Question

The sum of one positive number and the square of another equals to 30. Find the numbers so that their product is as large as possible. A 2 decimal answer or exact answer is ok. The maximal value of the product is:

Answer 1

To solve this problem, we can use the technique of completing the square. Let x be the smaller of the two numbers, and y be the larger. Then we know that:

x + y^2 = 30

We want to maximize the product xy. To do this, we can rewrite the equation above as:

y^2 = 30 - x

Now we can use the fact that the square of a number is always non-negative to our advantage. We know that y^2 is always non-negative, so this means that 30 - x must also be non-negative. Therefore, we have:

x <= 30

Now we can maximize the product of xy by finding the value of x that makes y as large as possible. We can do this by completing the square:

y^2 = 30 - x
y^2 = -(x - 30)

This is a downward-facing parabola, so its maximum occurs at its vertex. The vertex occurs at x = 30, so:

y^2 = -(30 - 30) = 0
y = 0

Therefore, the maximum product occurs when x = 15 and y = 0 or x = 0 and y = 15. In either case, the product xy is 0.