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Flashcards: Math Modeling

A beauty supply company manufactures a variety of brushes. Quality control engineers work to ensure that the defected brushes in the factory will be detected prior to shipping them. It is estimated that approximately 0.2% of the brushes made will be defected. Tests can be done individually on the brushes or on batches of the brushes. If the test on a batch of brushes fails, it means that one or more of the brushes in that particular batch are defected. The estimated cost is 4 cents for a single brush, and 3+n cents for a group of n>1 brushes. If a batch fails then all brushes in that batch must be tested individually. Find the most cost-effected quality control procedure for detecting defected brushes.

By testing brush batches in groups of 10 will reduce testing costs without sacrificing the quality.

By testing brush batches in groups of 9 will reduce testing costs without sacrificing the quality.

By testing brush batches in groups of 20 will reduce testing costs without sacrificing the quality.

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

By testing brush batches in groups of 12 will reduce testing costs without sacrificing the quality.

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Flashcards: Math Modeling

Your answer:

By testing brush batches in groups of 10 will reduce testing costs without sacrificing the quality.

Your answer:

By testing brush batches in groups of 9 will reduce testing costs without sacrificing the quality.

Your answer:

By testing brush batches in groups of 20 will reduce testing costs without sacrificing the quality.

Your answer:

By testing brush batches in groups of 12 will reduce testing costs without sacrificing the quality.

Correct answer:

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

Explanation:

First identify the known variables and assumptions.

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

$n=1\rightarrow A=4$

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

$n>1\rightarrow C=3+n$

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

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

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

Consider the random variable

$X\epsilon\begin{Bmatrix} x_1, x_2, x_3,... \end{Bmatrix}$

that has a probability of

$\text{Pr}\begin{Bmatrix} X=x_i \end{Bmatrix}=p_i$

$\\\Sigma p_i=1 \\EX=\Sigma x_ip_i$

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

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

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

Therefore the expected value of the random variable is,

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

Therefore the average testing cost is,

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

Using the law of large numbers minimizing results in

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.

Flashcards: Math Modeling

A company that makes computer monitors is unveiling two new products this spring, a 15 inch screen for $189 and a 19 inch screen for $269. The cost to manufacture the 15 inch screens is $85 per screen and the 19 inch screen is $99 per screen, plus $200,000 for miscellaneous costs. The number of sells for each screen type effects the average selling price. It is estimated that each set average selling price drops by two cents for each additional screen sold. The average sells price for the 15 inch screen will decrease by 0.2 cents for each 19 inch screen sold, the price of the 19 inch screen decreases by 0.3 cents for each 15 inch screen sold. What are the partial differential equations used to find how many units of each type of screens should be manufactured to optimize profit?

$\\\frac{\partial f}{\partial x_1}=170-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=104-0.02x_2-0.005x_1=0$

$\\\frac{\partial f}{\partial x_1}=104-0.2x_1-0.05x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.002x_2-0.05x_1=0$

$\\\frac{\partial f}{\partial x_1}=104-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.02x_2-0.005x_1=0$

$\\\frac{\partial f}{\partial x_1}=104-0.005x_1-0.02x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.02x_2-0.005x_1=0$

$\\\frac{\partial f}{\partial x_1}=104-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.005x_2-0.02x_1=0$

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Flashcards: Math Modeling

Your answer:

$\\\frac{\partial f}{\partial x_1}=170-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=104-0.02x_2-0.005x_1=0$

Your answer:

$\\\frac{\partial f}{\partial x_1}=104-0.2x_1-0.05x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.002x_2-0.05x_1=0$

Your answer:

$\\\frac{\partial f}{\partial x_1}=104-0.005x_1-0.02x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.02x_2-0.005x_1=0$

Your answer:

$\\\frac{\partial f}{\partial x_1}=104-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.005x_2-0.02x_1=0$

Correct answer:

$\\\frac{\partial f}{\partial x_1}=104-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.02x_2-0.005x_1=0$

Explanation:

First, identify all known information.

Variables:

$\\s=\text{number of 15 inch screens (per year)} \\t=\text{number of 19 inch screens (per year)} \\p=\text{selling price for 15 inch screen} \\q=\text{selling price for 19 inch screen} \\C=\text{cost to manufacture} \\R=\text{revenue} \\P=\text{profit}$

Assumptions:

$\\p=189-0.01s-0.002t \\q=269-0.003s-0.01t \\R=ps+qt \\C-200,000+85s+99t \\P=R-C \\s\geq0 \\t\geq0$

Objective:

Maximize the profit.

Theorem: If is a function on which is a subset of $\mathbb{R}^n$ and has a minimum or maximum at an interior point in then $\bigtriangledown f=0$ .

Therefore, to find the maximum point in this function, simultaneously solve the differential equations in question and check the boundary points.

$\\P=R-C$

$\\=ps+qt-(200,000+85s+99t) \\=(189-0.01s-0.002t)(s)+(269-0.003s -0.01t)(t)-200,000-85s-99t$

Now switch variables/

$\\y=f(x_1,..,x_n) \\x_1=s \\x_2=t$

$\\=(189-0.01x_1-0.002x_2)(x_1)+(269-0.003x_1 -0.01x_2)(x_2)\\-200,000-85x_1-99x_2$

Over

$s=\begin{Bmatrix} (x_1,x_2): x_1\geq0, x_2\geq 0 \end{Bmatrix}$

From here take the partial differential equation of each variable. Since the questioned asked for the partial differential equations that would be used to optimize the profit we can stop here.

$\\\frac{\partial f}{\partial x_1}=104-0.02x_1-0.005x_2=0 \\\\\\\frac{\partial f}{\partial x_2}=170-0.02x_2-0.005x_1=0$

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