Critical Points — Question 2

Question 2 - Solution

We are given $$P(x) = -2x^3 + 15x^2 + 36x.$$

(a) Find $P'(x)$: $$P'(x) = -6x^2 + 30x + 36.$$

Set $P'(x)=0$ to find critical points: $$-6x^2 + 30x + 36 = 0 \Rightarrow x^2 - 5x - 6 = 0 \Rightarrow (x-6)(x+1)=0.$$

Critical points: $x=-1,\;6$

(b) First Derivative Test:

Evaluate the sign of $P'(x)$ on intervals determined by the critical points.

• On $(-\infty,-1)$, pick $x=-2$: $$P'(-2)=-6(4)+30(-2)+36=-24-60+36=-48<0.$$

• On $(-1,6)$, pick $x=0$: $$P'(0)=36>0.$$

• On $(6,\infty)$, pick $x=7$: $$P'(7)=-6(49)+30(7)+36=-294+210+36=-48<0.$$

Conclusion:

• At $x=-1$, $P'(x)$ changes from negative to positive, so $P$ has a local minimum.

• At $x=6$, $P'(x)$ changes from positive to negative, so $P$ has a local maximum.

(c) Maximizing profit:

Since $x=6$ is a local maximum, the profit is maximized at $x=6$ (hundreds of units), which is $$6 \times 100 = \boxed{600 \text{ units}}.$$

Compute the maximum profit: $$P(6)=-2(6)^3+15(6)^2+36(6)=-432+540+216=\boxed{324}.$$

This represents $\boxed{\$324{,}000}$ since $P$ is in thousands of dollars.

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