1. **Problem statement:** Maximize the function $$f(x,y) = x^{\frac{1}{2}} y^{\frac{1}{3}}$$ subject to the constraint $$3x + 4y \leq 25$$. 2. **Method:** We use the method of Lagrange multipliers for inequality constraints (Kuhn-Tucker conditions). Since the problem is to maximize with a linear constraint, the maximum will occur on the boundary $$3x + 4y = 25$$. 3. **Set up the Lagrangian:** $$\mathcal{L}(x,y,\lambda) = x^{\frac{1}{2}} y^{\frac{1}{3}} - \lambda (3x + 4y - 25)$$ 4. **Find partial derivatives and set to zero:** $$\frac{\partial \mathcal{L}}{\partial x} = \frac{1}{2} x^{-\frac{1}{2}} y^{\frac{1}{3}} - 3\lambda = 0$$ $$\frac{\partial \mathcal{L}}{\partial y} = \frac{1}{3} x^{\frac{1}{2}} y^{-\frac{2}{3}} - 4\lambda = 0$$ $$3x + 4y = 25$$ 5. **From the first two equations, express $$\lambda$$:** $$\lambda = \frac{1}{6} x^{-\frac{1}{2}} y^{\frac{1}{3}} = \frac{1}{12} x^{\frac{1}{2}} y^{-\frac{2}{3}}$$ 6. **Equate the two expressions for $$\lambda$$:** $$\frac{1}{6} x^{-\frac{1}{2}} y^{\frac{1}{3}} = \frac{1}{12} x^{\frac{1}{2}} y^{-\frac{2}{3}}$$ 7. **Simplify:** Multiply both sides by 12: $$2 x^{-\frac{1}{2}} y^{\frac{1}{3}} = x^{\frac{1}{2}} y^{-\frac{2}{3}}$$ Rewrite powers: $$2 \frac{y^{\frac{1}{3}}}{x^{\frac{1}{2}}} = x^{\frac{1}{2}} y^{-\frac{2}{3}}$$ Multiply both sides by $$x^{\frac{1}{2}} y^{\frac{2}{3}}$$: $$2 y^{\frac{1}{3} + \frac{2}{3}} = x^{\frac{1}{2} + \frac{1}{2}}$$ $$2 y = x$$ 8. **Substitute $$x = 2y$$ into the constraint:** $$3(2y) + 4y = 25 \implies 6y + 4y = 25 \implies 10y = 25 \implies y = 2.5$$ 9. **Find $$x$$:** $$x = 2y = 5$$ 10. **Calculate the maximum value:** $$f(5, 2.5) = 5^{\frac{1}{2}} \times 2.5^{\frac{1}{3}} = \sqrt{5} \times 2.5^{\frac{1}{3}}$$ Approximate: $$\sqrt{5} \approx 2.236, \quad 2.5^{\frac{1}{3}} \approx 1.357$$ $$f_{max} \approx 2.236 \times 1.357 = 3.034$$ **Final answer:** The maximum value of $$x^{\frac{1}{2}} y^{\frac{1}{3}}$$ under the constraint $$3x + 4y \leq 25$$ is approximately $$3.034$$ at $$x=5$$ and $$y=2.5$$.