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....

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~ 
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. 
HI,
Thanks for the explanation, that makes a lot of sense, I don't know why it didn't occur to me that mass can't be negative... but I typed those answers and I still get them both wrong... I don't know why...