Answer:
$36
$6466
Explanation:
For each 1 dollar increase, 5 fewer cars are rented. If y is added to the daily rate. They will rent 210 - 5y at 30 + y dollars.
Hence 210 - 5y × 30 + y = income.
F(y) = 210 - 5y × 30 + y
F(y) = 6300 + 210y - 150y -5y^2
F(y) = -5y^2 + 60y + 6300
To maximize income, we get the first derivative
F(y) = -5y^2 + 60y + 6300
F'(y) = -10y + 60
-10y + 60 = 0
10y = 60
y = 6
Therefore, the should increase their rate by $6, hence they should charge $36.
To find max income, put 7 into y
F(y) = -5×7^2 + 60×7 + 6300
= -254 + 420 + 6300
= $6466
Max income = $6466
