+885 At a certain amusement park, there is a bulk discount for tickets. If you buy up to 60 tickets in one order, the price for each ticket is $70. However if you buy more than 60 tickets in a single order, the price of every ticket is reduced by $1 for each additional ticket bought. If t is the number of tickets bought in bulk at one time, what is the largest t which will bring the amusement park a profit greater than $4200?
+129852 Let x be the number of tickets above 60 = number of $1 decreases from $70
So
Revenue = Quantity * Price .....so we have....
Revenue = (60 + x) ( 70 - x) simplify
R = -x^2 + 10x + 4200
The number of $1 decreases that will maximize revenue is given by
-10 / (2 * -1 ] = -10 / -2 = 5
So.....the number of bulk tickets that will maximize the revenue is given by (60 + 5) = 65 tickets = t
+4622 We see that there is a price of \((70-(t-60))\) for each of the tickets, because there a price of 70 minus 1 for each ticket after the 60th ticket. So the total price is \((130-t)(t)\)and we want this to be greater than \(4200\). Rewriting this, we see that \(t^2-130t+4200<0.\) This means that \((t-65)^2<25\), so \(-5 Thus \(60 , and the largest possible \(t\) is \(\boxed{69}.\).
+37146 Agree with tertre.....MAXIMUM t would be 69 tix at $ 61 each to clear $4200.....
...Note that Chris found the number of tix which will maximize REVENUE @ 4225 (65 tix at $65 each)
