Optimization - Running/Rowing

[reardear]

Malcolm is 6km offshore on a boat at point B and wants to reach a coastal village 4km down the straight shoreline at point D from point A on the shore nearest to the boat. He can row at 5km/hr and run 7km/hr. Point C is where he should land his boat to reach the village in the least amount of time. Let \(\displaystyle x\) be the distance between A and C.

From my diagram, I can see that:
AD is \(\displaystyle x + (4-x)\)
AC is \(\displaystyle x\)
BC is \(\displaystyle \sqrt{36 + x^2}\) using Pythagorean theorem

But I don't know how to incorporate the running/rowing rates.

a) What two assumptions should be made about the wind and water?
Perhaps one would be that neither are moving?

b) State the following:

• running distance = \(\displaystyle 4-x\)
• rowing distance = \(\displaystyle \sqrt{x^2+36}\)
• running time = \(\displaystyle \large \frac{4-x}{7}\)
• rowing time = \(\displaystyle \large \frac{\sqrt{x^2+36}}{5}\)

c) How far away is point C from point A to 2 decimal places?

 

[MarkFL]

a) I would say your assumptions are valid, that there is no wind or water current.

b) From your diagram, what is the running distance? You have already stated the rowing distance.

To find the times of these distances, use the relationship between distance, constant velocity and time.

c) State the total time as a function of x, then use the technique of optimization you have been taught.

Be sure to demonstrate that you have indeed minimized the time function.

 

[reardear]

a) I would say your assumptions are valid, that there is no wind or water current.

b) From your diagram, what is the running distance? You have already stated the rowing distance.

To find the times of these distances, use the relationship between distance, constant velocity and time.

c) State the total time as a function of x, then use the technique of optimization you have been taught.

Be sure to demonstrate that you have indeed minimized the time function.

Ok, I got..

• running distance = \(\displaystyle 4-x\)
• running time = \(\displaystyle \large \frac{4-x}{7}\)
• rowing distance = \(\displaystyle \sqrt{x^2+36}\)
• rowing time = \(\displaystyle \large \frac{\sqrt{x^2+36}}{5}\)
• total time = \(\displaystyle \large \frac{\sqrt{x^2+36}}{5} + \frac{4-x}{7}\) on domain \(\displaystyle D = [0,4]\)

• \(\displaystyle T\prime(x) = \Large \frac{7x-5\sqrt{x^2+36}}{35\sqrt{x^2+36}}\)
• \(\displaystyle T\prime(x) = 0 \:at\: \large x = \frac{5\sqrt{x^2+36}}{7}\)

From there I'm missing a page of notes so I'm not sure what to do beyond that :S

 

[MarkFL]

You have correctly differentiated (except for a minor typo where you write 35 instead of 36 under the radical), so you want to equate the numerator of the derivative to zero, and solve for x:

\(\displaystyle 7x-5\sqrt{x^2+36}=0\)

\(\displaystyle 7x=5\sqrt{x^2+36}\)

What do you think you need to do next?

 

[reardear]

You have correctly differentiated (except for a minor typo where you write 35 instead of 36 under the radical), so you want to equate the numerator of the derivative to zero, and solve for x:

\(\displaystyle 7x-5\sqrt{x^2+36}=0\)

\(\displaystyle 7x=5\sqrt{x^2+36}\)

What do you think you need to do next?

Thanks for pointing that out. I found x (I forgot to paste it in my post); but my notes end where I find the derivative and that's where I'm stuck now. After the missing page, there are several endpoints to find where the absolute minimum is. I only found one critical point though. Oh. I have the domain [0,4].. nevermind

\(\displaystyle T(0) = \frac{62}{35}\)
\(\displaystyle T(...\) am I supposed to use that critical point?...
\(\displaystyle T(4) = \frac{7\sqrt{40}}{35} = \frac{\sqrt{40}}{5}\)

 

[MarkFL]

You want to solve:

\(\displaystyle 7x=5\sqrt{x^2+36}\)

Now, since we have a radical on one side, we should square both sides to get (and keep in mind we want \(\displaystyle 0\le x\le4\)):

\(\displaystyle 49x^2=25(x^2+36)\)

\(\displaystyle 24x^2=900\)

\(\displaystyle x^2=\dfrac{75}{2}\)

Now, take the positive root to get your critical value. You can easily confirm using the first-derivative test for relative extrema that this is at a relative minima.

What must we do though if the critical value is outside of the restricted domain?

