1. **State the problem:** We want to maximize the function $f(x,y) = x^2 y$ subject to the constraint $x^2 + 5xy = 108$ with $x > 0$. 2. **Use the constraint to express $y$ in terms of $x$:** From the constraint, $$x^2 + 5xy = 108 \implies 5xy = 108 - x^2 \implies y = \frac{108 - x^2}{5x}.$$ 3. **Substitute $y$ into $f(x,y)$:** $$f(x) = x^2 \cdot \frac{108 - x^2}{5x} = \frac{x(108 - x^2)}{5} = \frac{108x - x^3}{5}.$$ 4. **Find critical points by differentiating $f(x)$ with respect to $x$ and setting to zero:** $$f'(x) = \frac{108 - 3x^2}{5} = 0 \implies 108 - 3x^2 = 0 \implies 3x^2 = 108 \implies x^2 = 36 \implies x = 6 \text{ (since } x > 0).$$ 5. **Find corresponding $y$ value:** $$y = \frac{108 - 6^2}{5 \cdot 6} = \frac{108 - 36}{30} = \frac{72}{30} = \frac{12}{5}.$$ 6. **Calculate the maximum value:** $$f(6) = \frac{108 \cdot 6 - 6^3}{5} = \frac{648 - 216}{5} = \frac{432}{5} = 86.4.$$ **Final answer:** The maximum value is $\frac{432}{5}$. It occurs when $x = 6$ and $y = \frac{12}{5}$.