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Flashcards: Math Modeling

An oil spill occurs at a factory 12 kilometers up stream from a town. One hour after the spill happens it reaches the stream and a 1600 meters long patch of oil begins to flow towards the town at a rate of 2 kilometers per hour. The maximum concentration of the oil in the water is 15 times the acceptable level. What is the maximum concentration expected in town and when will it arrive?

P=6.15, t=6.7

P=6.10, t=5.4

P=6.12, t=6

P=6.12, t=5

P=6.14, t=6.3

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Flashcards: Math Modeling

Your answer:

P=6.15, t=6.7

Your answer:

P=6.10, t=5.4

Your answer:

P=6.12, t=5

Your answer:

P=6.14, t=6.3

Correct answer:

P=6.12, t=6

Explanation:

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

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

This function has been normalized resulting in the following.

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

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

$\frac{\partial C}{\partial t}=-\frac{\partial q}{\partial x}$

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

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

The diffusion equation is,

$\frac{\partial C}{\partial t}=\frac{D}{2}\frac{\partial ^2C}{\partial x^2}$

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

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

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

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

For this particular function

$\mu =2, \sigma=0.500$

so the interval will be

$\mu + 2\sigma\rightarrow s=4\sigma$

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

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

Calculate given P_0

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

Now find where the maximum occurs in time.

P=6.12, t=6

Flashcards: Math Modeling

A farm has 750 acres of land available for planting crops. The possible crops are soy, corn, and oats. There is 1,200 acre-ft of water for irrigation and the crew will be able to work of the farm 575 hours per week.

Find the amount of each crop that should be planted to maximize the profit.

The table below includes some additional, helpful information.

$\begin{tabular}{|c|c|c|c|} \hline requirements (per acre) &soy &corn& oats\\ \hline Irrigation&2.2&1.1&1.2\\ Labor & 0.9&0.1&0.2\\ Money Made&400&250&300 \\ \hline\end{tabular}$

$\\x_1=500 \\x_2=65 \\x_3=15\\y=\$233,500$

$\\x_1=150 \\x_2=70 \\x_3=65\\y=\$243,500$

$\\x_1=65 \\x_2=0 \\x_3=510\\y=\$223,500$

$\\x_1=210 \\x_2=100 \\x_3=65\\y=\$253,300$

$\\x_1=510 \\x_2=0 \\x_3=65\\y=\$223,500$

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Flashcards: Math Modeling

Find the amount of each crop that should be planted to maximize the profit.

The table below includes some additional, helpful information.

$\begin{tabular}{|c|c|c|c|} \hline requirements (per acre) &soy &corn& oats\\ \hline Irrigation&2.2&1.1&1.2\\ Labor & 0.9&0.1&0.2\\ Money Made&400&250&300 \\ \hline\end{tabular}$

Your answer:

$\\x_1=500 \\x_2=65 \\x_3=15\\y=\$233,500$

Your answer:

$\\x_1=150 \\x_2=70 \\x_3=65\\y=\$243,500$

Your answer:

$\\x_1=65 \\x_2=0 \\x_3=510\\y=\$223,500$

Your answer:

$\\x_1=210 \\x_2=100 \\x_3=65\\y=\$253,300$

Correct answer:

$\\x_1=510 \\x_2=0 \\x_3=65\\y=\$223,500$

Explanation:

The objective of this particular problem is to maximize the profit.

Maximize the function:

y=f(x_1,...,x_n)=c_1x_1+...+c_nx_n

over the region defined by the constraints.

First identify the variables and assumptions of the problem.

$\\x_1=\text{acres of soy planted} \\x_2=\text{acres of corn planted} \\x_3=\text{acres of oats planted} \\l=\text{labor required} \\w=\text{water required} \\t=\text{total acreage planted} \\y=\text{total profit}$

$\\w=2.2x_1+1.1x_2+1.2x_3 \\l=0.9x_1+0.1x_2+0.2x_3 \\t=x_1+x_2+x_3 \\y=400x_1+250x_2+300x_3 \\w\leq 1200 \\l\leq 750 \\t\leq 575 \\x_1\geq0, x_2\geq0, x_3\geq 0$

Now using the constraints formulate the linear programming model.

$\\2.2x_1+1.1x_2+1.2x_3\leq1200 \\0.9x_1+0.1x_2+0.2x_3\leq750 \\x_1+x_2+x_3\leq575 \\x_1\geq0, x_2\geq0, x_3\geq 0$

From here use computer implementation to solve using the simplex method.

$\\x_1=510 \\x_2=0 \\x_3=65$

Therefore, the optimal solution will occur when there are 510 acres of soy planted and 65 acres of oats.

The maximum profit at this solution will be, $223,500.

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All Math Modeling Resources

Flashcards: Math Modeling

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