Calculus
posted by Katrina on .
In a chemical reaction, substance A combines with substance B to form subtance Y. At the start of the reaction, the quantity of A present is a grams, and the quantity of B present is b grams. Assume a is less than b. At time t seconds after the start of the reaction, the quantity of Y present is y grams. For certain types of reaction, the rate of the reaction, in grams/sec, is given by
Rate = k(ay)(by), k is a positive constant.
a. For what values of y is the rate nonnegative?
Give your answer as a union of intervals, e.g., (infinity,a] U (a, 2b)
y E_______________________
b. Find the value of y at which the rate of the reaction is fastest.
y= _________________________
I thought that in part A all nonnegative values were going to be anything less than a and everything larger than b so I typed (my homework is online) (INF, a] U [b, INF) and in b the answer I got was (1/2)(a+b) but they are both wrong... please help....
Calculus  Jai, Sunday, April 10, 2011 at 12:48am
for (a) since a < b , for a certain value of y , the value of (a  y) will become negative first, and thus the Rate becomes negative,, then after some time (b  y) will become negative too, and the Rate becomes positive. let's look at some points:
at 0 <= y < a , Rate > 0
at y = a , Rate = 0
at a < y < b , Rate < 0
at y = b , Rate = 0
at y > b , Rate > 0
thus, rate is nonnegative (but may be equal to zero) at values of y which is
[0 , a] U [b , +infinity)
*note that we start at 0 since quantity/mass can never be negative. another, the +infinity will only possible if a and be is continuously supplied or fed to the reactor. otherwise, at a finite value of a and b, y will only reach a certain maximum value (y,max) when the reaction is complete (or at infinite time)
for (b), we take the derivative of Rate = k(ay)(by) with respect to y, and equate Rate to zero since maximum rate (slope is zero):
R = k(ay)(by)
R = k(ab  by  ay + y^2)
0 = k[b  a + 2y]
0 = b  a + 2y
y = (a+b)/2
*we got the same answer. are you sure it's wrong?
hope this helps~
Calculus  Jai, Sunday, April 10, 2011 at 1:24am
ahh i think i know why y = (a+b)/2 is wrong,, it's actually the MINIMUM, not the maximum~ ^^;
i tried assigning some values to the variables,, and from the graph, rate > infinity at y > inifinity , or at a finite value of a and b, when the reaction is at completion (time at infinity), the rate is max at y = y,max , provided that this y,max is greater than b.
Sorry to post this one again but there are things that I still don't understand... Thanks for the explanation btw, I don't know why it didn't occur to me that mass can't be negative =P... I now get why the answer for B can't be (a+b)/2 but still don't get what the max is, sorry.... and I typed the new answer for part A and for some reason it still says it's wrong.

If we expect the rate to be nonnegative, we must have 0 ≤ y ≤ a and 0 ≤ y ≤ b.
Since we assume a < b, we restrict y to 0 ≤ y ≤ a.
In fact, the expression for the rate is nonnegative for y greater than b, but these
values of y are not meaningful for the reaction.
Rate
0
(b) From the graph, we see that the maximum rate occurs when y = 0; that is, at the
start of the reaction. 
a) y E (0,a)
b) y = 0 
Why are you posting under so many different names  and answering your own question??
?? 
Hi, I wasn't posting under different names, I have just used my name and I was just posting the explanation I received from other tutor, I didn't understand some of it so I posted it again because right now it must be on page 5 or something like that... I'm not answering my own questions... I was truly confused...
Thank you helplessHelper...