Please help me i'm not sure how to solve this

Question

Suppose Orchestra Stratford can sell out a 300-seat theatre when tickets are sold for $48. For each $6 increase, 15 less tickets will be sold. Use a quadratic relation to determine the necessary ticket price to earn maximum revenue.

Answer 1

N = no. tickets; P = price Then revenue (R) = NP

Let's use a constant k for the no. of price increases (and ticket sale reductions).

If you lose 15 seats for every increase, then N = 300 - 15k. The equation for P is P = 48 + 6k

Then Revenue is (300 - 15k)(48 + 6k) = 14400 + 1080k - 90k². That works because when k = 0 (no price change) R = $14,400 (300 x $48).

If you take the first derivative and set result to = 0 for maximum revenue, you get -180k + 1080 = 0, k = 6

So revenue is maximized after price is increased by 6($6) or $36, to $84. At that price point, you'll sell 6(15) = 90 fewer tickets, or 210 tickets.

Why not make a table to prove it?

k # tickets $ Price $ Revenue

0 300 48 14,400

1 285 54 15,390 etc. You'll see that revenue tops out at $17,640 at k = 6.