Assignment 29

course Mth 272

žþ»¹¥ãwêùærv•šEˆ]šûpÚÂ×Äassignment #029

029.

Applied Calculus II

11-26-2008

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14:15:04

What is the slope in the x direction at the given point? Describe specifically how you obtained your result.

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RESPONSE -->

z= x^2 - y^2

hold ""y constant, differentiate x

f(x,y) = 2x

f(-2,1)= -4

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confidence assessment: 3

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14:15:47

What is the slope in the y direction at the given point? Describe specifically how you obtained your result.

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RESPONSE -->

hold ""x"" constant, differentiate y

f(x,y)= -2y

f(-2,1)= -2

confidence assessment: 3

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14:16:45

Query problem 7.4.65 (was 7.4.61) all second partials of ln(x-y) at (2,1)

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RESPONSE -->

ln (x-y)

ln(x)-ln(y)

confidence assessment: 3

ln(x-y) is not equal to ln(x) - ln(y). Be sure to review the definitions and rules related to logarithms, to guard against similar errors on tests.

** The first x derivative is found by the Chain Rule to be (x-y)' * 1/(x-y), where the ' is derivative with respect to x. We get fx = 1 * 1 / (x-y) = 1 / (x-y), or if you prefer (x-y)^-1, where fx means the first x derivative.

The x derivative of this expression is the derivative of (x-y)^-1, which by the Chain Rule is fxx = (x-y)' * -1 (x-y)^-2 = 1 * -1 * (x-y)^-2 = -1/(x-y)^2; here fxx means second x derivative and the ' means derivative with respect to x.

fxy is the y derivative of fx, or the y derivative of (x-y)^-1, which by the Chain Rule is fxy = (x-y)' * -1 (x-y)^-2 = -1 * -1 * (x-y)^-2 = 1/(x-y)^2; here fxy means the y derivative of the x derivative and the ' means derivative with respect to y.

The first y derivative is found by the Chain Rule to be (x-y)' * 1/(x-y), where the ' is derivative with respect to y. We get fy = -1 * 1 / (x-y) = -1 / (x-y), or if you prefer -(x-y)^-1, where fy means the first y derivative.

The y derivative of this expression is the derivative of -(x-y)^-1, which by the Chain Rule is fyy = -(x-y)' * -1 (x-y)^-2 = -[1 * -1 * (x-y)^-2] = 1/(x-y)^2; here fyy means second y derivative and the ' means derivative with respect to y.

fyx is the x derivative of fy, or the x derivative of -(x-y)^-1, which by the Chain Rule is fyx = -(x-y)' * -1 (x-y)^-2 = -[1 * -1 * (x-y)^-2] = 1/(x-y)^2; here fyx means the x derivative of the y derivative and the ' means derivative with respect to x.

When evaluated at (2, 1) the denominator (x - y)^2 is 1 for every second partial. So we easily obtain

fxx = -1

fyy = -1

fxy = fyx = +1. **

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14:17:41

What is fxx at the given point?

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RESPONSE -->

1- ln(y)

1-ln(1)= 1

confidence assessment: 2

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14:18:39

What is fyx at the given point?

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RESPONSE -->

ln(x)- 1/y

1- 1/1= 0

confidence assessment: 2

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14:19:22

What is fxy at the given point?

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RESPONSE -->

1/x - 1 = 1/2 - 1 = -0.5

confidence assessment: 2

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14:19:46

What is fyy at the given point?

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RESPONSE -->

ln (2)- 1/1 = -0.31

confidence assessment: 2

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14:20:08

What is fx at the given point?

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RESPONSE -->

1/x - ln(X)

confidence assessment: 3

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14:20:27

What is fy at the given point?

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RESPONSE -->

LN(X)- 1/Y

confidence assessment: 3

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14:20:43

Query problem 7.4.68 R = 200 x1 + 200 x2 - 4x1^2 - 8 x1 x2 - 4 x2^2; R is revenue, x1 and x2 production of plant 1 and plant 2

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RESPONSE -->

confidence assessment:

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14:21:25

What is the marginal revenue for plant 1?

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RESPONSE -->

r= 200 + 200X2 - 8X1 - 8X2 -4X2^2'

R= 1896

confidence assessment: 3

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14:22:01

What is the marginal revenue for plant 2?

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RESPONSE -->

200x1 + 200 - 4x1^2 - 8x1 - 8x2

R= 808

confidence assessment: 3

Close, but be careful about the details.

** The derivative of R with respect to x1 is 200 + 0 - 4 (2 x1) - 8 x2 - 0; All all derivatives treat x1 as the variable, x2 as constant. Derivatives of 200 x2 and -4 x2^2 do not involve x1 so are constant with respect to x1, hence are zero.

So the marginal revenue with respect to plant 1 is 200 - 8 x1 - 8 x2.

The derivative of R with respect to x2 is 0 + 200 - 0 - 8 x1 - 4 ( 2 x2) = 200 - 8 x1 - 8 x2; All all derivatives treat x2 as the variable, x1 as constant.

So the marginal revenue with respect to plant 2 is 200 - 8 x1 - 8 x2. **

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14:22:46

Why should the marginal revenue for plant 1 be the partial derivative of R with respect to x1?

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RESPONSE -->

x1 corresponds to plant 1. to find its revenue it is necessary to leave x2 alone and find the derivative of all x1 parts

confidence assessment: 3

** Marginal revenue is the rate at which revenue changes per unit of increased production. The increased production at plant 1 is the change in x1, so we use the derivative with respect to x1. **

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14:23:36

Why, in real-world terms, might the marginal revenue for each plant depend upon the production of the other plant?

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RESPONSE -->

Because both plants produce the dsame medicine, if one plant is earning more revenue than thew other, it would make sense to produce more of the most profitable plant.

confidence assessment: 3

The marginal revenues for each plant may depend on the other if one plant awaits shipment of a part from the other, or if one plant is somewhat slow resulting in a bottleneck.

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14:24:18

What is is about the function that ensures that the marginal revenue for each plant will depend on the production of both plants?

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RESPONSE -->

The function contains both variables, so no matter what plant you're trying to find the marginal revenue fore, the answer depends on both values.

confidence assessment: 3

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&#Your work looks very good. Let me know if you have any questions. &#