1. **State the problem:** We want to maximize the monthly profit given a total advertising budget of 20000 dollars, where $x$ is the amount spent on newspapers and $y$ on television. 2. **Define the profit function:** Sales are given by $$S = 80x^{\frac{1}{4}}y^{\frac{3}{4}}$$ Profit is 10% of sales minus the advertising cost: $$P = 0.1S - (x + y) = 0.1 \times 80x^{\frac{1}{4}}y^{\frac{3}{4}} - (x + y) = 8x^{\frac{1}{4}}y^{\frac{3}{4}} - x - y$$ 3. **Constraint:** The total advertising budget is fixed: $$x + y = 20000$$ 4. **Set up the Lagrangian:** $$\mathcal{L} = 8x^{\frac{1}{4}}y^{\frac{3}{4}} - x - y + \lambda(20000 - x - y)$$ 5. **Find partial derivatives and set to zero:** $$\frac{\partial \mathcal{L}}{\partial x} = 8 \times \frac{1}{4} x^{-\frac{3}{4}} y^{\frac{3}{4}} - 1 - \lambda = 0$$ $$\frac{\partial \mathcal{L}}{\partial y} = 8 \times \frac{3}{4} x^{\frac{1}{4}} y^{-\frac{1}{4}} - 1 - \lambda = 0$$ $$\frac{\partial \mathcal{L}}{\partial \lambda} = 20000 - x - y = 0$$ 6. **Simplify the first two equations:** $$2 x^{-\frac{3}{4}} y^{\frac{3}{4}} - 1 - \lambda = 0$$ $$6 x^{\frac{1}{4}} y^{-\frac{1}{4}} - 1 - \lambda = 0$$ 7. **Set the two expressions for $1 + \lambda$ equal:** $$2 x^{-\frac{3}{4}} y^{\frac{3}{4}} = 6 x^{\frac{1}{4}} y^{-\frac{1}{4}}$$ 8. **Divide both sides by 2:** $$x^{-\frac{3}{4}} y^{\frac{3}{4}} = 3 x^{\frac{1}{4}} y^{-\frac{1}{4}}$$ 9. **Rewrite powers:** $$\frac{y^{\frac{3}{4}}}{x^{\frac{3}{4}}} = 3 \frac{x^{\frac{1}{4}}}{y^{\frac{1}{4}}}$$ 10. **Multiply both sides by $x^{\frac{3}{4}} y^{\frac{1}{4}}$ to clear denominators:** $$y^{\frac{3}{4}} y^{\frac{1}{4}} = 3 x^{\frac{1}{4}} x^{\frac{3}{4}}$$ 11. **Simplify exponents:** $$y^{1} = 3 x^{1}$$ $$y = 3x$$ 12. **Use the budget constraint:** $$x + y = 20000$$ Substitute $y = 3x$: $$x + 3x = 20000$$ $$4x = 20000$$ $$x = \frac{20000}{4} = 5000$$ 13. **Find $y$:** $$y = 3 \times 5000 = 15000$$ 14. **Calculate maximum profit:** $$P = 8 x^{\frac{1}{4}} y^{\frac{3}{4}} - x - y = 8 \times 5000^{\frac{1}{4}} \times 15000^{\frac{3}{4}} - 5000 - 15000$$ Calculate powers: $$5000^{\frac{1}{4}} = \sqrt{\sqrt{5000}} \approx 8.41$$ $$15000^{\frac{3}{4}} = (15000^{\frac{1}{4}})^3 \approx (11.14)^3 = 1383.5$$ So, $$P \approx 8 \times 8.41 \times 1383.5 - 20000 = 93144 - 20000 = 73144$$ **Final answer:** Allocate 5000 dollars to newspaper advertising and 15000 dollars to television advertising to maximize profit, which is approximately 73144 dollars.