Single variable optimization problem

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SUMMARY

The optimization problem involves Jane, who is 2 miles offshore and needs to reach a coastal village 6 miles down the shoreline. She rows at 5 mph and walks at 3 mph. The analysis reveals that the total time function T(x) is defined as T(x) = √(x² + 4)/5 + (6 - x)/3. Upon taking the derivative T'(x) and evaluating it, it is confirmed that the optimal landing point is at x = 6, as T(6) is less than T(0), indicating that rowing directly to the village minimizes travel time.

PREREQUISITES
  • Understanding of basic calculus, specifically derivatives and critical points.
  • Familiarity with the Pythagorean theorem for distance calculations.
  • Knowledge of optimization problems in mathematics.
  • Ability to analyze continuous functions and their behavior over intervals.
NEXT STEPS
  • Study optimization techniques in calculus, focusing on finding minima and maxima.
  • Learn about the application of the Pythagorean theorem in real-world problems.
  • Explore monotonic functions and their implications in optimization.
  • Investigate the use of endpoints in determining optimal solutions in constrained problems.
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Students studying calculus, particularly those focusing on optimization problems, as well as educators looking for practical examples of applying derivatives in real-life scenarios.

Daveyboy
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Homework Statement


Jane is 2 miles offshore in a boat and wishes to reach a coastal village 6 miles down a straight shoreline from the point nearest the boat. She can row her boat at 5 mph and can walk at 3 mph. Where should she land her boat to reach the village in the least amount of time.

Homework Equations


I don't know how to make a diagram so I'll try to describe it as carefully as possible.
There are two legs of a right triangle one 2 miles and the other 6 miles. (distance to shore and shore to village).
Let x be the distance from where the two legs (from above) meet to where the boat lands.
Then the distance the boat travels is \sqrt{x^{2}+4} by the Pythagorean theorem.
Then the distance walked is 6-x miles.
Now use time = distance/rate
also the times can be added together to find the total time it takes for the trip. I'll take the derivative and try to solve for 0 but the solution is not real.
I'm sure I have the strategy correct for this problem, and I'm very confident I have the derivative and algebra correct. I think I need to interpret my system differently though, because I feel like I am only off my a minus sign somewhere.

The Attempt at a Solution



T(x) = \frac{\sqrt{x^{2}+4}}{5} +\frac{6-x}{3}

T'(x) = \frac{x}{5(\sqrt{x^{2}+4}} - \frac{1}{3}

When I try to solve T'(x) = 0 I do not get real answers. This is a problem.
 
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If she can row faster than she can walk, wouldn't it make sense to go directly to the village from her current location (in which case x=6)?
 
Daveyboy said:

Homework Statement


Jane is 2 miles offshore in a boat and wishes to reach a coastal village 6 miles down a straight shoreline from the point nearest the boat. She can row her boat at 5 mph and can walk at 3 mph. Where should she land her boat to reach the village in the least amount of time.

Strange, are you certain that you read/copied the question correctly? Perhaps you reversed the rates for walking and rowing.
 
Yes, I copied the question down correctly. (I'm looking at the question right now and she can indeed row faster than she can walk.) Do you have any idea as to why my set up would give no real solutions? As Dango said x=6 should be a solution, so when I take the derivative I should get 6 as a critical point?
I now recall that one always has to check the endpoints of the interval which x belongs to. Namely, [0,6]. Upon evaluation I find that T(6)<T(0).
Since T is continuous and T'(x) is negative on [0,6]. T is monotonic decreasing there, so T(6) is the smallest time possible.

Is this a reasonable solution to the problem?
The teacher wrote this question on the fly I guess, so I'll attempt the problem with the rates changed.
Thanks for the responses guys.
 

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