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Chapter 3: Q. 19 (page 166)

Chemical Reactions: A self-catalytic chemical reaction results in the formation of a compound that causes the formation ratio to increase. If the reaction rate $V$ is modeled by $V (x) = k x (a - x), 0 \leq x \leq a$
where $k$ is a positive constant, $a$ is the initial amount of the compound and $x$ is the variable of the compound for what value of $x$ is the reaction rate is maximum?

Short Answer

Expert verified

The value of $x$ is $\frac{a}{2}$ , the reaction rate is maximum.

Step by step solution

01

Step 1. Given Information

It is given that the reaction rate $V$ is modeled by $V (x) = k x (a - x), 0 \leq x \leq a$ , we need to determine the value of $x$ in which the reaction rate is maximum.

02

Step 2. Simplify

$V (x) = k x (a - x)$

$= - k x^{2} + k a x$

The vertex of the function is calculated by the formula,

$x = \frac{- k a}{- 2 \cdot (- k)}$

$x = \frac{a}{2}$

For value of $x = \frac{a}{2}$ the reaction rate is maximum.

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Most popular questions from this chapter

If $p (x) = - x + 6,$ then

(a) Determine the slope and y-intercept of each function.

(b) Use the slope and y-intercept to graph the linear function.

(c) Determine the average rate of change of each function.

(d) Determine whether the linear function is increasing, decreasing, or constant.

In Problems 7–22, solve each inequality.

$4 x^{2} + 9 < 6 x$

Find the average rate of change of $f (x) = 3 x^{2} - 2$ from 2 to 4.

Graph the function $f$ by starting with the graph of $y = x^{2}$ and using transformations (shifting, compressing, stretching, and/or reflection). Verify your results using a graphing utility.

Hint: If necessary, write $f$ in the form $f (x) = a (x - h)^{2} + k$ .

$f (x) = 3 x^{2} + 6 x$

If $f (x) = 2 x + 3$ , then

(a) Determine the slope and y-intercept of each function.

(b) Use the slope and y-intercept to graph the linear function.

(c) Determine the average rate of change of each function.

(d) Determine whether the linear function is increasing, decreasing, or constant

See all solutions

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