Hi,         

P(t)= -16t² + 800t – 4000

the vertex form of a parabola,  where(h,k) is the vertex

P(t)= -16(t^2 -50t)-4000)   Completing the square to find the vertex

P(t)= -16[(t-25)^2 -625] - 4000 

P(t)= -16(t-25)^2 + 10,000 - 4000

P(t)= -16(t-25)^2 + 6000   vertex is (25,6000) OR ordered pair(t,P(t))

parbola opens downward (a<0), vertex is the maximum point for P(t)

a. What ticket price gives the maximum profit? $25

b. What is the maximum profit?  $6000

c. What ticket price(s) would generate a profit of $5424? 

P(t)= -16t² + 800t – 4000

5424 = -16t² + 800t – 4000

 -16t^2 + 800 - 9424 = 0







   t = $31

   t = $19   

Answer by CharlesG2(834)   (Show Source): You can put this solution on YOUR website!
 Each year a school’s booster club holds a dance to raise funds. In the past, the profit the club makes after paying for the bank and other costs has been modeled by the function P(t)= -16t² + 800t – 4000, where t represents the ticket price in dollars. 

a.	What ticket price gives the maximum profit?

b.	What is the maximum profit?

c.	What ticket price(s) would generate a profit of $5424? 





P(t) = -16t^2 + 800t - 4000

standard form parabola is y = ax^2 + bx + c,

here a = -16, b = 800, c = -4000,

vertex x-coordinate = -b/2a = -800/(2 * -16) = -800/-32 = 25

P(25) = -16 * 25^2 + 800 * 25 - 4000

P(25) = -16 * 625 + 20000 - 4000

P(25) = -10000 + 20000 - 4000

P(25) = 10000 - 4000 = 6000

$25 dollar ticket price gives maximum profit of $6000

for $5424 profit:

5424 = -16t^2 + 800t - 4000

0 = -16t^2 + 800t - 9424

0 = t^2 - 50t + 589









t = 62/2 or t = 38/2

t = 31 or t = 19

either $19 or $31 dollar ticket prices



 

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