 

[reardear]

You want to solve:

\(\displaystyle 7x=5\sqrt{x^2+36}\)

Now, since we have a radical on one side, we should square both sides to get (and keep in mind we want \(\displaystyle 0\le x\le4\)):

\(\displaystyle 49x^2=25(x^2+36)\)

\(\displaystyle 24x^2=900\)

\(\displaystyle x^2=\dfrac{75}{2}\)

Now, take the positive root to get your critical value. You can easily confirm using the first-derivative test for relative extrema that this is at a relative minima.

What must we do though if the critical value is outside of the restricted domain?

Ignore it? \(\displaystyle 5\sqrt{\frac{3}{2}}\) is outside of it (6.1237). Do I need to use limits?

 

[MarkFL]

Recall the method you used probably a few lessons back to find absolute extrema on a given interval.

The critical value is outside the interval because there isn't a great enough difference between the running and rowing velocities.

As a follow-up, can you give the range of running velocities (assuming the same rowing velocity as given in the problem) that would put the critical number within the given interval?

 

[reardear]

Recall the method you used probably a few lessons back to find absolute extrema on a given interval.

The critical value is outside the interval because there isn't a great enough difference between the running and rowing velocities.

As a follow-up, can you give the range of running velocities (assuming the same rowing velocity as given in the problem) that would put the critical number within the given interval?

\(\displaystyle T(0) = \frac{62}{35} = 1.77\)<-- absolute maximum
\(\displaystyle T(5\sqrt{\frac{3}{2}}) = \Large \frac{\sqrt{(5\sqrt{\frac{3}{2}})^2+36}}{5} + \frac{4-5\sqrt{\frac{3}{2}}}{7} = \frac{20 - 20\sqrt{\frac{3}{2}}\cdot7\sqrt{73.5}}{35} = 1.41\)

\(\displaystyle T(4) = \frac{7\sqrt{40}}{35} = \frac{\sqrt{40}}{5} = 1.26\) <-- absolute minimum

 

[MarkFL]

You should find (in hours):

\(\displaystyle T(0)=\dfrac{6}{5}+\dfrac{4}{7}=\dfrac{62}{35} \approx1.77\)

\(\displaystyle T(4)=\dfrac{\sqrt{4^2+36}}{5}=\dfrac{2\sqrt{13}}{5}\approx1.44\)

So, we find that the boat should be landed at the village, i.e., Malcolm should row the entire distance.

Only if Malcolm's running speed is greater than \(\displaystyle \dfrac{5\sqrt{13}}{2}\,\dfrac{\text{km}}{\text{hr}}\) would he have to run part way to minimize the time.

 

[reardear]

You should find (in hours):

\(\displaystyle T(0)=\dfrac{6}{5}+\dfrac{4}{7}=\dfrac{62}{35} \approx1.77\)

\(\displaystyle T(4)=\dfrac{\sqrt{4^2+36}}{5}=\dfrac{2\sqrt{13}}{5}\approx1.44\)

So, we find that the boat should be landed at the village, i.e., Malcolm should row the entire distance.

Only if Malcolm's running speed is greater than \(\displaystyle \dfrac{5\sqrt{13}}{2}\,\dfrac{\text{km}}{\text{hr}}\) would he have to run part way to minimize the time.

How do we know he should row the entire way? What happened with the abs. min and max?

 

[MarkFL]

Since the function we found tor the total time only makes sense for \(\displaystyle 0\le x\le 4\) and we found the critical point is outside that restricted domain we then look at the end-points of the domain to see which gives the smaller value, and this will be our absolute minimum. Because we found there are no stationary points within the interval, we know the function is monotonic there, and so the absolute extrema will occur at the end-points.

So, we find the time to get to the village is minimized when Malcolm rows straight to the village. Malcolm doesn't run fast enough to make up for the increase in total distance that occurs if he lands to the west of the village.

 

[reardear]

Since the function we found tor the total time only makes sense for \(\displaystyle 0\le x\le 4\) and we found the critical point is outside that restricted domain we then look at the end-points of the domain to see which gives the smaller value, and this will be our absolute minimum. Because we found there are no stationary points within the interval, we know the function is monotonic there, and so the absolute extrema will occur at the end-points.

So, we find the time to get to the village is minimized when Malcolm rows straight to the village. Malcolm doesn't run fast enough to make up for the increase in total distance that occurs if he lands to the west of the village.

Wouldn't T(4) be the minimum we use (1.44) then? Also I completely forgot that the question asks for distance..

Edit: never mind I got it. Thank you for your help, Mark!