Let x = the number of $1 decreases in price. The number of items produced will be 235 + 5x. The price will be $35 - x. The revenue, R(x), will be:
R(x) = (235 + 5x)($35-x) = 5875 - 60x - 5x2
To optimize, take the derivative of R(x) wrt x, set it to zero, and solve for x. That will give you the number of $1 decreases in price that will optimize revenue. Plug that value of x into the Revenue equation above to find the maximum revenue. The optimal price will be $35-x.