Find the largest possible revenue from the demand equation "Q = - 2p + 1000

Revenue=quantity x price = (-2p+1000) (p) = - 2p^2+1000p

The maximum revenue will occur when the first derivative is zero so when 2 (-2p) + 1000=0; p=250

Which generates 125,000 in revenue

Try prices of 245 and 255 and you will see they both are less than 250 thereby proving the max revenue is 250