Minimal Path Problems

Lesson 10-4

Guidelines1) Read and understand the problem. 2) Draw and label a sketch.3) Decide which quantity is to be optimized and

express it as a function of one or more other variables.

4) Use information given to express the function in terms of just one variable.

5) Determine the domain and draw its graph.6) Find the global extrema of the function.7) Answer the question asked.

Example 1: Autumn is in her canoe, located 3 miles

from the nearest point on the shore of Lake Mead. She wants to get to her campsite, which is 5 miles from the nearest point, as quickly as possible because she is hungry. She can paddle her canoe 2 mph and walk 4 mph. How far from the nearest point should Autumn head in order to minimize her trip time?

Example 1: Step 2, 3 and 4

Distance = Rate • TimeT = Distance/Rate

Total time = paddling time + walking time

Domain is [ 0,5 ]

Shore

Lake

Autumn

3 mi

5 miles

Campx 5 – x

2 23 52 4 mph mph

x xt

Example 1: Step 5Domain is [0,5]

Windowx[-1, 8]

y[-1, 10]

2 23 52 4 mph mph

x xt

Example 1: Step 6

1

2 21 1 19 52 4 4

t x x

1

2 21 192 4x x

1

2 21 1 1' 9 22 2 4

t x x

1

2 21 10 92 4x x

1

2 21 1 94 2

x x

1

2 21 92

x x

Example 1: Step 6&7

2 3x

2 24 9x x

1

2 21 92

xx

23 9x

1

2 2

1 12

9x

x

1

2 22 9x x

3 1.732x

Autumn should head for 1.7 miles down the shore!!!

Example 1b:

1

2 21 19 12 4

t x x

How would the solution to the example be affected if the camp were only 1 mile from the nearest point on the shore?

For [0,1]

1

2 21 1 1' 9 22 2 4

t x x

Since the derivatives are the same the minimal path would be the same…1.7 miles.But since this is outside the domain the minimal occurs at the endpoint (x = 1).So it would be quicker to head straight for camp.

Example 2: A new subdivision is being built and pipes

for the water must be installed and connected to the city lines. The main water connector to the subdivision is to be in a desert lot 50 yards from the nearest street, and the main connector to the city water line is 400 yards down the street. Installing water lines across the lot costs $75/yd, while installing them along the street is $50/yd. How should the water line be laid out in order to minimize its total cost?

Example 2:

2 250y x

Cost = Distance • price per yd

Total cost = y(75) + (400 – x)50Since,

Street

lot

connector

50 yd

400 yds

Main connectorx 400 – x

2 275 50 (400 )50c x x

To minimize the cost we want to find where the derivative is equal to zero.

y

Example 2:

12 2 21' 75 50 2 50

2c x x

1

2 2 20 75 50 50x x

1

2 2 250 75 50x x

2 2

2 13 50x x

2 2

2 23 502x x

Example 2:

44.721x

22 29 50

4x x

2 29 4 25004 4x x

25 25004x

2 2000x

So, the water line should be about 44.7 yds from the foot of the perpendicular to minimize the cost.