Let's break down the two parts of your question: determining the domain of the function $A(r)$ and solving for $r$ when the surface area is 125 square inches.

Part 1: Domain of $A(r)$

The function you provided is: $A(r) = 2\pi r^2 + 16\pi r$ This is the formula for the surface area of a cylinder, where:

• $r$ is the radius of the base.
• The term $2\pi r^2$ represents the area of the two circular bases.
• The term $16\pi r$ represents the lateral surface area, which depends on the height (given as 8 inches, hence the 16).

For the domain, we are interested in the values of $r$ for which $A(r)$ is defined.

• The radius $r$ must be a non-negative real number, since a negative radius does not make sense geometrically.
• Therefore, the domain is $r \geq 0$.

So, the domain of $A(r)$ is: $r \in [0, \infty)$

Part 2: Solving for $r$ when $A(r) = 125$

Now, we are given that the surface area is 125 square inches and need to find $r$.

We have the equation: $A(r) = 2\pi r^2 + 16\pi r = 125$ Dividing both sides by $\pi$ to simplify: $2r^2 + 16r = \frac{125}{\pi}$

Now, let's solve this quadratic equation for $r$. I will calculate the value for $r$.The solutions to the equation are approximately:

$r \approx -9.99 \quad \text{or} \quad r \approx 1.99$

Since the radius $r$ must be a non-negative value, we discard the negative solution. Thus, the radius is approximately:

$r \approx 1.99 \, \text{inches}$

Summary:

• The domain of $A(r)$ is $[0, \infty)$.
• The radius $r$ for a surface area of 125 square inches is approximately $1.99$ inches.

Do you want further details or have any questions?

Here are 5 relative questions to explore:

1. How is the surface area formula of a cylinder derived?
2. What happens to $A(r)$ as $r$ approaches infinity?
3. Can the formula $A(r)$ be rearranged to solve for the height $h$ instead of radius?
4. How would the surface area change if the height of the cylinder were different?
5. What is the significance of the different parts of the formula $A(r)$?

Tip: When solving quadratic equations, always consider the physical meaning of the variable to decide if both solutions are valid.