Chapter 3: Applications of Differentiation
Section 3.8: Optimization
Example 3.8.8
A hiker in a forest is at point A, five miles from B, the nearest point along a straight road bounding the forest. The hiker wishes to get to point C, some eight miles down the road from B. If the hiker can walk at 2 mph in the woods, and 4 mph along the road, where should the hiker emerge from the woods to complete the journey in least time?
Solution
Analysis
As depicted in Figure 3.8.8(a), let x be the distance from point B that the hiker emerges from the woods. Then, 8−x is the distance the hiker walks along the road, and x2+25 is the distance the hiker walks through the woods.
Since rate × time = distance, the total time for the walk from A to C is
Tx=x2+252+8−x4
Minimize Tx subject to the implied constraint x∈0,8.
p1:=plot([[0,8],[0,0],[5,0],[0,2.5]],style=line,color=black): p2:=plots:-textplot({[.4,1.25,typeset(x)],[1.1,5.75,typeset(8-x)],[2,.3,5],[3.5,2,typeset(sqrt(x^2+25))]},align=left,font=[default,12]): p3:=plots:-textplot({[5,.4,typeset(A)],[.5,.3,typeset(B)],[.5,8,typeset(C)]},font=[default,14],align=left): plots:-display(p1,p2,p3,axes=none,scaling=constrained);
Figure 3.8.8(a) Schematic of hiker's path
Graphical Solution
The total travel time Tx is graphed in Figure 3.8.8(b). There appears to be a minimum of approximately 4 between x=2 and x=3.
Of course, the graph can be probed to get a refined estimate of the minimum, but that is left to the reader.
That the "dip" in graph of Tx is so shallow is significant, suggesting that the travel time does not vary greatly with x. In fact, the difference in travel time between the optimal, and the "least effort" one with x=0, is small.
Figure 3.8.8(b) Graph of Tx
Numeric Solution
Control-drag the expression for the total time T.
Context Panel: Optimization≻Minimize (local)
x2+252+8−x4→minimize4.16506350946109638,x=2.88675134590030
A minimal travel time of 4.17 hours is possible if the road is reached some 2.89 miles from point B.
Analytic Solution
Define the objective function Tx
Control-drag Tx=… Context Panel: Assign Function
Tx=x2+252+8−x4→assign as functionT
Obtain the critical number
Write T′x=0 and press the Enter key.
Context Panel: Solve≻Solve
T′x=0
12xx2+25−14=0
→solve
x=533
Write the critical number.
Context Panel: Approximate≻5 (digits)
5/3→at 5 digits2.8869
Apply the Second-Derivative test
Context Panel: Simplify≻Simplify (The extremum is a relative minimum.)
T″5/3= simplify 3803
Obtain the relative minimum time
Evaluate T at the critical number. Press the Enter key.
Context Panel: Simplify≻Simplify
T5/3
161003+2−5123
= simplify
543+2
→at 5 digits
4.1651
Check the times at the endpoint values of x
Evaluate T at x=0. Context Panel: Evaluate and Display Inline
T0= simplify 92→at 5 digits4.5000
Evaluate T at x=8. Context Panel: Evaluate and Display Inline
T8= simplify 1289→at 5 digits4.7170
The relative minimum time is the absolute minimum time, T5/3=2+53/4≐4.1651 hrs.
The most toilsome path (x=8) takes some 4.72 hours; but the optimal path, just under 4.17. The difference is about 33 minutes. The least toilsome path (x=0) takes 4.5 hours, differing from the optimal path by just 20.1 minutes. The least toilsome path requires hiking through the woods for 5 miles; the optimal path, 10/3≐5.8 miles. Only the hiker knows if saving 20 minutes is worth hiking through an additional 0.8 miles of woods.
<< Previous Example Section 3.8 Next Example >>
© Maplesoft, a division of Waterloo Maple Inc., 2024. All rights reserved. This product is protected by copyright and distributed under licenses restricting its use, copying, distribution, and decompilation.
For more information on Maplesoft products and services, visit www.maplesoft.com
Download Help Document
