Expert: Paul Klarreich - 3/6/2007

Question
For my chemical engineering homework assignment, we were given an equation we need to differentiate in order to prove when a certain product recovery is maximized. I am familiar with how to find the critical points and points of inflection etc. but the equation we were given for this homework assignment was rather difficult and I'm not sure how to go about differentiating.

Here is the whole problem: Please don't let the chemical engineering context bother you!

#2: The previous problem is based on the assumption that the wash water is divided equally among the various washes.  It is legitimate to ask if this is the best way to distribute the wash water.  Let us assume the total amount of solvent is fixed and the fraction of solvent fed to a particular extraction is Fi, show that X1/Xn+1 = (Product) i [1/(1+Ei)], where Ei = KFiL/H. and Xn/Xn+1 reaches minimum if Fi = Fn = 1/N.  Therefore, you just prove that the recovery will be maximized when the wash water is divided uniformly among the various washes.  

I used lowercase to represent subscripts. i hope this isnt terribly confusing. the "(Product)" is one of those big uppercase Pi symbols that i think represents a product. My main problem is how to differentiate a (Product). If you could show me how to do this i would really appreciate it! Thanks,
Chelsea

Answer
Questioner:   Chelsea
Category:  Advanced Math
Private:  No
 
Subject:  Differentiating an Equation
Question:  For my chemical engineering homework assignment, we were given an equation we need to differentiate in order to prove when a certain product recovery is maximized. I am familiar with how to find the critical points and points of inflection etc. but the equation we were given for this homework assignment was rather difficult and I'm not sure how to go about differentiating.

Here is the whole problem: Please don't let the chemical engineering context bother you!

>> IT DOES, but that's a personal matter and certainly not your fault.

#2: The previous problem is based on the assumption that the wash water is divided equally among the various washes.  It is legitimate to ask if this is the best way to distribute the wash water.  Let us assume the total amount of solvent is fixed and the fraction of solvent fed to a particular extraction is Fi, show that X1/Xn+1 = (Product) i [1/(1+Ei)], where Ei = KFiL/H. and Xn/Xn+1 reaches minimum if Fi = Fn = 1/N.  

Therefore, you just prove that the recovery will be maximized when the wash water is divided uniformly among the various washes.  

I used lowercase to represent subscripts. i hope this isnt terribly confusing. the "(Product)" is one of those big uppercase Pi symbols that i think represents a product. My main problem is how to differentiate a (Product). If you could show me how to do this i would really appreciate it! Thanks,
Chelsea
..................................
Hi, Chelsea,

This is a bit out of my league, and it does not look like a calculus problem.  But:

I gather that what you are doing is trying to dissolve some stuff and you have a fixed amount of some expensive solvent like acetone or petroleum ether and want to use it most effectively; i.e. dissolve as much stuff as you can.

And the rules are something like:

1. If you use more solvent for a single rinse, you take away a greater  fraction of stuff.

2. Given an amount of solvent for the rinse, the fraction will be the same regardless of how much stuff is present.  That is, if I use 2 liters of acetone and it takes away 40% of the stuff, the same will be true of the next 2-liter rinse, even though there is less stuff present for that next rinse.

(I will use brackets for subscripts unless it is obvious.  And I'm going to iterate using j, not i, because it looks clearer in the text.)

WARNING: THIS DISCUSSION MAY CONTAIN FRACTIONS AND OTHER MATERIAL DIFFICULT TO VIEW ON CERTAIN COMPUTING SYSTEMS.  VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

So I guess that each time you rinse with  F[j] liters of solvent, you reduce the existing amount  x[j] to x[j+1] and their ratio is:

x[j+1]      1
------ = -------
x[j]    1 + KF[j]

I know you have an L and an H there, but I have no idea what they are, and since you didn't subscript them, I must assume they are constants.  In that case, they can 'fold' into the K.

Then the fraction after two successive rinses would be:

x[j+1] x[j+2]      1           1
------ ------ = --------- -----------
x[j]  x[j+1]   1 + KF[j] 1 + KF[j+1]

Obviously when you do this N times, the first ratio is  x[1]/x[0] and the

last ratio is  x[N]/x[N-1], and the product 'telescopes' with everthing canceling except x[N] and x[0], so that:
x[N]    N       1
---- = PROD ---------
x[0]   k=1  1 + KF[j]

Now you want to minimize that final ratio, and you want to prove that the choice of F[j] = 1/N does that.

So you have:

x[N]    N       1            1
---- = PROD --------- = --------------
x[0]   k=1  1 + KF[j]   PROD (1+KF[j])

which you want to minimize.  {As I recall, if you minimize this, you are getting the most stuff out with your rinses.}  And you could minimize the fraction by MAXimizing the denominator, so let's work on that.

This is not, as I see it, a calculus problem.  Instead it is a case of the Arithmetic-Geometric Mean inequality.  

[As Casey Stengel used to say: You could look it up.]

That inequality says that given a set of positive numbers:

P[1]..P[N]

their geometric mean is less than or equal to their arithmetic mean.  

You get the arithmetic mean by adding them up and dividing by N.  In this case, the F[j] all add up to 1, so we compute the mean:

1 + KF[1] +
1 + KF[2] +
...
1 + KF[N]
-----------
N + K(sum of F[j]) = N + K

The sum is  N + K and their arithmetic mean is 1 + K/N

You get the geometric mean by multiplying them and taking the N-th root of the product.  In this case the geometric mean is:

Nth root of(Prod (1 + KF[j])

So the A-GMI says that

Nth root of(Prod (1 + KF[j]) <= 1 + K/N

Now what if all the F[j] are equal to 1/N?  The GM is:

Nth root of(Prod (1 + K/N)) =
Nth root of((1 + K/N)^N) =
1 + K/N,

which is the SAME as the AM.

Therefore equal values of F[j] achieve the maximum product of the denominators, therefore the minimum of the product of the fractions.

WHEW!