Slow going at the pretzel factory

Classical Mechanics Level 5

A chocolate covered, peanut butter filled, pretzel factory has three worker pods, and that each pod produces one important part of the snack: pod 1 prepares the chocolate, pod 2 the peanut butter, and pod 3 the pretzel shell.

The net production coming from each pod is balanced so that any given pod produces $\lambda = N_i s_i$ snacks worth of their part per unit time, where $N_i$ is the number of workers in pod $i$ , and $s_i(0)$ is the speed at which a worker in pod $i$ can produce things.

When everything is working smoothly the factory can output 100 chocolate covered, peanut butter filled pretzels per unit time. One night the workers in pod 1 have a party late into the night and mistakenly dump a thickening agent into the chocolate supply. This makes the chocolate harder to handle, so that the rate at which workers can prepare chocolate drops to $s_i(T) < s_i(0)$ . However, the speed of production in the other pods remains constant. The drop in $s_1$ causes the overall rate of pretzel production to fall to $\lambda_T = 60$ per unit time.

What is the new speed of chocolate production per worker, $s_1(T)$ (in snack equivalents per unit time)?

Assumptions and details

The $N_T = \sum_i N_i = 500$ factory workers always distribute themselves across the pods such that the rate of snack production is maximal, given their current working conditions.
When everything is working smoothly, a worker in pod 1 can produce chocolate at the rate $s_i(0) = 10$ snack equivalents per unit time.
The company has a fixed amount of money each day $M$ to pay workers, and each worker gets paid fixed daily wage $p$ .

The answer is 0.291262.

1 solution

Mikhael Glen Lataza
May 4, 2015

We know that $\lambda = N_{i} s_{i}$ .

So,

$N_{i} = \frac{\lambda}{s_{i}}$

$\sum_{i} N_{i} = 500$

$\sum_{i} \frac{\lambda}{s_{i}} = 500$

$\lambda = \frac{500}{\frac{1}{s_1} + \frac{1}{s_2} + \frac{1}{s_3}}$

Since, $\lambda = 100, s_{1} = 10$ , we get

$\frac{1}{s_{2}} + \frac{1}{s_{3}} = 4.9$

Since $\lambda_{T} = 60$ , and $s_{2}$ and $s_{3}$ remain constant.

$\lambda_{T} = \frac{500}{\frac{1}{s_{1}(T)} + \frac{1}{s_{2}} + \frac{1}{s_{3}}}$

$60 = \frac{500}{\frac{1}{s_{1}(T)} + 4.9}$

Therefore,

$s_{1}(T) = \frac{30}{103} \approx \boxed{0.29}$

This is a nice solution, and I think simpler than mine!

Josh Silverman Staff - 6 years, 1 month ago

In effect, pod 2 and pod 3 act like a single pod.

Joe Mansley - 1 year, 12 months ago

Slow going at the pretzel factory

The answer is 0.291262.

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