I did this problem and I'm hoping it's correct, but I want to be sure as I have no answer key at hand. It's a fun little problem that involves Simpson's rule. Here it goes:
I am given that (b-a)/n = 200ft where n = 8.You are the game warden in your town and are responsible for stocking the local lake prior to the opening of the fishing season. The average depth of the lake is 20 feet. Your plan is to stock the lake with 1 fish per 1000 cubic feet, and have approximately 25% of the original fish population remaining at the end of the season.
What is the maximum number of fishing licenses that can be sold if the average catch is 20 fish per license?
So here is my calculus using Simpson's where I calculate the area of the lake:
(200/3) * (0 + 4(200) + 2(400) + 4(600) + 2(800) + 4(1000) + 2(1200) + 4(1400) + 1600) = 1,280,000 ft^2
Given average depth was 20ft. So...
20 * 1,280,000 = 25,600,000 ft^3 = volume
1 fish per 1000 cubic feet
25,600,000 / 1000 = 25,600 fish in the lake.
25% of 25,600 = 6400 = the amount of fish that need to remain.
75% of 25,600 = 19,200 = the amount of fish to be fished out.
Average catch is 20 fish per person, so..
19200 / 20 = 960 licenses per season allowed = answer to original question.
So again.. I'm wondering if I did this right. I hope I didn't screw up on the calc portion.