Answer:
a) $675000
b) $289000 profit,3300 set, $190 per set
c) 3225 set, $272687.5 profit, $192.5 per set
Step-by-step explanation:
a) Revenue R(x) = xp(x) = x(300 - x/30) = 300x - x²/30
The maximum revenue is at R'(x) =0
R'(x) = 300 - 2x/30 = 300 - x/15
But we need to compute R'(x) = 0:
300 - x/15 = 0
x/15 = 300
x = 4500
Also the second derivative of R(x) is given as:
R"(x) = -1/15 < 0 This means that the maximum revenue is at x = 4500. Hence:
R(4500) = 300 (4500) - (4500)²/30 = $675000
B) Profit P(x) = R(x) - C(x) = 300x - x²/30 - (74000 + 80x) = -x²/30 + 300x - 80x - 74000
P(x) = -x²/30 + 220x - 74000
The maximum revenue is at P'(x) =0
P'(x) = - 2x/30 + 220= -x/15 + 220
But we need to compute P'(x) = 0:
-x/15 + 220 = 0
x/15 = 220
x = 3300
Also the second derivative of P(x) is given as:
P"(x) = -1/15 < 0 This means that the maximum profit is at x = 3300. Hence:
P(3300) = -(3300)²/30 + 220(3300) - 74000 = $289000
The price for each set is:
p(3300) = 300 -3300/30 = $190 per set
c) The new cost is:
C(x) = 74000 + 80x + 5x = 74000 + 85x
Profit P(x) = R(x) - C(x) = 300x - x²/30 - (74000 + 85x) = -x²/30 + 300x - 85x - 74000
P(x) = -x²/30 + 215x - 74000
The maximum revenue is at P'(x) =0
P'(x) = - 2x/30 + 215= -x/15 + 215
But we need to compute P'(x) = 0:
-x/15 + 215 = 0
x/15 = 215
x = 3225
Also the second derivative of P(x) is given as:
P"(x) = -1/15 < 0 This means that the maximum profit is at x = 3225. Hence:
P(3225) = -(3225)²/30 + 215(3225) - 74000 = $272687.5
The money to be charge for each set is:
p(x) = 300 - 3225/30 = $192.5 per set
When taxed $5, the maximum profit is $272687.5
