Question

A chemical reaction converts substance A to substance Y; the prensence of Y catalyzes the reaction, the quantity of A present is in grams. At time t seconds later, the quantity of Y present is y grams. The rate of the reaction, in grams/sec, is given by Rate=ky(a-y), k is a positive constant.

(a)For what values of y is the rate a maximum. A chemical reaction converts substance A to substance Y; the prensence of Y catalyzes the reaction, the quantity of A present is in grams. At time t seconds later, the quantity of Y present is y grams. The rate of the reaction, in grams/sec, is given by Rate=ky(a-y), k is a positive constant.

(a)For what values of y is the rate a maximum. @Mathematics

Answer 1

To find the maximum of the rate, take the derivative and set it to zero. So,
\[R(t) = ky(a-y)\]
\[R'(t) = k(a-y) - ky = 0 \rightarrow a-y = y \rightarrow y= \frac{a}{2} \]
In this context you can be sure it is a maximum, but just in case you can take the second derivative to get
\[R''(t) = -k-k = -2k <0\]
which means that the extremum at y=a/2 is indeed a maximum.