Step-by-step explanation:
the original revenue is
4×300 = $1200
revenue(n) = (4 + n×0.1)×(300 - n×5) = 1200 - 20n + 30n - 0.5×n² =
-0.5×n² + 10n + 1200
with n being the number of units of $0.10 increasing the price.
now, where is the extreme (maximum, we hope) value of the function ?
it is the 0 point of the first derivative :
revenue'(n) = -n + 10 = 0
n = 10
so, the maximum is when increasing the price by 10 times of $0.10 (that means by increasing it by a full $1).
that means the best is selling the slices for 4+1 = $5.
and the max. revenue is then
(4+1)(300-50) = 5×250 = $1250.