First identify the known variables and assumptions.

\(\displaystyle \\n=\text{number of brushes per test batch} \\C=\text{testing cost for one group(cent)} \\A=\text{average testing cost (cents/brush)}\)

If 

\(\displaystyle n=1\rightarrow A=4\)

If a batch of brushes is tested and if the test shows that all the brushes are good then,

\(\displaystyle n>1\rightarrow C=3+n\)

If the batch test shows that there is a defected brush in the batch then,

\(\displaystyle A=\text{(Average value of C)}/n\)

Use a discrete probability model to find the most cost-effected quality control procedure for detecting defected brushes.

Consider the random variable

\(\displaystyle X\epsilon\begin{Bmatrix} x_1, x_2, x_3,... \end{Bmatrix}\)

that has a probability of

\(\displaystyle \text{Pr}\begin{Bmatrix} X=x_i \end{Bmatrix}=p_i\)

\(\displaystyle \\\Sigma p_i=1 \\EX=\Sigma x_ip_i\)

If the probability of a brush being good is \(\displaystyle p\) then the probability of a brush being defected is \(\displaystyle 1-p\). Then the average expected value of \(\displaystyle C\) is as follows:

\(\displaystyle EC=(3+n)p+[(3+n)+4n](1-p)\)

Now there are \(\displaystyle n\) brushes and the probability that one brush is defected is \(\displaystyle 0.002\) thus assuming independence, the probability of all \(\displaystyle n\) brushes in one test group are good is \(\displaystyle p=0.998^n\).

Therefore the expected value of the random variable \(\displaystyle C\) is,

\(\displaystyle \\EC=(3+n)0.998^n+[(3+n)+4n](1-0.998^n) \\EC=(3+n)+4n(1-0.998^n) \\EC=3+5n-4n(0.998^n)\)

Therefore the average testing cost is,

\(\displaystyle A=\frac{3}{n}+5-4(0.998)^n\)

Using the law of large numbers minimizing \(\displaystyle A\) results in

\(\displaystyle A=1.31, n=15\)

Now answer the question.

By testing brush batches in groups of 15 will reduce testing costs without sacrificing the quality. 

Identify what is known and the assumptions.

This is a diffusion problem and thus a diffusion equation will be used with a term of relative concentration. 

The relative concentration is denoted as 

\(\displaystyle \\C(x,t) \\x=\text{location of oil} \\t=\text{time}\)

This function has been normalized resulting in the following.

\(\displaystyle \int C(x,t)dx=1\)

The law of conservation of mass will also be useful is solving this problem.

\(\displaystyle \frac{\partial C}{\partial t}=-\frac{\partial q}{\partial x}\)

\(\displaystyle \\t=\text{Time since release of oil} \\\mu=\text{Distance travelled by the patches center} \\x=\text{Distance between patches center and town} \\s=\text{Patches spread at time t} \\\\ \mu=2t \\x=12-\mu \\P=15, t=1 \\s=1600, t=1\)

The goal is to calculate the maximum pollution level in town.

The diffusion equation is,

\(\displaystyle \frac{\partial C}{\partial t}=\frac{D}{2}\frac{\partial ^2C}{\partial x^2}\)

Using the Fourier transforms to solve the diffusion equation is as follows.

\(\displaystyle \hat C(k,t)=\int_{-\infty}^{\infty} e^{-ikx}C(x,t)dx\)

\(\displaystyle \frac{d \hat C}{dt}=\frac{D}{2}(ik)^2 \hat C=-\frac{D}{2} k^2\hat C\)

\(\displaystyle C(x,t)=\frac{1}{\sqrt{2\pi Dt}} e^{-\frac{x^2}{2Dt}}\)

For this particular function

\(\displaystyle \mu =2, \sigma=0.500\)

so the interval will be

\(\displaystyle \mu + 2\sigma\rightarrow s=4\sigma\)

\(\displaystyle D=\sigma^2=0.25\)\(\displaystyle C(x,t)=\frac{1}{\sqrt{0.5 \pi t}}e^{-\frac{x^2}{0.5t}}\)

\(\displaystyle P=15=P_0\frac{1}{\sqrt{0.5\pi }}e^{-\frac{0}{0.5}}\)

Calculate \(\displaystyle P\) given \(\displaystyle P_0\)

\(\displaystyle \\P=P_0C \\ P_0=15\sqrt{0.5\pi} \\\\P=\frac{15}{\sqrt{t}}e^{-\frac{(12-2t)^2}{(0.5t)}}\)

Now find where the maximum occurs in time.

\(\displaystyle P=6.12, t=6\)