Critical Points — Question 4

Question 4

Problem

A company models its revenue $R(x)$, in thousands of dollars, as a function of the number of items $x$ (in hundreds) it produces and sells: $$R(x) = -x^3 + 12x^2 + 9x.$$

(a) Find all critical points of $R(x)$.

(b) Use the First Derivative Test to classify each critical point as a local maximum, local minimum, or neither.

(c) How many items should be produced to maximize revenue?

Question 4 - Solution

We are given $$R(x) = -x^3 + 12x^2 + 9x.$$

(a) Find the derivative: $$R'(x) = -3x^2 + 24x + 9.$$

Set $R'(x)=0$: $$-3x^2 + 24x + 9 = 0 \Rightarrow x^2 - 8x - 3 = 0.$$

Using the quadratic formula: $$x = \frac{8 \pm \sqrt{64 + 12}}{2} = \frac{8 \pm \sqrt{76}}{2} = 4 \pm \sqrt{19}.$$

Critical points: $x = 4 - \sqrt{19},\; 4 + \sqrt{19}$

(b) First Derivative Test

Since $R'(x)$ is a downward-opening parabola, it is negative outside its roots and positive between them.

• At $x = 4 - \sqrt{19}$, $R'(x)$ changes from negative to positive, so $R(x)$ has a local minimum.

• At $x = 4 + \sqrt{19}$, $R'(x)$ changes from positive to negative, so $R(x)$ has a local maximum.

(c) Maximizing revenue

Revenue is maximized at $$x = 4 + \sqrt{19} \approx 8.36.$$

Since $x$ is measured in hundreds of items, this corresponds to $$\boxed{836 \text{ items}}.$$

Graph of $R(x)$ with critical points marked