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....

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 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~

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 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.

I apologize for any confusion. Let me explain the correct answers for both parts (a) and (b) in more detail.

(a) For what values of y is the rate nonnegative?

To determine the values of y for which the rate is nonnegative, we need to consider the factors (a - y) and (b - y) in the rate equation: Rate = k(a-y)(b-y).

Let's analyze the possible intervals for y:
1. If y < a, then (a - y) > 0, and (b - y) can be either positive or negative. In this case, the rate is nonnegative.
2. If a ≤ y ≤ b, then both (a - y) and (b - y) are negative, leading to a negative rate.
3. If y > b, then both (a - y) and (b - y) are negative. In this case, the rate is still nonnegative.

Combining these intervals, the values of y that make the rate nonnegative are [0, a] ∪ (b, +∞).

(b) Find the value of y at which the rate of the reaction is fastest.

To find the value of y at which the rate is fastest, we need to find the maximum value of the rate function.

Let's take the derivative of the rate function with respect to y:
d(Rate)/dy = k[2y - (a + b)].

Setting this derivative equal to zero and solving for y:
0 = 2y - (a + b)
2y = a + b
y = (a + b)/2.

So, the value y = (a + b)/2 is the critical point where the rate reaches its maximum value. However, note that this is the case only if y > b. If y ≤ b, the maximum rate occurs at y = b.

I hope this clarifies the correct answers for both parts (a) and (b) of the problem.