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# help

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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
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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   Dec 29, 2018
#2
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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  \(\boxed{69}.\).  Dec 29, 2018
edited by tertre  Dec 29, 2018
#3
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Ugh, my Latex is not working well.

tertre  Dec 29, 2018
#4
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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
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Yep...go with tertre's answer.....I  didn't interpret this one properly...!!!!!   CPhill  Dec 29, 2018