Answer:
Explanation:
Given the profit a club makes after paying for the bank and other costs modeled by the function
p(t) = -16t^2 - 800t - 4000
t is the ticket price
The maximum profit occurs when dp/dx = 0.
dp/dx = -32t-800
since dp/dx
0 = -32t-800
32t = -800
t = -800/32
t = - 25
Hence ticket price gives the maximum profit is $-25
b) to get the maximum profit, substitute t = -25 into the modelled function
p(t) = -16t^2 - 800t - 4000
p(-25) = -16(-25)² - 800(-25)- 4000
p(-25) = -10000+20000-4000
p(-25) = 6000
Hence maximum profit is $6000
c) To get the ticket prices would generate a profit of $5000, we will substitute p(t) = 5000 and find t
p(t) = -16t^2 - 800t - 4000
5000 = -16t^2 - 800t - 4000
-16t^2 - 800t - 4000 -5000 = 0
16t^2 + 800t + 9000 = 0
Answer:
Explanation:
Given the profit a club makes after paying for the bank and other costs modeled by the function
p(t) = -16t^2 - 800t - 4000
t is the ticket price
The maximum profit occurs when dp/dx = 0.
dp/dx = -32t-800
since dp/dx
0 = -32t-800
32t = -800
t = -800/32
t = - 25
Hence ticket price gives the maximum profit is $-25
b) to get the maximum profit, substitute t = -25 into the modelled function
p(t) = -16t^2 - 800t - 4000
p(-25) = -16(-25)² - 800(-25)- 4000
p(-25) = -10000+20000-4000
p(-25) = 6000
Hence maximum profit is $6000
c) To get the ticket prices would generate a profit of $5000, we will substitute p(t) = 5000 and find t
p(t) = -16t^2 - 800t - 4000
5000 = -16t^2 - 800t - 4000
-16t^2 - 800t - 4000 -5000 = 0
t^2 + 50t + 562.5= 0
Factorize
-50±√2500-4(562.5)/2
t = -50±√2500-2250)/2
t = -50±15.81/2
t = -65.81/2 and -34.19/2
t = -32.91 and -17.09