Question 1028198
x = 100 - 2p ==> {{{p = (100 - x)/2}}}

a)  Revenue = {{{R(x) = xp = (x(100 - x))/2 = -x^2/2 +50x}}}

b)  Marginal revenue is R'(x) = -x + 50.
==> -x + 50 = 25 ==> -x = -25 ==> x = 25 units must be sold for the marginal revenue to be 25.

c)  Profit ={{{ R(x) - TC(x) = -x^2/2 +50x - x^2/2-10x - 100 = -x^2 + 40x -100 }}}.
 
d)  Maximum number of units is obtained from {{{x = -b/(2a) = -40/(2*-1) = 20}}} units.

e)  maximum profit = {{{p(20) = -20^2 +40*20 - 100 = -400 +800 - 100 = 300}}}.
Thus the maximum profit is $300.