Question

The Purely Sweet Candy Company has just released its latest candy. It is a chocolate sphere filled with a candy surprise. If the sphere is 6 centimeters in diameter, what is the minimum amount of foil paper it will take to wrap the sphere?

Short answer

The minimum foil paper needed is \( 36\pi \) cm² or about 113.04 cm².

Step-by-step solution

Understanding the Problem

The problem asks for the surface area of a sphere, which represents the amount of foil paper needed to cover it. We are given the diameter of the sphere as 6 centimeters.

Calculate the Radius

The radius of a sphere is half the diameter. Hence, the radius \( r \) is calculated as follows: \( r = \frac{6}{2} = 3 \) centimeters.

Use the Surface Area Formula

The formula for the surface area of a sphere is \( 4\pi r^2 \). Substitute \( r = 3 \) into the formula to find the surface area.

Calculate the Surface Area

Substitute the calculated radius into the surface area formula: \[ 4\pi (3)^2 = 4\pi (9) = 36\pi \] square centimeters.

Final Answer

The minimum amount of foil paper required to wrap the sphere is \( 36\pi \) square centimeters. Optionally, approximate \( \pi \) as 3.14 for a numerical answer, which results in approximately \( 113.04 \) square centimeters.

Key concepts

Diameter and Radius

To begin with, understanding the terms "diameter" and "radius" is essential when working with spheres. The diameter is the total distance across a sphere, going through the center. It is like measuring a straight line spanning the widest part of the ball. In contrast, the radius is the distance from the center of the sphere to any point on its surface. We can consider it a single "spoke" from the center to the side of the sphere.

When you know the diameter, you can easily find the radius by dividing the diameter by two. This is because the radius is half of the sphere's diameter. For example, if a sphere has a diameter of 6 centimeters, then its radius would be:

• 
  \( r = \frac{6}{2} = 3 \) centimeters

Using the radius provides a simpler measure to work with, especially in formulas where only half of the full span is needed.

Surface Area Calculation

Calculating the surface area of a sphere is a necessary step to determine the amount of material needed to cover it, like foil for wrapping. The surface area formula for a sphere uses the radius and is written as:

• \( 4\pi r^2 \)

To find the surface area, plug in the radius value into this formula. Continuing with our example, if the radius is 3 centimeters, the calculation becomes:

• First, square the radius: \( r^2 = 3^2 = 9 \)
• Then multiply by 4 and \( \pi \): \( 4\pi r^2 = 4\pi \times 9 = 36\pi \) square centimeters

This multiplied result tells you how much foil or covering material is needed in terms of area. You have the option to approximate \( \pi \) as 3.14 if a numerical answer is desired, which gives you about 113.04 square centimeters as the surface area.

Mathematical Problem Solving

Solving mathematical problems, like calculating the surface area of a sphere, follows a logical process of understanding, applying formulas, and computing. Here's how you can tackle such problems efficiently:
1. **Understanding the Problem:** Start by identifying what is being asked. In our case, finding the amount of foil needed means calculating the sphere's surface area.

• Make note of the given values, such as the diameter or radius.

2. **Applying the Formula:** Use the appropriate mathematical formula. For spheres, the surface area formula \( 4\pi r^2 \) is essential.

• Ensure you've correctly identified and calculated the necessary measurements, like the radius.

3. **Performing Calculations:** Carefully perform each mathematical operation:

• Substitute values into the formula.
• Calculate using multiplication and exponentiation as required.

4. **Providing the Solution:** Present your final answer clearly, whether exact or approximated. Offering both options, as in leaving terms with \( \pi \) for exactness or using 3.14 for estimates, can be helpful.

These steps allow you to systematically approach and resolve mathematical exercises, ensuring clarity and accuracy.