Assignment 26

course Mth 272

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16:44:37

Query problem 7.3.38 level curves of z = e^(xy), c = 1, 2, 3, 4, 1/2, 1/3, 1/4.

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16:44:56

What is the level curve z = c for the given function?

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The level of curves are hyperbolas.

confidence assessment: 1

** The z = c level curve of e^(xy) occurs when e^(xy) = c.

We solve e^(xy) = c for y in terms of x. We first take the natural log of both sides:

ln(e^(xy)) = ln(c), or

xy = ln(c). We then divide both sides by x:

y = ln(c) / x.

For c = 1 we get y = ln(1) / x = 0 / x = 0. Thus the c = 1 level curve is the x axis y = 0.

For c = 2 we get y = ln(2) / x = .7 / x, approximately. This curve passes through the points (1,.7) and (-1, -.7), and is asymptotic to both the x and y axes.

For c = 3 we get y = ln(3) / x = 1.1 / x, approximately. This curve passes through the points (1,1.1) and (-1, -1.1), and is asymptotic to both the x and y axes.

For c = 4 we get y = ln(4) / x = 1.39 / x, approximately. This curve passes through the points (1,1.39) and (-1, -1.39), and is asymptotic to both the x and y axes.

For c = 1/2 we get y = ln(1/2) / x = -.7 / x, approximately. This curve passes through the points (-1,.7) and -1, -.7), and is asymptotic to both the x and y axes.

For c = 1/3 we get y = ln(1/3) / x = -1.1 / x, approximately. This curve passes through the points (-1,1.1) and -1, -1.1), and is asymptotic to both the x and y axes.

For c = 1/4 we get y = ln(1/4) / x = -1.39 / x, approximately. This curve passes through the points (-1,1.39) and -1, -1.39), and is asymptotic to both the x and y axes.

The c = 2, 3, 4 level curves form similar hyperbolas in the first and third quadrant which progressively 'bunch up' closer and closer together. Similar behavior is observed for the c = 1/2, 1/3, 1/4 hyperbolas, which occur in the second and fourth quadrants.

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16:45:05

Describe how the level curves look for the given values of c, and how they change from one value of c to another.

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16:45:09

Query problem 7.3.46 queuing model W(x,y) = 1 / (x-y), y < x (y = ave arrival rate, x = aver service rate).

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16:45:22

What are the values of W at (15, 10), (12, 9), (12, 6) and (4,2)?

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W(15, 10) = 1/(15/10) = 1/5

W(12, 9) = 1/(12-9) = 1/3

W(12, 6) = 1/(12-6) = 1/6

W4,2) = 1/(4-2) = 1/2

confidence assessment: 2

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16:45:30

You may take extra time with the following: What is the nature of the worst combination of x and y, and why is this bad--both in terms of the behavior of the function and in terms of the real-world situation?

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confidence assessment:

Good answer by student:

x=service, y=arrival rate

therefore the worst senerio would be when the arrival rate was almost as long as the service rate...its worse to have a customer waiting a long time than having a customer being served for a long time

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16:45:32

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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confidence assessment:

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Your work looks good. See my notes. Let me know if you have any questions. &#