please help!! this is due today! shortest distance problem..

[jackie.s]

A mathematician runner starts at point A and runs to the brick wall at some point C. The runner must then run to point B. The mathematician is lazy and wants to run the shortest distance. Find the shortest distance and the ratio of CD to CE.

see attachment for the diagram

 

[jackie.s]

this is what I have done so far, and im stuck. this could also be completely wrong! lol
Let x= EC
ED = 1200ft, so CD is (1200-x)

using Pythagorean theorem, I get
AC^2= AE^2 + EC^2
AC^2= 600^2 + x^2
AC= sq.rt. (360000+x)

BC^2= BD^2+ CD^2
BC^2= 300^2+ (1200-x)^2
BC^2=90000+1440000-2400x+x^2
BC= sq.rt. (x^2-2400x+1530000) ?reduce further?

total distance d=AC+BC.......but this gets messy with the square roots, and I am not sure what to do next. I feel like I should be graphing it somehow and looking for a minimum, but not sure what to graph...

 

[Quaid]

A mathematician runner starts at point A and runs to the brick wall at some point C. The runner must then run to point B. The mathematician is lazy and wants to run the shortest distance. Find the shortest distance and the ratio of CD to CE.

Hi Jackie:

Did you miss reading the forum guidelines? You have not posted anything, other than an exercise statement.

You already understand that you need to first write a distance function (i.e., sum of the two hypotenuses involved) and then calculate its derivative, yes? So, what symbols/expressions have you assigned for the unknowns?

We need to see the point in this exercise where you're stuck. If you're not able to do anything, then you need to explain what confuses you.


Also, your image is way blurry. (In the future, please preview your posts before submitting them, to ensure that everything's jake. You can enlarge and crop images before uploading them, yes?)

Is the distance AE given as 500 feet?

It looks like BD is 300 feet and ED is 1200 feet, but I'm not 100% on that.


Once we see what you're doing, we can begin to provide guidance.

Cheers :cool:

 

[Quaid]

We just cross-posted. I'm looking over your last post …

 

[jackie.s]

ok, lol yes I apologize I read the guidelines after I posted this. but I put my work thus far. and AE is 600 feet.

 

[Quaid]

this is what I have done so far

AC= sq.rt. (360000+x)

BC= sq.rt. (x^2-2400x+1530000) ?reduce further?

total distance d=AC+BC

Your work looks good, up to this point.

You do not need to simplify the radicals.

Let's call the distance function f.

f(x) = (x^2 + 360000)^(1/2) + (x^2 - 2400x + 1530000)^(1/2)

I feel like I should be graphing it somehow and looking for a minimum, but not sure what to graph

We're on the calculus board; did this exercise come from a calculus course?

I would expect that you're supposed to find some local minimum of f(x), using what you've learned about f'(x)

Did you receive specific instructions to solve by graphing?

 

[jackie.s]

this problem is from a pre-calc course. there were not specific instructions to use a graph...all of the info I was given is the problem and diagram. but he did say if we used a graph we didn't need to draw it out in our final answers. so we can solve it any way we choose, I am just stuck. I thought if I could figure out how to graph it, it would be the easiest way

 

[Quaid]

this problem is from a pre-calc course

[instructor] did say if we used a graph we didn't need to draw it out

I thought if I could figure out how to graph it, [that] would be the easiest way

We may graph by hand, after building a sufficient table of (x,y) values. (You already have some familiarity with this method, yes?)

We may use a graphing calculator, to plot the graph for us.

We may use one of many free, online graphers.


Let y = runner's distance

Let x = distance from point E to point C

y = (x^2 + 360000)^(1/2) + (x^2 - 2400x + 1530000)^(1/2)


Here's what comes from an online plot. (Click thumbnail, for larger image.)

Cheers :cool:

 

[jackie.s]

wow. thank you so much for your help. I was almost there...! but I get it now, and I can complete my other problems too....thanks again for your time! so appreciated!