Let's break down and solve the two parts of the problem.

Problem 3a:

The map of the park has dimensions of 4 inches by 6 inches, and the scale is $1 \text{ inch} = 30 \text{ miles}$.

1. Length of the park:

   • In the real world, the length is $4 \text{ inches} \times 30 \text{ miles/inch} = 120 \text{ miles}$.

2. Width of the park:

   • In the real world, the width is $6 \text{ inches} \times 30 \text{ miles/inch} = 180 \text{ miles}$.

Now, to find the actual area:

• The area is $\text{Length} \times \text{Width}$.
• $\text{Area} = 120 \text{ miles} \times 180 \text{ miles} = 21,600 \text{ square miles}$.

Problem 3b:

The map needs to fit in a brochure with an area of 6 square inches. We need to find a new scale.

• The current area of the map on paper is $4 \times 6 = 24 \text{ square inches}$.
• The scale factor is given by the ratio of the new area to the old area:
  $\text{Scale factor} = \sqrt{\frac{\text{New area}}{\text{Old area}}} = \sqrt{\frac{6}{24}} = \frac{1}{2}.$

Thus, the new scale should be half the current scale. Since the current scale is $1 \text{ inch} = 30 \text{ miles}$, the new scale will be: $1 \text{ inch} = 60 \text{ miles}.$

Let me know if you'd like further clarification or details!

Here are 5 related questions to deepen your understanding:

1. How would the area change if the scale were 1 inch = 15 miles?
2. If the brochure only allowed 3 square inches, what would the new scale be?
3. What if the map dimensions were 5 inches by 7 inches with the same scale of 1 inch = 30 miles?
4. How does changing the map's scale affect the perception of distance?
5. Can you calculate the perimeter of the park at the given scale?

Tip: Always double-check your units when working with different scales in maps to avoid confusion!