SOLUTION: 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(

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Question 389250: 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?

Found 2 solutions by ewatrrr, CharlesG2:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi,         

P(t)= -16t² + 800t – 4000

the vertex form of a parabola, y=a%28x-h%29%5E2+%2Bk 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+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

t+=+%28-800+%2B-+sqrt%2836864+%29%29%2F%28-32%29+

t+=+%28-800+%2B-+192%29%29%2F%28-32%29+

   t = $31

   t = $19

Answer by CharlesG2(834) About Me  (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+=+%2850+%2B-+sqrt%28+%28-50%29%5E2+-+4%2A589+%29%29%2F2+
t+=+%2850+%2B-+sqrt%28+2500+-+2356+%29%29%2F2+
t+=+%2850+%2B-+sqrt%28+144+%29%29%2F2+
t+=+%2850+%2B-+12%29%2F2+
t = 62/2 or t = 38/2
t = 31 or t = 19
either $19 or $31 dollar ticket prices