F.4 Maths quadratic equation

(a) Let $p be the price of each ticket and q be the number of tickets sold. Express q in terms of p.
For every increase of $1 in the ticket price, the number of tickets sold will decrease by 50.
So if the price is p, then it is $(p - 160) above $160 so the number of ticket sold
q = 12000 - (p - 160)*50
q = 12000 - 50p + 8000
q = 20000 - 50p
(b) What should the price of each ticket be in order to get the max. income?
Income = (20000 - 50p)p
= -50(p^2 - 400p + 40000 - 40000)
= -50(p - 200)^2 + 2000000 which is a max when p = 200
The price should be set at $200

(a) Let q=A+Bp. Since p increase=>qecrease 50, wehave B=50.Also,when p=160,q=12000. 12000=A-50*160=>A=20000. So q=20000-50p

(b) The total income is pq=-50p^2+20000p which can be rearranged as -50(p-200)^2+2000000

So when the price is 200, Andy can get the max. income 2000000