# Homework Help: Work rate (word) problem

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1. Nov 22, 2017

### nightingale123

1. The problem statement, all variables and given/known data
A factory has gotten an order for $N$ products. For this job it has $n$ available machines.If all the machines would work simultaneously the order would have been completed in 24 hours. However due some complications there exist only two possible way to build the products:

$a)$ The machines begin working one after the other like this:

In the first hour only one machines works
In the second hour two machines work
3 hour tree machines are working
$\vdots$
n-th hour and all next hours n machines are working

$b)$ $n-5$ machines are working all the time

in case $b)$ the products are made 6 hours faster than in case $a)$ How many machines are in the factory

2. Relevant equations

3. The attempt at a solution
I was asked this question by some students I was helping out and I was really surprised when I was unable to answer it .
The way I tried to tackle this was the following:

First I tried to calculate the amount of work 1 machine does in 1 hour .
I know that $n$ machines complete $N$ in 24 hours therefore one machines does $\frac{N}{n}$ of work in 24 hours. therefore the amount of work a machine does in 1 hour should be $\frac{N}{24n}$( not sure if this is correct)

then I started with case $a)$
1 hour->1 machine->$\frac{N}{24n}$ of total work done
2 hour ->2 machines->$\frac{N}{24n}+\frac{2N}{24n}$ of total work done
3 hour -> 3 machines ->$\frac{N}{24n}+\frac{2N}{24n}+\frac{3N}{24n}$ of total work done
here is where I started noticing a recursion
n-th hour ->n machines ->$\sum\limits_{i=1}^{n} \frac{i \cdot N}{24n}=\frac{N}{24n}(1+2+3+4+\cdots+n)=\frac{N}{24n}\frac{n(n+1)}{2}=\frac{N(n+1)}{48}$
and every next hour we would add another $\frac{N}{24}$ of work

then I tried using case $b)$
we know that every hour we get $\frac{(n-5)\cdots N}{24n}$ work done
1 hour ->$\frac{(n-5)\cdot N}{24n}$ work done
2 hour -> $\frac{(n-5)\cdot N}{24n}+\frac{2(n-5)\cdot N}{24n}$
3 hour-> $\frac{(n-5)\cdot N}{24n}+\frac{2(n-5)\cdot N}{24n}+\frac{3(n-5)\cdot N}{24n}$
$\vdots$
t-th hour $\frac{(n-5)\cdot N}{24n}\sum\limits_{i=1}^{t} i=\frac{(n-5)\cdot N}{24n}\frac{t(t+1)}{2}$

however after (if it even is correct ) I'm not completely sure how to continue:
I know I still haven't used that case $b)$ takes 6 less hours to complete than $a)$ but i'm now exactly sure how to use that here. I know there is something obvious that i'm missing but I just cant find it

Any help is greatly appreciated

2. Nov 22, 2017

### StoneTemplePython

I'd assume time taken $t \geq n$ to avoid awkward corner cases. Key assumption is that output scales linearly with number of machines, from here we have the key unit to follow is "machine hours" which are the same on each side of each equation (i.e. your amount of stuff made N is some fixed number of machine hours, whether N machines for 1 hour or 1 machine for N hours or some mixture -- the product is the same).

This seems about right, but if you break it into smaller pieces it should be easier I'd think.

Your left hand side is a triangular number up to n, and then linear scaling of $(t-n) * n$ for residual time.

RHS = $24n$

You should be able to solve for $n$ in terms of some scalars and $t$.

Part 2:
Plug this into a second equation where $24_{hours} * n_{machines} = (n-5)t$ and solve for $t$. (Again follow the units...)

I believe you may have double roots, so you use the clue "the products are made 6 hours faster than in case a)" to throw one out.

- - - -
I haven't actually implemented this solution though, so I could be missing something.

3. Nov 22, 2017

### nightingale123

thanks I understand what you mean and how you got the second equation however could you please write out your first equation because for some reason I just don't get it

4. Nov 22, 2017

### StoneTemplePython

again assuming $t \geq n$,

$\big(\frac{n^2 + n}{2}\big) + x = 24n$

$x = (t-n)*n$
- - - -
to make sure I don't double count, consider a simple example, where n = 3 and t =4.

hour 1: 1
hour 2: 2
hour 3: 3
this is our triangular number and has value = 6.

Now $x = (4-3)*3 = 3$ i.e. you produced 3 items in hour 4.

The total amount of stuff produced in this example would be $6 + 3 = 9$.
- - --
Hopefully this doesn't go against forum rules as too explicit of a solution. I've seen you ask considerably more advanced questions elsewhere and take it on face value that this isn't your own homework here, and you're trying to help other students.

5. Nov 22, 2017

### Ray Vickson

As with "StoneTemplePython" I hope I am not violating PF policy by giving too much away (especially as you say you are a teacher rather than a student).

Taking the production rate to be 1 item per hour per machine, the total number of items is $N = 24 \, n$ when all $n$ machines are used constantly, and the production time is 24 (hours) in that case.

In the variable-machine portion (regime A), the first $T_{a} = n$ hours produce $N_a = \sum_{i=1}^n i$ items. That leaves $N_b = N - N_a$ items to be produced using all $n$ machines consntanly (regime B), and so taking $T_{b} = N_b/n$ hours. The total production time is $T_1 = T_a+T_b$.

The other possibility is to produce the $N = 24 \, n$ items at a steady rate of $n-5$ per hour, so getting the needed production time $T_2$ is easy enough. You are told that $T_1 = T_2 + 6$, so you have an equation for $n$.

6. Nov 23, 2017

### nightingale123

Thanks I finally think I got what you mean you two were great help.

But I would like to clarify something.

I'm not a teacher I'm a student myself (2. year uni ) however I am a tutor (we are are basicaly older students who help the first years or any other students with problems / definitions or anything they are interested in) and this was one of the Problems a student from the first year asked me to help him solve.

I'm sorry if I caused any confusion