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Flashcards: Math Modeling

Using Newton's Method as the computational method, optimize the follow problem.

A cow that weighs 175 pounds gains 3 pounds per day, and it costs 30 cents per day to keep that cow. The market price for cows is 75 cents per pound, but the price is falling one cent per day. When should the cow be sold to make the maximum profit?

It is recommended to wait 6 or 7 days to sell the cow. This should result in a net profit of $130.

It is recommended to wait 2 or 5 days to sell the cow. This should result in a net profit of $131.90.

It is recommended to wait 7 or 8 days to sell the cow. This should result in a net profit of $131.90.

It is recommended to wait 9 or 10 days to sell the cow. This should result in a net profit of $132.

It is recommended to wait 8 or 9 days to sell the cow. This should result in a net profit of $132.90.

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Flashcards: Math Modeling

Using Newton's Method as the computational method, optimize the follow problem.

Your answer:

It is recommended to wait 6 or 7 days to sell the cow. This should result in a net profit of $130.

Your answer:

It is recommended to wait 2 or 5 days to sell the cow. This should result in a net profit of $131.90.

Your answer:

It is recommended to wait 9 or 10 days to sell the cow. This should result in a net profit of $132.

Your answer:

It is recommended to wait 8 or 9 days to sell the cow. This should result in a net profit of $132.90.

Correct answer:

It is recommended to wait 7 or 8 days to sell the cow. This should result in a net profit of $131.90.

Explanation:

Assume that the growth rate of the cow is proportional to its weight.

In mathematical terms:

$\frac{dw}{dt}=cw$

Looking at the known information,

Variables:

$\\t=\text{time (days)} \\w=\text{weight (lbs)} \\p=\text{proft (}\$\text{/lbs)} \\C=\text{cost to keep the cow} \\R=\text{revenue} \\P=\text{profit}$

Therefore,

$\frac{dw}{dt}=3\text{ lbs/day}$

when

$w=175\text{ lbs}$

$\frac{dw}{dt}=cw, w(0)=175$

$c=\frac{3}{175}=0.017$

Thus the differential equation to solve for is as follows.

$\frac{dw}{dt}=0.017w, w(0)=175$

Solving by separation of variables results in the following equation.

$w=175e^{0.017t}$

From here, formulate the model with the new weight equation.

$\\y=f(x) \\y=(0.75-0.01t)(175e^{0.017t})-0.30t$

Recall that the goal is to optimize the question over

$S=\begin{Bmatrix} x:x\geq0 \end{Bmatrix}$ .

Use a graphing calculator or computer technology to graph the function.

Screen shot 2016 04 27 at 9.31.27 am

Observing the graph it appears the maximum profit for the cow occurs around x=7 and y=131.9 .

From here, answer the question. Taking into account the growth rate of the cow is still increasing, it is recommended to wait 7 or 8 days to sell the cow. This should result in a net profit of $131.90.

Flashcards: Math Modeling

A computer company has one service repair man and has space for 32 computers in the shop at one time. Last year the shop repaired 71 computers with an average repair time of 3 days per computer. Formulate a Markov process model for X_t which represents the number of computers in the shop for repair at time months and calculate the EX_t .

$EX_t\approx 4.176$

$EX_t\approx 4.716$

$EX_t\approx 4.761$

$EX_t\approx 4.236$

$EX_t\approx 4.426$

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Flashcards: Math Modeling

Your answer:

$EX_t\approx 4.716$

Your answer:

$EX_t\approx 4.761$

Your answer:

$EX_t\approx 4.236$

Your answer:

$EX_t\approx 4.426$

Correct answer:

$EX_t\approx 4.176$

Explanation:

Using Markov Process for this question assume that the space is finite.

$X_t\epsilon\begin{Bmatrix} 1,2,3,4,...,m \end{Bmatrix}$

The stochastic process which determines the future of the process from a probabilistic standpoint is defined as follows.

Let

$T_i=\text{time spent in state }i$

$Pr{\begin{Bmatrix} T_i>t+s|T_i>s \end{Bmatrix}}=Pr\begin{Bmatrix} T_i>t \end{Bmatrix}$

Recall that the exponential distribution, and thus the T_i could have the density function,

$F_i(t)=\lambda_ie^{-\lambda_it}$

Now formulate the Markov Process specifically for this problem.

$X_t\epsilon\begin{Bmatrix} 0,1,2,..., 32\end{Bmatrix}$

The transitions of computers in and out of the shop are,

X_t=i to X_t=i+1 or X_t=i-1

Now calculate the rates up and down.

To calculate $\lambda$ take the total number of computers fixed last year and divide it by the total months in the year. To calculate $\mu$ assume there are 22 working days in a month and it takes on average, two days to fix a computer.

$\\\lambda=5.9 \\\mu=7.3$

Its key to know that at zero we cannot move down a state and like wise when at 29, we cannot move up a state.

$[\text{Rate In}]=[\text{Rate Out}]$

$\\\lambda P_0=\mu P_1 \\(\mu +\lambda)P_1=\lambda P_0+\mu P_1 \\(\mu +\lambda)P_2=\lambda P_2+\mu P_3\\ \begin{matrix} .\\ .\\ . \end{matrix} \\\mu P_{32}=\lambda P_{31}$

Solving with

$\sum P_i=1$

$EX_t=\sum iP_i$

Continuing in this fashion results in the following.

$\\P_n=(\lambda / \mu)P_{n-1} \\P_n=(\lambda / \mu)^nP_0 \\\\\sum_{n=0}^{32}P_n=P_0\sum_{n=0}^{32}\left(\frac{\lambda}{\mu} \right )^n=1 \\\\P_0=\frac{1-\rho }{1-\rho^{33}}$

Now, writing the sum of a finite geometric series is

$P_n=\rho^nP_0=\frac{\rho^n(1-\rho)}{1-\rho^{30}}$

Now

$\rho=\frac{\lambda}{\mu}=\frac{5.9}{7.3}=0.808$

therefore,

$1-\rho^{33}=1-0.808^{33}\approx 0.9991198131$

Next,

$Pr\begin{Bmatrix} X_t>0 \end{Bmatrix}=1-P_0=\rho\approx 0.808$

$\\EX_t=\sum_{n=0}^{32}nP_n \\EX_t=\sum_{n=0}^{32}n\rho^n(1-\rho) \\EX_t\approx 4.176$

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