Question

Find $x$ for which the total revenue function is maximum where $R = 2x^{3}-63x^{2}+648x+300$.

Answer 1

For any function $f(x)$, the point at which the function is maximum or minimum is given by the following steps:
Find the derivative of the function $f'(x)$.
Equate the derivative to 0 to determine the critical points. Suppose $a$ be the critical point of the function $f(x)$.
Check the nature of the derivative near the point $a$.
If $f'(x) < 0$ for $x < a$ and $f'(x) > 0$ for $x > a$, then $x = a$ is the point of minimum. If $f'(x) > 0$ for $x < a$ and $f'(x) < 0$ for $x > a$, then $x = a$ is the point of maximum.

Complete step-by-step answer:
Given the revenue function is $R = 2x^{3}-63x^{2}+648x+300$.
The maximum value of the function at a point can be determined by observing the nature of the derivative of the function near the point.
Differentiate the function with respect to $x$ as,
\begin{align*}\dfrac{dR}{dx} &= \dfrac{d}{dx}(2x^{3}-63x^{2}+648x+300)\\\ R' &= 6x^{2}-126x+648\end{align*}
Equate the above equation with 0 to determine the critical points, i.e., the points where the function may be maximum or minimum.
\begin{align*}6x^{2}-126x+648 &= 0\\\ x^{2}-21x+108 &= 0\\\ x^{2}-12x-9x+108 &= 0\\\ x(x-12)-9(x-12) &= 0\\\ (x-12)(x-9) &= 0\end{align*}
Thus the critical points are $x = 12,9$.
Now, check the nature of the derivative near these points. If the derivative changes from positive to negative at a point, then the point is the point of maximum and if the derivative changes from negative to positive at a point, then the point is the minimum point.
For $x > 9$, the derivative $R'$ is greater than 0, i.e. positive.
For $9 < x< 12$, the derivative $R'$ is less than 0, i.e. negative.
For $x > 12$, the derivative $R'$ is greater than 0, i.e. positive.
Since the derivative first increases and then decreases at $x = 9$, this point is the maximum point for the function.

Therefore, the value of $x$ for which the revenue function $R = 2x^{3}-63x^{2}+648x+300$ is maximum is 9.

Note: The point of maximum and minimum is also given using the second derivative test. If the second derivative is positive at a point, then the point is the point of relative minima and if the second derivative is negative at a point, then the point is the point of relative maxima for the function.