A self-catalytic chemical reaction results in the formation of a product that causes its formation rate to increase. The reaction rate $V$ of one of such reactions is given by $V=1.2 x(14-x), 0 \leq x \leq 14$, where 14 is the initial amount of the chemical and $x$ is the variable amount of the chemical. For what value of $x$ is the reaction rate maximum?
(Use symbolic notation and fractions where needed.)
Answer
\(\boxed{x=7}\) is the value at which the reaction rate is maximum.
Step 1 ：Rewrite the function in standard form: \(V=16.8x - 1.2x^2\)
Step 2 ：Take the derivative of the function: \(V' = 16.8 - 2.4x\)
Step 3 ：Set the derivative equal to zero and solve for \(x\): \(16.8 - 2.4x = 0\) which gives \(x = 7\)
Step 4 ：Check the second derivative to confirm that this is a maximum: \(V'' = -2.4\). Since the second derivative is negative, \(x=7\) gives a maximum value for the function.
Step 5 ：\(\boxed{x=7}\) is the value at which the reaction rate is maximum.
