This is inverse linear relationship between price and demand (P, D) defined with slope for values (30,200) and (45,175)
(200-175)/(30-45)=-25/15=-5/3
D=-5P/3 + C, where C is constant aka demand intercept
200=-5*30/3 + C, C=250
We should maximize revenue, D*P ---> P*(-5P/3+250).
Derivative of \(-5P^2/3+250P\) equated to 0 will maximize the revenue.
\(-10P/3+250=0\), \(10P=750\), \(P=75\) and answer is
D Carcass wrote:
Two hundred people will attend a university fund-raiser if tickets cost $30 per person. For each $15 increase in ticket price, 25 fewer people will attend. What ticket price will yield the maximum amount of money for the university?
(A) $30
(B) $45
(C) $60
(D) $75
(E) $90