# Simpson's rule, did I do it right?

• Jun 20th 2010, 05:21 PM
nautica17
Simpson's rule, did I do it right?
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:

Quote:

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?
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 = 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.
• Jun 21st 2010, 12:00 AM
CaptainBlack
Quote:

Originally Posted by nautica17
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.

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.

CB
• Jun 21st 2010, 09:15 AM
nautica17
That is pretty much the whole question. What else do I need to add? The only other thing I was given was a graph that looked like a parabola, and it went from x=0 to x=8.
• Jun 21st 2010, 02:09 PM
CaptainBlack
You appear to be calculating the volume of a solid but don't tell us what the shape is. How can we say if your application of Simpson's rule is correct if we don't know the integrand?

CB
• Jun 21st 2010, 02:24 PM
nautica17
The thing is, I wasn't given any function. This question is part of a programming problem that I'm doing that just happens to have Simpson's rule involved to help in finding the volume of a lake as part of the problem. I was only given what I mentioned in the original post. The problem is workable without having to do any programming so I figured I try to solve it by hand first to make sure I'm on the right track. So as for the shape of the object in this case, I really don't know.
• Jun 21st 2010, 04:00 PM
skeeter
Quote:

Originally Posted by nautica17
The thing is, I wasn't given any function.

then where did this dimension come from?

Quote:

I am given that (b-a)/n = 200ft where n = 8.
I don't see that anywhere in the problem statement.

Quote:

(200/3) * (0 + 4(200) + 2(400) + 4(600) + 2(800) + 4(1000) + 2(1200) + 4(1400) + 1600) = 1,280,000 ft^2
also, where did these values come from?

you have a picture we can't see?
• Jun 21st 2010, 04:34 PM
nautica17
Attachment 17950
Here is the only picture I have of the graph.

And again, I don't know why there is no function, but there just isn't.

Things I was given:
(b-a)/n = 200
average depth of lake = 20ft
average catch per person is 20 fish

Definition of Simpson's:
integral from a to b of f(x) dx = (h/3)(y0 + 4*y_1 + 2*y_2 + 4*y_3 + ...+ 2*y_(n−2) + 4*y_(n−1) + y_n )

Basically I tried to go by the example on this page, (example 1): Pauls Online Notes : Calculus II - Approximating Definite Integrals

So for the y values, I plugged 0 to 1600 since n = 8.

Sorry for the confusion everyone. This is basically all I have.