >= nashut, and if d1,d2 > 0, then m > nashut.
# The proof will be by picture.

plot3d(m-nashut,d[1]=0..1,d[2]=0..1);

# The plot shows that m - nashut is nonnegative
# and only equal to 0 when d1 = 0 or d2 = 0

# Therefore, the nash equilibrium does not maximize combined utility
# unless d1 = 0 or d2 = 0.