You are the game warden in your town and are responsible for stocking the local

Question

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?

Input

A data file containing space-delimited integers representing the y values of the coordinate

points in Figure 1. Notice that the file may or may not have a newline character
before EOF.

Output

The annotated, maximum number of fishing licenses that can be sold if the average catch is 20 fish per license.

Hint
You will have to compute the volume of the lake, since the criterion for stocking is 1 fish per 1000 ft3
Volume can be computed as Area

Explanation / Answer

I am given that (b-a)/n = 200ft where n = 8.

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

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