Question

A calculus student plans to leave for school from his house at 8 am. His house and the school are located directly across from each other on the edge of a circular cornfield of diameter 2 km. Suppose the following:

• The student walks at a constant rate of 2 km/h when he is in the cornfield.
• The student walks at a constant rate of 4 km/h when he is not in the cornfield.
• The student has three options for his path to school.
1. Walk around the cornfield to school.
2. Walk directly to school through the cornfield.
3. Walk a straight path through the cornfield to a position on the cornfield’s boundary, and then walk the remaining distance to school around the cornfield
What path should the student take to maximize his walking time? If school starts at 9:15 am, will he be late?

Answer 1

to maximize his walking time; he should take path 3

He will reach his school by (8 hours + 1 hour 6 minutes) = 9 : 06 am ;

Thus , he will not be late by taking this path.

Explanation:

We have three case study for this problem.

1. Walk around the cornfield to school.
2. Walk directly to school through the cornfield.
3. Walk a straight path through the cornfield to a position on the cornfield’s boundary, and then walk the remaining distance to school around the cornfield

Let take a look at the first case study:

Walk around the cornfield to school.

Here, the distance traveled = circumference of a circle /2 (since he is walking in a circular path)

Distance traveled = πD/2

Given that the diameter is 2 km

Distance traveled = 2π/2

Distance traveled = π km = 3.14 km

Travel time = 3.14/4 hour = 0.785 hour

Travel time = (0.785 × 60) = 47.1 minutes

Second case study:

Walk directly to school through the cornfield.

Distance traveled D = 2 km

Travel time = 2/2 hour = 1 hour

Travel time = 1 × 60 = 60 minutes

Third case study:

Walk a straight path through the cornfield to a position on the cornfield’s boundary, and then walk the remaining distance to school around the cornfield

Here; we have a illustration of what represent a form of right angled triangle.

Using Pythagoras rule:

(the distance traveled)² = [ (D/2)² + (D/2)² ] + π D/4

(the distance traveled) =
`√(D^2/4 +D^2/4)+ \pi D /4`

(the distance traveled) =
`2/√(2) + \pi /2 \ km`

The travel time =
`(2/√(2) )/2 + (\pi /2 )/4`

The travel time =
`1/√(2) * \pi/8`

The travel time = 1.099 hours

The travel time = 1 hour 6 minutes

The travel time = 65.94 minutes

Therefore; to maximize his walking time; he should take path 3

He will reach his school by (8 hours + 1 hour 6 minutes) = 9 : 06 am ;

Thus , he will not be late by taking this path.