1. **Problem statement:** A boat tour company charges 11 for a harbour tour and averages 450 passengers on Saturdays. Every increase of 1 in the price decreases the number of customers by 25. We want to find: a) An equation to represent the daily revenue. b) The maximum daily revenue. 2. **Define variables:** Let $x$ be the number of 1-dollar price increases above 11. Then the price per ticket is $11 + x$. The number of passengers decreases by 25 for each increase, so the number of passengers is $450 - 25x$. 3. **Write the revenue function:** Revenue $R$ is price times number of passengers: $$R = (11 + x)(450 - 25x)$$ 4. **Expand the revenue function:** $$R = 11 \times 450 - 11 \times 25x + x \times 450 - 25x^2$$ $$R = 4950 - 275x + 450x - 25x^2$$ $$R = 4950 + 175x - 25x^2$$ 5. **Rewrite in standard quadratic form:** $$R = -25x^2 + 175x + 4950$$ 6. **Find the value of $x$ that maximizes revenue:** Since the quadratic coefficient is negative, the parabola opens downward and the vertex gives the maximum. The vertex $x$-coordinate is given by: $$x = -\frac{b}{2a} = -\frac{175}{2 \times (-25)} = -\frac{175}{-50} = 3.5$$ 7. **Calculate the maximum revenue:** Substitute $x=3.5$ into the revenue function: $$R = -25(3.5)^2 + 175(3.5) + 4950$$ $$R = -25 \times 12.25 + 612.5 + 4950$$ $$R = -306.25 + 612.5 + 4950 = 5256.25$$ 8. **Calculate the price that maximizes revenue:** $$\text{Price} = 11 + x = 11 + 3.5 = 14.5$$ **Final answers:** - a) The revenue function is $$R = (11 + x)(450 - 25x) = -25x^2 + 175x + 4950$$ where $x$ is the number of 1-dollar increases. - b) The maximum daily revenue is $$5256.25$$ when the price is $$14.5$$.