Calculus

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(a-y)(b-y), 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....

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  1. 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 non-negative (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(a-y)(b-y) with respect to y, and equate Rate to zero since maximum rate (slope is zero):
    R = k(a-y)(b-y)
    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~

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  2. 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.

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  3. 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...

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