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This was a problem I did with my students in Calculus 151c (that's differential calculus at OSU). The students seemed to like it... It's a fairly basic rate problem, I think it's a really good problem because it ties together the relationship between rates, derivatives, and integrals. Here goes...

## The Beer Bong Problem

Jim is a college student, and like many others, he likes to spend his weekend drinking at parties. At the party in question, Jim is invited to drink from a beer bong, and he accepts. Before he begins, the bong is filled with 12 ounces of beer. Jim can drink beer at an initial rate of 1.5 ounces per second. However, as he drinks and runs out of breath, the rate at which he can drink decreases by a constant factor. Of course, Jim's friends will be adding more beer to the bong at a rate of 2 ounces per second. Assume the bong can hold a maximum of 22 ounces of beer.

a) Give an equation for the amount of beer in the bong at any time t
b) Assuming the constant factor of decreasing drinking rate is .1 ounces per second, will the bong overflow before Jim is finished drinking?
c) Find a constat that that will leave the bong empty just as Jim is finished drinking (to minimize waste of beer)

Note: I'll post the solution to this tomorrow, time for bed now :)

Okay, well, since the rate of consumption is 1.5-kt, the total consumption (assuming 0 initial consumption) is 1.5t-kt2/2, and it gets filled at a rate of 2t, so the answer to a) is b(t)=12+.5t-kt2/2.

The answers for b) and c), however, rely on common sense - if his drinking rate starts at 1.5oz/sec and only decreases, then his drinking rate will NEVER overtake the filling rate of 2.0oz/sec. So the bong will certainly overflow no matter what - unless you have a negative constant (which defies the setup of the problem).

Also, how is "done drinking" defined, anyway? The time at which the bong is empty? Then the answer to c) is circular anyway. Or perhaps it's the time at which Jim's ability to drink has been expended (i.e. 1.5-kt=0)? Who knows?

In any case, this problem has many, many holes in it. I'd like to see your solution, ccunning, if you don't mind... :)

Well, it seems this node's author has no intention of providing a solution to this problem of extreme importance to frat boys worldwide, here I go:

b(t) = 12

Then we add 2 ounces per second:
b(t) = 12 + 2t

Then, we need to add the basic rate of Jim's drinking, being 1.5:
b(t) = 12 + 2t - 1.5t

Lastly, we add the rate at which he slows drinking, being c ounces per second per second. However, since this is a continual slide, not a series of steps down in speed, the actual figure we must insert is the integral of ct (constant * time), which is (ct2/2):
b(t) = 12 + 2t - 1.5t + (ct2/2)

This holds up to the point that Jim stops drinking the beer, but it does have a problem in that once t exceeds the number of seconds Jim spends chugging, his rate of drinking becomes negative (backwash!). However, for the purposes of this question I take it that the exercise ends once Jim stops drinking, so this shouldn't be a problem. Therefore the solution is:
b(t) = 12 + t/2 + ct2/2

b) It doesn't take a genius to see that if he starts drinking the beer at 1.5 ounces per second, and slows at .1 per second, he will stop after 15 seconds. So we simply substitute t = 15 into our function and see if it exceeds 22:

b(15) = 12 + 15/2 + 0.1*152/2
b(15) = 12 + 7.5 + 11.25
b(15) = 30.75

Therefore, as you can see, the beer bong will overflow before Jim finishes.

The exact time at which the beer bong overflows can be found by putting the equation into general form as a quadratic equation:
22 = 12 + t/2 + 0.1t2/2
0 = 0.05t2 + 0.5t - 10
this allows the use of the quadratic formula:

-0.5 +- sqrt((0.5)2 - 4*0.05*(-10))
t = ------------------------------------------------
0.1

This resolves down to:
t = 10, -20

Since the negative value is meaningless for the purposes of this question, we can safely say the beer bong will overflow after 10 seconds.

c) This question has no answer. Seeing as the rate of drinking is always slower than the rate of filling, it is clearly impossible for the beer bong ever to empty if the rate of drinking only falls. Of course, it would be possible for the beer bong to empty if the rate of drinking were to be increasing, but if that were the case, then Jim would never finish.

Indeed, even if Jim's drinking rate started higher than the rate of slowing, it would still be impossible for the beer bong to be empty when finished, as finishing requires he reduce his drinking to 0 ounces per second, while the filling rate is always constant. As such, unless Jim were to go from a drinking rate above 2 ounces per second down to zero instantly (which he can't do under the circumstances of the question), the rate of filling has to be higher than the rate of drinking for at least some period of time prior to finishing. In other words, no matter what the constants in this question, this problem has no solution.

However, just to prove it mathematically, here is the simultaneous equation for this problem:
0 = 12 + t/2 + ct2/2
t = 1.5/c (since that is how long it takes for Jim to finish)
which can be substituted in to form:
0 = 12 + 0.75c + 0.5c(1.5/c)2
0 = 12 + 0.75c + 0.5c*2.25/c2
0 = 12 + 0.75c + 1.125/c
0 = 0.75c2 + 12c + 1.125
which is another quadratic formula, so we again use the quadratic equation:

-12 +- sqrt(122 - 4*0.75*1.125)
c = -----------------------------------------
1.5

which resolves to:
c = -0.0943058, -15.9057

Both of these answers are totally meaningless for the purposes of the problem we have been given since they give forever increasing rates of drinking. Just in case you're curious, they are "valid" answers if we could accept negative time values, however this problem does not allow that. As such, it has been proven that there is no value of the constant which will result in Jim finishing drinking the beer bong at the same time that it empties.

So there you have it frat boys, next time you're chugging from a beer bong, you know how to calculate if it is going to overflow before you finish. After all, it's not that hard to remember and apply the quadratic formula while drunk is it?

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