1. **State the problem:** We are given the total revenue function $R=400+25x-x^2$ and the total cost function $C=50+50x$. We need to find the minimum number of units $x$ to produce and sell to avoid loss, write the profit function, and find the profit when $x=20$. 2. **Formulas and rules:** - Profit function $P(x) = R(x) - C(x)$. - To avoid loss, profit must be non-negative: $P(x) \geq 0$. 3. **Find the profit function:** $$P(x) = (400 + 25x - x^2) - (50 + 50x) = 400 + 25x - x^2 - 50 - 50x = 350 - 25x - x^2$$ 4. **Find when profit is zero (break-even points):** Set $P(x) = 0$: $$350 - 25x - x^2 = 0$$ Rewrite as: $$-x^2 - 25x + 350 = 0$$ Multiply both sides by $-1$: $$x^2 + 25x - 350 = 0$$ 5. **Solve quadratic equation:** Use quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=25$, $c=-350$: $$x = \frac{-25 \pm \sqrt{25^2 - 4 \times 1 \times (-350)}}{2} = \frac{-25 \pm \sqrt{625 + 1400}}{2} = \frac{-25 \pm \sqrt{2025}}{2}$$ $$\sqrt{2025} = 45$$ So, $$x = \frac{-25 \pm 45}{2}$$ 6. **Calculate roots:** - $x = \frac{-25 + 45}{2} = \frac{20}{2} = 10$ - $x = \frac{-25 - 45}{2} = \frac{-70}{2} = -35$ (not valid since units cannot be negative) 7. **Interpretation:** The break-even point is at $x=10$. For profit to be non-negative, $x \geq 10$ units must be produced and sold. 8. **Calculate profit when $x=20$:** $$P(20) = 350 - 25 \times 20 - 20^2 = 350 - 500 - 400 = -550$$ So, the profit at 20 units is $-550$, which means a loss. **Final answers:** - Minimum units to avoid loss: $10$ - Profit function: $P(x) = 350 - 25x - x^2$ - Profit at $x=20$: $-550$ (loss)