Find two non-negative numbers whose sum is 30 that make the maximum product of the square of one and the cube of the other.

Let's denote the two numbers as x and y . We need to find the maximum product of x^2 \times y^3 subject to the constraint that x + y = 30 . We can rewrite the constraint as y = 30 - x .

Now, we can express the product x^2 \times y^3 in terms of one variable:
P = x^2 \times (30 - x)^3

To find the maximum product, we can differentiate P with respect to x and set the derivative equal to 0:
\frac{dP}{dx} = 2x \times (30 - x)^3 - 3x^2 \times 3 \times (30 - x)^2 = 0
Simplify:
2x \times (30 - x)^3 - 9x^2 \times (30 - x)^2 = 0

Solving the above equation will give us the value of x , and then we can find the corresponding value of y using y = 30 - x .

Finally, we can calculate the maximum product:
\text{Product} = x^2 \times y^3

\textbf{Answer:} The two non-negative numbers that make the maximum product are x = 10 and y = 20 . The maximum product is 10^2 \times 20^3 = 8,000,000 .