Exercise 15e Fitzpatrick (1 Viewer)

[acevipa]

I'm having a lot of trouble with exercise 15e from 2 unit Fitzpatrick. I can't seem to get any of the questions, and my exam is next week. Could someone help me with these questions.

 

[Drongoski]

I'm having a lot of trouble with exercise 15e from 2 unit Fitzpatrick. I can't seem to get any of the questions, and my exam is next week. Could someone help me with these questions.

I don't have a copy of the book handy. Post the question now.

 

[acevipa]

Well, to tell you the truth, I'm having trouble with the majority of the questions in exe 15e.

1. A man in a boat is 4km from the nearest pt. O of a straight beach; his destination is 4km along the beach from O. If he can row at 4km/h and walk at 5km/h, how should he proceed in order to reach his destination in the least possible time.

 

[Timothy.Siu]

firstly, u would write out as the relationships of the things in the question,

maybe area=basexlength if they wanted to find the max area

but u'd have to write other relationships and then isolate the variables into one equation, usually by substuting one for the other one

i'll do 15(e) 1.

Area=400cm
x is one side, let y be the other side
xy=400
y=400/x
perimeter=2x+2y
perimeter=2x+800/x (substituting)
then u find the derivative
P'=2-800/x^2
then to find max/min point (this one we're finding minimum)
we let P'=0 and solve
i.e. 2x-800/x=0
x^2-400=0
x=20 (since x>0)

 

[Timothy.Siu]

Well, to tell you the truth, I'm having trouble with the majority of the questions in exe 15e.

1. A man in a boat is 4km from the nearest pt. O of a straight beach; his destination is 4km along the beach from O. If he can row at 4km/h and walk at 5km/h, how should he proceed in order to reach his destination in the least possible time.

D=Speedxtime

Time=distance/Speed

let x be where he rows to
rowing distance=root(x^2+16)
walking distance=4-x
rowing time=root(x^2+16)/4
walking time=(4-x)/5
total time=root(x^2+16)/4 + (4-x)/5
T'=x/4root(x^2+16)-1/5
T'=0 when x/4root(x^2+16)=1/5
5x=4(x^2+16)
25x^2=16(x^2+16)
9x^2=256
x=16/3=5 1/3
but 0=<x<=4

therefore x=4...i.e. he rows to his destination only
lol, maybe i did something wrong in the working out, but u shud also check if its a max/min/watever

 

[Drongoski]

Well, to tell you the truth, I'm having trouble with the majority of the questions in exe 15e.

1. A man in a boat is 4km from the nearest pt. O of a straight beach; his destination is 4km along the beach from O. If he can row at 4km/h and walk at 5km/h, how should he proceed in order to reach his destination in the least possible time.

I recall attempting this question about 6 months ago. It is, to me, a non-trivial excercise. I remember writing out the solution and I'm not sure where I've left it.

I personally do not think you need worry about this particular question because it is not well-defined.

 

[Timothy.Siu]

I recall attempting this question about 6 months ago. It is, to me, a non-trivial excercise. I remember writing out the solution and I'm not sure where I've left it.

I personally do not think you need worry about this particular question because it is not well-defined.

what do u mean, it makes perfect sense.

 

[Drongoski]

what do u mean, it makes perfect sense.

In my old analysis I got x = 16/3 for T-min and I was troubled as to what this might mean. If he rows direct to destination, the time = sqrt(2) hrs. I made an error in my earlier analysis where i got the time when rowing directly = 1.4 hrs (by subst. x = 16/3 in the original equation for T(x) when it should have been 20//(3x4) hrs.= 1.6666 hrs.

So looks like you have sorted out an error which led to my getting conflicting results. Thank you Timothy.Siu.

I then analysed the case of the distance from O to be say 8 km and found that we end up with x=16/3 still. In fact with any distance greater than 16/3 km, T(min) corresponds to x = 16/3 km and that makes sense since he can walk faster than he can row.

Acevipa . . . my apologies.

 

[Timothy.Siu]

sorry i didn't mean it in a bad way

 

[JayoeZ]

I'm having problems with this exercise too.

Can anyone help me with the following questions.

11. The running cost (cost of fuel) for a certain ship is $3 per hour when the ship is not moving, and this cost increases by an amount that is proportional to the cube of its speed, Vkm/h. If the running cost per hour is % 6.75 when the speed is 15 km/h obtain a formula for the running cost per hour at speed V, and calculate the value of V for which the total running cost for a journey of 450km is a minimum

14. A printed page of a book is to have side margins of 1cm, a top margin of 2 cm and a bottom margin of 3cm. It is to contain 200cm^2 of printed matter. Find the dimensions of the page if the area of paper used is to be a minimum.

Solutions on the back says... 11. 11 14.

Thank you

 

[Timothy.Siu]

C=3+kv^3 (given) C=cost/hour v=speed per hour
when C=6.75, v=15
therefore k=1/900

we want to find the minimum cost for 450km
450km/v=time taken
C=450/v (3+v^3/900)
C=1350/v+v^2/2
C'=-1350/v^2+v
C'=0 when v=1350v^2
v^3=1350
v=11.05

 

[Timothy.Siu]

let LW=200
but there are margins, so for the area of the page including margins, it is

A=(W+2)(L+5) where w is the width and L is the length
L=200/W
A=(W+2)(200/w+5)
=210+5w+400/w
A'=5-400/w^2
A'=0 when 400/w^2=5
when w^2=80,
w=root80 since w>0
=4root5
then L=10root5

so the dimensions of the page is (4root5+2)x(10root5+5)

note: ur supposed to check if its max/min point and instead of letting LW=200 probably should have done (l-5)(w-2)=200 but i guess it doesn't really matter

 

[Drongoski]