A woman is standing at the edge of a slow-moving river which is one mile wide, and she wishes to return to her campground on the opposite side of the river. Assume that the woman can walk at 55 miles per hour and swim at 33 miles per hour, and that she will first swim to cross the river and then walk the remaining distance to the campground, which is 77 miles downstream from the point directly across the river from the woman's starting point. What route will take the least amount of time?

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Answer:

The route that take the least amount of time is the one that she swimms 1.25 miles across the river, to go out of the river on the other shore, 0.25 miles horizontally from the starting point, where she walks 76.25 miles to the campground.

x=0.25 (see figure attached).

Step-by-step explanation:

We have to minimize the time to make the route to the camp.

The time can be expressed as the sum of the time walking and the time swimming:

[tex]t=t_w+t_s=\dfrac{d_w}{v_w}+\dfrac{d_s}{v_s}=\dfrac{d_w}{55}+\dfrac{d_s}{33}[/tex]

The distance swimming can be expressed using the Pithagoras theorem, using an auxiliary variable x (see picture attached):

[tex]d_s^2=x^2+w^2=x^2+1\\\\d_s=\sqrt{x^2+1}[/tex]

Using the same auxiliary variable x, the distance walking can be expressed as:

[tex]d_w=L-x=77-x[/tex]

Then, we can minimize t in function of x by deriving t and equal to 0:

[tex]t=\dfrac{d_w}{55}+\dfrac{d_s}{33}=\dfrac{77-x}{55}+\dfrac{\sqrt{x^2+1}}{33}\\\\\\\dfrac{dt}{dx}=-\dfrac{1}{55}+\dfrac{1}{33}\dfrac{x}{\sqrt{x^2+1}}=0\\\\\\\dfrac{x}{\sqrt{x^2+1}}=\dfrac{33}{55}\\\\\\55x=33\sqrt{x^2+1}\\\\55^2x^2=33^2(x^2+1)=33^2x^2+33^2\\\\3025x^2=1089x^2+1089\\\\(3025-1089)x^2=1089\\\\1936x^2=1089\\\\x=\sqrt{\dfrac{1089}{1936}}=\sqrt{0.5625}=0.75[/tex]

The distance swimming is then:

[tex]d_s=\sqrt{x^2+1}=\sqrt{0.5625+1}=\sqrt{1.5625}=1.25[/tex]

The distance walking is:

[tex]d_w=77-0.75=76.25[/tex]

Then, the route that take the least amount of time is the one that she swimms 1.25 miles across the river, to go out of the river on the other shore, 0.25 miles horizontally from the starting point, where she walks 76.25 miles to the campground.

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