Quickest route Jane is 2 mi offshore in a boat and wishes to reach a coastal village 6 mi down a

Quickest route Jane is 2 mi offshore in a boat and wishes to...

Key Concepts

Optimization

Optimization involves finding the best solution among a set of feasible options, typically by minimizing or maximizing a particular function. This concept is central to many problems where resources such as time or cost need to be efficiently allocated or reduced.

Calculus-Based Optimization

Calculus-based optimization uses derivatives to identify the minimum or maximum values of a continuous function. By setting the derivative of a function to zero and solving for the critical points, one can determine the points at which the function is optimized.

Rate, Time, and Distance Relationships

This concept relates to understanding how time is affected by the speed at which different segments of a journey are completed. It involves setting up and analyzing equations where distance, speed, and time interact, often requiring adjustments when different segments are completed at different rates.

Trade-Off Analysis

Trade-off analysis is the process of balancing competing factors, such as different speeds or costs associated with various segments of a problem. It requires an examination of the benefits and drawbacks of different choices in order to determine the strategy that minimizes overall cost or time.

Transcript

00:01 Okay, so jane is, here's, all right, here's the shore, and the jane is out here in her boat.

00:14 Okay, and she is two miles offshore.

00:22 So this is shore.

00:25 All right, i'm going to try not to be too artistic.

00:29 Yeah, so this is two miles, and she wants to reach a point.

00:35 So here's this village right here, and the village is, six miles down the shore.

00:53 Okay, and she can row two miles per hour and can walk five miles per hour.

00:58 Where should she land her boat to reach the village in the least amount of time? okay, so basically she has kind of two extremes.

01:11 So she can go straight down here, row straight to the shore, and then walk, or she doesn't have to walk at all.

01:23 She can just row straight to the village.

01:32 Now presumably the best thing to do would be to row some of the way down and then walk the rest.

01:46 Okay, so here's kind of the two extremes.

01:49 You could go straight in and straight down.

01:52 You could just row all the way or you can do something in between.

01:57 Okay, so so how are you going to figure this out? all right, so we're trying to minimize the time.

02:04 So let's say that the distance she goes in the water is w, and the distance on the land is l.

02:19 And how long is it going to take her to get to the village? well, it's going to be w and then divided by her speed.

02:30 So the distance over speed will give time so w over two so this is time plus the distance she travels on land divided by how fast she can walk okay good and then where she lands let's call this okay so we'll try to put t in terms of x and then so she can land uh when x is zero so that would be going straight or she she can land when x is 6 that's like going straight to the village so x is going to be between 0 and 6 it's just where she lands um on the shore right here so what we want to do now is express w and l in terms of so w we have a little pythagorean theorem here so w squared is going to be 2 squared plus x squared and then l is going to be 6 minus x and so t as a function of x will be w over 2 so square root of 4 plus x squared over 2 plus l is 6 minus x over and so we need to consider what is t of 0 if we go down and then down that's going to be so just one plus six over five which is 2 .2 that'll be an hours and then what is t of six that's going to be let's see 36 plus four so square a 40 over 2 which is going to be square root of 10 which is greater than 2.

05:22 So so far, this will be longer, and this will be shorter.

05:31 Okay, but maybe the optimal is somewhere in between, at a critical point in between x equals 0 and x equals 6.

05:38 So let's take the derivative, and that's going to be, let's see, we'll have the 2x of x over 2, square root of 4 plus x squared, and then minus this will just be the derivative of this is just 1 5th.

06:00 And so when is t prime equal to zero? well, when 5x equals 2 square root of 4 plus x square...

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