text: q=100,000 – 1.25p = what price should you charge in order to maximize revenues?   the previous answer is correct, however, you can get it much easier by applying calculus.  as stated by the previous poster,   r = qp = 10000p - 1.25 p^2  this function will have a maximum where its slope, (more properly, the slope of its tangent line) = 0.  that is, when a change in price causes the change in revenue to go from positive to negative, that's where revenue is highest.  thus, we want  dq/dp = 0  dq/dp = 100,000 - 2 (1.25)p  0 = 100,000 - 2.5p 100,000 = 2.5 p  p = 40,000.