Question

the rate of a chemical reaction varies, depending on the initial amount of a certain compound. in general, the reaction rate, v, can be expressed by the following formula, where k is a positive constant, and a is the initial amount of the compound.
v(x)=kx(a - x)
the positive constant, k, for one such reaction is 0.7, and the initial amount of the compound, a, is 14. for what value of x is the rate a maximum?
the rate is a maximum when x is

Explanation:

Step1: Expand the function

Given $V(x)=k x(a - x)=k(ax - x^{2})$. Substitute $k = 0.7$ and $a = 14$, we get $V(x)=0.7(14x - x^{2})=9.8x-0.7x^{2}$.

Step2: Find the derivative

The derivative of $V(x)$ with respect to $x$ is $V^\prime(x)=\frac{d}{dx}(9.8x - 0.7x^{2})$. Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $V^\prime(x)=9.8-1.4x$.

Step3: Set the derivative equal to zero

To find the critical points, set $V^\prime(x) = 0$. So, $9.8-1.4x = 0$.

Step4: Solve for x

Add $1.4x$ to both sides: $9.8=1.4x$. Then divide both sides by $1.4$, we get $x=\frac{9.8}{1.4}=7$.

Step5: Confirm it's a maximum

The second - derivative of $V(x)$ is $V^{\prime\prime}(x)=\frac{d}{dx}(9.8 - 1.4x)=-1.4<0$. Since $V^{\prime\prime}(x)<0$ when $x = 7$, the function $V(x)$ has a maximum at $x = 7$.

Answer:

7