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....
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~
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...
To determine the values of y for which the rate is nonnegative, we need to find the intervals where the expression (a-y)(b-y) is greater than or equal to 0.
A nonnegative number is any number that is greater than or equal to zero. Therefore, we need to find the intervals where the expression (a-y)(b-y) is greater than or equal to zero.
We can solve this by examining the factors (a-y) and (b-y) individually.
1. If (a-y) > 0 and (b-y) > 0, then the rate will be nonnegative. This occurs when y is less than a and y is less than b. The interval for this condition is (-infinity, min(a, b)).
2. If (a-y) < 0 and (b-y) < 0, then the rate will also be nonnegative. This occurs when y is greater than a and y is greater than b. The interval for this condition is (max(a, b), infinity).
Therefore, the combined interval for nonnegative values of y is (-infinity, min(a, b)) U (max(a, b), infinity).
For part B, to find the value of y at which the rate of the reaction is fastest, we need to maximize the expression (a-y)(b-y). We can find the maximum using calculus or completing the square.
Using calculus:
- Take the derivative of the expression with respect to y: (a-y)(b-y).
- Set the derivative equal to zero and solve for y.
- This will give us the value of y at which the rate is maximized.
Using completing the square:
- Expand the original expression to obtain y^2 - (a+b)y + ab.
- Complete the square by adding and subtracting (a+b)^2/4 within the square term: y^2 - (a+b)y + (a+b)^2/4 - (a+b)^2/4 + ab.
- Simplify the expression: (y - (a+b)/2)^2 - (a+b)^2/4 + ab.
- The value of y that maximizes the expression will be (a+b)/2.
In conclusion:
a) The nonnegative interval for y is (-infinity, min(a, b)) U (max(a, b), infinity).
b) The value of y that maximizes the reaction rate is (a+b)/2.