A man wishes to walk home

Question

A man wishes to walk home, which is 10 km away down a main road and then a further 4 km down a smaller road perpendicular to the first. The man can walk at 5 km/hr along the road and can walk across the fields surrounding the area at 3 km/hr. Suppose the man was to walk along the main road for some time and then cut diagonally across the fields straight to his house (that is, without ever going on to the smaller road).

 

1.The minimum time that it takes for the man to get home by taking the main road and cutting across the field is __hr

2.The total time that it would take for the man to walk along the main road and the smaller road (without cutting across the field) is __hr

3.True or False:

The fastest route for the man to get home involves taking the main road followed by the smaller road (without cutting across the field).

Answer 1

Suppose the man walks x km by taking the main road. The time of this walking is x/5.

Then he cut diagonally across the field. By the Pythagorean theorem the distance is

sqrt((10-x)² + 4²) = sqrt(x² - 20x + 116).

The time of walking across the field is (sqrt(x² - 20x + 116)) / 3.

So the total time as function of x is

T(x) = x/5 + (sqrt(x² - 20x + 116))/3.

Find the minimum of this function.

The first derivative

T '(x) = 1/5 + (1/3) (1/2) (1/(sqrt( x² - 20x + 116)) (x² - 20x + 116)' =

= 1/5 + (1/6) ((1/(sqrt( x² - 20x + 116)) (2x - 20) = 1/5 + 1/3 (x-10) / sqrt(x² - 20x + 116)

Find for what x T '(x) = 0.

1/5 + 1/3 (x-10) / sqrt( x² - 20x + 116) = 0

1/3 (x-10) / sqrt(x² - 20x + 116) = -1/5

5(x-10) = -3 sqrt(x² - 20x + 116)

25 (x-10)² = 9 (x² - 20x + 116)

25x² - 500x + 2500 = 9x² - 180x + 1044

16x² - 320x + 1456 = 0

x² - 20x + 91 = 0

(x-7)(x-13) = 0

Solutions: x =7 and x =13.

Exclude x = 13, because the man walks no more than 10km taking the main road (Actually, this is extraneous solution and T '(13) ≠ 0)

So, x = 7. Is it minimum or maximum of T(x)?

Easy to check that:

(a) T '(0) = 1/5 - (1/3) 10 / sqrt(116) ≈ -0.11 < 0

(b) T '(10) = 1/5 > 0

So, T '(x) < 0 on (0,7) and T '(x) > 0 on (7,10).

T(x) has a minimum at x=7.

Find the minimum value: T(7) = 7/5 + sqrt((10-7)² + 4²) / 3 = 7/5 + 5/3 = 46/15 = 3 h 4 min

Now, what is the total time for the man to walk along the main road and the smaller road (without cutting across the field) ?

T = 10/5 + 4/5 = 2.8 = 2 h 48 min

3 is true: The fastest route for the man to get home involves taking the main road followed by the smaller road (without cutting across the field).