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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?

 Dec 29, 2018
 #1
avatar+98005 
+2

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

 

 

 

cool cool cool

 Dec 29, 2018
 #2
avatar+3987 
+2

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  \(\boxed{69}.\).

 

smileysmiley

 Dec 29, 2018
edited by tertre  Dec 29, 2018
 #3
avatar+3987 
0

Ugh, my Latex is not working well. 

tertre  Dec 29, 2018
 #4
avatar+17163 
+1

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)

ElectricPavlov  Dec 29, 2018
 #5
avatar+98005 
+1

Yep...go with tertre's answer.....I  didn't interpret this one properly...!!!!!

 

 

cool cool cool

CPhill  Dec 29, 2018

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