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(

Algebra ->  Quadratic Equations and Parabolas -> 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(      Log On


   

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