Let's begin by interpreting and deriving the function [tex]\(r(s)\)[/tex] from the given problem statement. The given formula is:
[tex]\[ f = \frac{1}{2} \sqrt{\frac{T}{\pi}} \][/tex]
We need to model the function [tex]\(r(s)\)[/tex] where [tex]\(s\)[/tex] is the surface area of a spherical balloon. In this context, [tex]\(T = s\)[/tex], so the formula translates into:
[tex]\[ r(s) = \frac{1}{2} \sqrt{\frac{s}{\pi}} \][/tex]
This function [tex]\(r(s)\)[/tex] gives us the radius of the balloon as a function of its surface area [tex]\(s\)[/tex].
Next, we predict how the radius of the balloon changes as the balloon is inflated, given that the maximum surface area is 1,500 square centimeters.
1. For very small surface area [tex]\(s \approx 0\)[/tex]:
[tex]\[
r(0) = \frac{1}{2} \sqrt{\frac{0}{\pi}} = 0
\][/tex]
When the surface area is very small (close to zero), the radius is also close to zero.
2. For the maximum surface area of [tex]\(s = 1500\)[/tex] square centimeters:
[tex]\[
r(1500) = \frac{1}{2} \sqrt{\frac{1500}{\pi}} = \frac{1}{2} \sqrt{\frac{1500}{3.14159265359}} = 10.925484305920792
\][/tex]
So, when the surface area reaches its maximum of 1500 square centimeters, the radius is approximately 10.93 cm.
Now, let's summarize the behavior of the radius as the surface area increases:
- When the surface area is zero, the radius is zero.
- As the surface area increases, the radius increases accordingly.
- When the surface area is at its maximum (1500 square centimeters), the radius reaches approximately 10.93 cm.
- There is no indication that the radius will decrease as the surface area increases.
Given the above analysis, the correct option is:
[tex]\[ \boxed{\text{A. As the surface area of the balloon increases, the radius of the balloon increases until the maximum surface area is reached.}} \][/tex]
