1. **State the problem:** We want to find the maximum and minimum profit values for the profit function $$f(x) = 4x^5 - 20x^4 + 20x^3 + 20$$ where $x$ is the number of units sold (in hundreds). 2. **Find the first derivative $f'(x)$:** This derivative gives the rate of change of profit and helps locate critical points where maxima or minima may occur. $$f'(x) = \frac{d}{dx}(4x^5 - 20x^4 + 20x^3 + 20) = 20x^4 - 80x^3 + 60x^2$$ 3. **Find critical points by solving $f'(x) = 0$:** $$20x^4 - 80x^3 + 60x^2 = 0$$ Divide both sides by 20: $$x^4 - 4x^3 + 3x^2 = 0$$ Factor out $x^2$: $$x^2(x^2 - 4x + 3) = 0$$ Set each factor to zero: - $x^2 = 0 \implies x = 0$ - $x^2 - 4x + 3 = 0$ Solve quadratic: $$x = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2}$$ So, $$x = 1 \text{ or } x = 3$$ Critical points are at $x = 0, 1, 3$. 4. **Find the second derivative $f''(x)$:** This helps determine the nature of each critical point. $$f''(x) = \frac{d}{dx}(20x^4 - 80x^3 + 60x^2) = 80x^3 - 240x^2 + 120x$$ 5. **Evaluate $f''(x)$ at each critical point:** - At $x=0$: $$f''(0) = 0$$ (inconclusive, test further or use first derivative test) - At $x=1$: $$f''(1) = 80(1)^3 - 240(1)^2 + 120(1) = 80 - 240 + 120 = -40 < 0$$ Since $f''(1) < 0$, $x=1$ is a local maximum. - At $x=3$: $$f''(3) = 80(27) - 240(9) + 120(3) = 2160 - 2160 + 360 = 360 > 0$$ Since $f''(3) > 0$, $x=3$ is a local minimum. 6. **Check $x=0$ with first derivative test:** - For $x$ slightly less than 0, say $-0.1$, $f'(x)$ is positive (since $x^2$ dominates and is positive). - For $x$ slightly greater than 0, say $0.1$, $f'(x)$ is positive. No sign change, so $x=0$ is neither max nor min (possible inflection or flat point). 7. **Calculate profit values at critical points:** - $f(0) = 4(0)^5 - 20(0)^4 + 20(0)^3 + 20 = 20$ - $f(1) = 4(1) - 20(1) + 20(1) + 20 = 4 - 20 + 20 + 20 = 24$ - $f(3) = 4(243) - 20(81) + 20(27) + 20 = 972 - 1620 + 540 + 20 = -88$ **Final conclusion:** - Local maximum profit is $24$ thousand dollars at $x=1$ (100 units). - Local minimum profit is $-88$ thousand dollars at $x=3$ (300 units). - At $x=0$, profit is $20$ thousand dollars but not a max or min.