1. **State the problem:** You sold 300 shirts at 20 each. For every 1 increase in price, sales drop by 5 shirts. Find the number of shirts to sell to maximize revenue and the maximum revenue. 2. **Define variables:** Let $x$ be the number of 1 increases in price. Price per shirt: $20 + x$ Number of shirts sold: $300 - 5x$ 3. **Write the revenue function:** Revenue $R(x) = (\text{price per shirt}) \times (\text{number of shirts sold})$ $$R(x) = (20 + x)(300 - 5x)$$ 4. **Expand the revenue function:** $$R(x) = 20 \times 300 - 20 \times 5x + x \times 300 - 5x^2$$ $$R(x) = 6000 - 100x + 300x - 5x^2$$ $$R(x) = 6000 + 200x - 5x^2$$ 5. **Rewrite in standard quadratic form:** $$R(x) = -5x^2 + 200x + 6000$$ 6. **Find the vertex to maximize revenue:** The vertex $x$-value is given by $$x = -\frac{b}{2a} = -\frac{200}{2 \times (-5)} = -\frac{200}{-10} = 20$$ 7. **Calculate the number of shirts sold at $x=20$:** $$300 - 5 \times 20 = 300 - 100 = 200$$ 8. **Calculate the maximum revenue:** $$R(20) = -5(20)^2 + 200(20) + 6000 = -5(400) + 4000 + 6000 = -2000 + 4000 + 6000 = 8000$$ **Final answer:** The number of shirts to sell to maximize revenue is **200**. The maximum revenue is **8000**.