1. **State the problem:** We need to find two numbers $x$ and $y$ such that their sum is 15, i.e., $x + y = 15$, and the product of the square of one number and the cube of the other, $x^2 y^3$, is maximized. 2. **Express one variable in terms of the other:** From $x + y = 15$, we get $y = 15 - x$. 3. **Write the function to maximize:** The function is $$f(x) = x^2 (15 - x)^3.$$ We want to find $x$ that maximizes $f(x)$ for $0 \\leq x \\leq 15$. 4. **Differentiate $f(x)$:** Use the product and chain rules: $$f(x) = x^2 (15 - x)^3$$ $$f'(x) = 2x (15 - x)^3 + x^2 imes 3(15 - x)^2 (-1)$$ Simplify: $$f'(x) = 2x (15 - x)^3 - 3x^2 (15 - x)^2$$ 5. **Factor $f'(x)$:** $$f'(x) = x (15 - x)^2 [2(15 - x) - 3x] = x (15 - x)^2 (30 - 2x - 3x) = x (15 - x)^2 (30 - 5x)$$ 6. **Find critical points:** Set $f'(x) = 0$: $$x = 0, \quad 15 - x = 0 \Rightarrow x = 15, \quad 30 - 5x = 0 \Rightarrow x = 6$$ 7. **Evaluate $f(x)$ at critical points and endpoints:** - At $x=0$: $$f(0) = 0^2 \times 15^3 = 0$$ - At $x=6$: $$f(6) = 6^2 \times (15 - 6)^3 = 36 \times 9^3 = 36 \times 729 = 26244$$ - At $x=15$: $$f(15) = 15^2 \times 0^3 = 0$$ 8. **Conclusion:** The maximum value of $f(x)$ is $26244$ at $x=6$, $y=15-6=9$. **Final answer:** The two numbers are $6$ and $9$, and the maximum value of $x^2 y^3$ is $26244$.