Let p(t) be the number attendees and t be the ticket price measured in units of 10s of rupees. When the price is t = 7 (i.e. Rs70), there were p(7) = 300 people in attendance.
For each unit increase in t (i.e. for each Rs10 increase in price), p(t) is expected to fall by 20, so that
p(t + 1) = p(t) - 20
Solve for p(t). Suppose t ≥ 7. By substitution,
p(t + 1) = (p(t - 1) - 20) - 20 = p(t - 1) - 2×20
p(t + 1) = (p(t - 2) - 20) - 2×20 = p(t - 2) - 3×20
p(t + 1) = (p(t - 3) - 20) - 3×20 = p(t - 3) - 4×20
and so on, down to
p(t + 1) = p(7) - (t - 6)×20 = 420 - 20t
or
p(t) = 420 - 20 (t - 1) = 440 - 20t
With t = price per ticket (Rs/ticket) and p(t) = number of attendees = number of tickets sold, it follows that the income made from ticket sales for some fixed ticket price t would be t×p(t). If one plots p(t) in the coordinate plane, the price that maximizes income and number of attendees is such that the area of a rectangle inscribed by the line p(t) and the coordinate axes is maximized.
Let A(t) be the area of this rectangle, so
A(t) = t p(t) = 440t - 20t²
Without using calculus, complete the square:
A(t) = 440t - 20t² = 2420 - 20 (t - 11)²
This is the equation of a parabola with vertex at (11, 2420), so the optiml ticket price is Rs11, and at this price the drama club can expect an income of Rs2420.
With calculus, differentiate A with respect to t and find the critical points:
A'(t) = 440 - 40t = 0 ⇒ 11 - t = 0 ⇒ t = 11
Differentiate A again and check the sign of the second derivative at this critical point:
A''(t) = -40 ⇒ A''(11) = -40 < 0
which indicates a local maximum at t = 11 of A(11) = 2420.
