text: rapid help, needed for exam...economists and math experts? if you have function of demand (x) which is in relation with price (p) given like p = sqr(x) – 10x + 25  1.  find the revenue function (r)  r = px   and find the specific demand and price for which we have r(max) - maximal revenue  2. find marginal revenue (r') for price p=2 and p=3  3. find revenue elasticity (er) for price p=2 and p=3  thnx in advance.... i'll help you with the first part, but unfortunately you were ambiguous with your terminology.  is r = p(x) [p is a function of x] or r = p*x, i.e., p(x)*x, i.e., the function p times x?  not only that, you say r = px, lower case p, which strictly means p is another variable you have not mentioned yet.  while that's what you wrote, i doubt that's what you meant.  i'm going to assume you meant r = p(x) - if not, use the same logic by multiplying through by x.  p(x) = sqr(x) - 10x + 25 r(max) = p'(x) for x = 0 (derivative of p(x) = 0 implies the maximum)  so take the derivative of p(x) and set p'(x) = 0.  find x.  replace this value for x in the original equation and you will have r(max).  2 and 3:  again, here you literally imply there is a new variable, p.  if that's what you mean there isn't enough info to help.  if you really mean p(x) = 2 and 3, then do the following:  solve for x in p(x) = 2 and 3  replace this variable in your marginal revenue, which you've defined as r' = p'(x) which you found in (1) (or you'll have to recalculate if i misinterpreted your revenue function).  do the same in the elasticity function, which you have not written down.