1. **State the problem:** A bakery sells 250 loaves of bread per day at price 20 each. For every 0.25 increase in price, 10 fewer loaves are sold. We want to find the price that maximizes revenue. 2. **Define variables:** Let $x$ be the number of 0.25 price increases. Then the new price per loaf is: $$p = 20 + 0.25x$$ The number of loaves sold is: $$q = 250 - 10x$$ 3. **Write the revenue function:** Revenue $R$ is price times quantity: $$R(x) = p \times q = (20 + 0.25x)(250 - 10x)$$ 4. **Expand the revenue function:** $$R(x) = 20 \times 250 - 20 \times 10x + 0.25x \times 250 - 0.25x \times 10x$$ $$R(x) = 5000 - 200x + 62.5x - 2.5x^2$$ $$R(x) = 5000 - 137.5x - 2.5x^2$$ 5. **Rewrite in standard quadratic form:** $$R(x) = -2.5x^2 - 137.5x + 5000$$ 6. **Find the vertex to maximize revenue:** For quadratic $ax^2 + bx + c$, vertex at $$x = -\frac{b}{2a} = -\frac{-137.5}{2 \times -2.5} = -\frac{-137.5}{-5} = -27.5$$ 7. **Interpretation:** Negative $x$ means price decreases, but problem context suggests $x \geq 0$. Check revenue at $x=0$ and $x=27.5$: Calculate revenue at $x=0$: $$R(0) = 5000$$ Calculate revenue at $x=27.5$: $$R(27.5) = -2.5(27.5)^2 - 137.5(27.5) + 5000$$ $$= -2.5 \times 756.25 - 3781.25 + 5000 = -1890.625 - 3781.25 + 5000 = -671.875$$ Revenue decreases at $x=27.5$, so maximum revenue is at $x=0$ (price 20). 8. **Check revenue at $x=1$ (price 20.25):** $$R(1) = -2.5(1)^2 - 137.5(1) + 5000 = -2.5 - 137.5 + 5000 = 4860$$ Revenue decreases as price increases. **Conclusion:** The bakery should charge the original price of 20 to maximize revenue. **Final answer:** $$\boxed{20}$$