Question

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 $$ V(x)=k x(a-x), \quad 0 \leq x \leq a $$ where \(k\) is a positive constant, \(a\) is the initial amount of the compound, and \(x\) is the variable amount of the compound, for what value of \(x\) is the reaction rate a maximum?

Short answer

The reaction rate is maximum when \( x = \frac{a}{2} \).

Step-by-step solution

Write down the given function

The reaction rate is given by the function:\[ V(x) = k x (a - x) \]where \( k \) is a positive constant, \( a \) is the initial amount of the compound, and \( x \) is the variable amount of the compound.

Find the derivative of the function

To find the maximum reaction rate, we first find the derivative of \( V(x) \). Using the product rule, where \( u = kx \) and \( v = a - x \), the derivative is:\[ V'(x) = k(a - x) - kx = k(a - 2x) \]

Set the derivative equal to zero

To find the critical points, set the derivative equal to zero:\[ k(a - 2x) = 0 \]

Solve for \( x \)

Solve the equation for \( x \):\[ a - 2x = 0 \ 2x = a \ x = \frac{a}{2} \]

Determine whether the critical point is a maximum

To determine if \( x = \frac{a}{2} \) is a maximum, we check the second derivative:\[ V''(x) = -2k \]Since \( V''(x) = -2k < 0 \) for positive \( k \), the function has a maximum at \( x = \frac{a}{2} \).

Conclusion

The reaction rate reaches its maximum value when \( x = \frac{a}{2} \).

Key concepts

product rule

The product rule is a fundamental concept in calculus used when taking the derivative of a product of two functions. If you have two differentiable functions, say \( u(x) \) and \( v(x) \), their product's derivative is given by:
\[ (uv)' = u'v + uv' \]
In our exercise, the reaction rate function was \( V(x) = k x (a - x) \). Here, we can identify \( u(x) = k x \) and \( v(x) = a - x \). Using the product rule, we get:
\[ V'(x) = k(a - x) - kx \]
This helps us find where the reaction rate reaches its maximum by further analyzing the critical points through the derivative.

derivative

The derivative of a function represents its rate of change. In this exercise, the reaction rate function is differentiated to find the maximum value. Calculating the derivative of the reaction rate \( V(x) \):
\[ V(x) = k x (a - x) \]
yields:
\[ V'(x) = k(a - x) - kx = k(a - 2x) \]

The derivative tells us how the reaction rate changes with respect to the amount of compound \( x \). Setting the derivative equal to zero finds critical points which might be maximum, minimum, or inflection points.

second derivative test

The second derivative test is a way to determine if a critical point is a maximum or a minimum. After finding the first derivative and its critical points, take the second derivative:
\[ V''(x) = -2k \]
Notice that \( V''(x) = -2k \) is always negative for positive \( k \). According to the second derivative test:

• If \( V''(x) > 0 \), the function has a local minimum.


• If \( V''(x) < 0 \), the function has a local maximum.

Since \( V''(x) = -2k < 0 \), it indicates that the reaction rate has a maximum value at the discovered critical point.

maximum value

To find the maximum value of the function, we evaluate the critical point found through the first and second derivatives. For the reaction rate given by:
\[ V(x) = k x (a - x) \]
We solved for the critical point:
\[ x = \frac{a}{2} \]
By verifying with the second derivative test, we concluded that this value of \( x \) provides a local maximum for the reaction rate. Hence, the reaction rate is maximized when \( x = \frac{a}{2} \).

critical points

Critical points are where the first derivative of a function is zero or undefined. These points are potential candidates for local maxima or minima. In this exercise, we derived:
\[ k(a - 2x) = 0 \]
Solving this equation gives:
\[ 2x = a \]
\[ x = \frac{a}{2} \]
This is the critical point. To determine if it is a maximum, we used the second derivative test. Eventually, we identified that \( x = \frac{a}{2} \) as the reaction rate's maximum point.