Question

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

Answer 1

Hint: 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.