1. **State the problem:** Find the value of $x$ that maximizes the daily profit for a firm with total cost function $$C(x) = x^3 - 6x^2 + 13x + 15$$ and total revenue function $$R(x) = 28x$$. 2. **Define profit function:** Profit $$P(x)$$ is revenue minus cost: $$P(x) = R(x) - C(x) = 28x - (x^3 - 6x^2 + 13x + 15)$$ 3. **Simplify profit function:** $$P(x) = 28x - x^3 + 6x^2 - 13x - 15 = -x^3 + 6x^2 + 15x - 15$$ 4. **Find critical points:** To maximize profit, find where derivative $$P'(x) = 0$$. $$P'(x) = \frac{d}{dx}(-x^3 + 6x^2 + 15x - 15) = -3x^2 + 12x + 15$$ 5. **Solve for critical points:** Set derivative to zero: $$-3x^2 + 12x + 15 = 0$$ Divide both sides by \cancel{-3}: $$\cancel{-3}x^2 + \cancel{-4}x + \cancel{-5} = 0 \implies x^2 - 4x - 5 = 0$$ 6. **Factor quadratic:** $$x^2 - 4x - 5 = (x - 5)(x + 1) = 0$$ So, $$x = 5 \text{ or } x = -1$$ 7. **Determine maximum:** Check second derivative: $$P''(x) = \frac{d}{dx}P'(x) = \frac{d}{dx}(-3x^2 + 12x + 15) = -6x + 12$$ Evaluate at $x=5$: $$P''(5) = -6(5) + 12 = -30 + 12 = -18 < 0$$ (concave down, local max) Evaluate at $x=-1$: $$P''(-1) = -6(-1) + 12 = 6 + 12 = 18 > 0$$ (concave up, local min) 8. **Conclusion:** The profit is maximized at $$x = 5$$. **Final answer:** C. 5