Assignment 29

course Mth 272

r򰇏yassignment #029

029.

Applied Calculus II

04-29-2009

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21:28:00

Query problem 7.4.50 (was 7.4.46) slope in x direction and y direction for z=x^2-y^2 at (-2,1,3)

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RESPONSE -->

first differentiate with respect to x fx(x,y)= 2x

fx(-2,1) = 2(-2) = -4

then for y

fy(x,y) = -2y

fy(-2,1)= -2(1)= -2

confidence assessment: 3

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21:28:07

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

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RESPONSE -->

answered

confidence assessment: 3

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21:28:14

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

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RESPONSE -->

answered

confidence assessment: 3

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21:29:15

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

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RESPONSE -->

taking derivative of fx and fy first

fx= 1/ x-y

fy= -1/x-y

confidence assessment: 3

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21:34:29

What is fxx at the given point?

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RESPONSE -->

fxx= (x-y)^-1 = -1(x-y)^-2 (1)

=-1(x-y)^-2 = -1/ (x-y)^2 plug in x and y

= -1

confidence assessment: 3

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21:35:16

What is fyx at the given point?

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RESPONSE -->

fyx= -(x-y)^-1 = 1(x-y)^-2

=1/(x-y)^2 = 1

confidence assessment: 3

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21:41:10

What is fxy at the given point?

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RESPONSE -->

fxy= -1(x-y)^-2 (-1) = 1/(x-y)^2 plug in values

= 1

confidence assessment: 3

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21:43:24

What is fyy at the given point?

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RESPONSE -->

fyy= -(x-y)^-1 = 1(x-y)^-2 (-1)

= -1/ (x-y)^2 = -1

confidence assessment: 3

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21:43:43

What is fx at the given point?

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RESPONSE -->

it is 1

confidence assessment: 3

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21:43:55

What is fy at the given point?

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RESPONSE -->

it's -1

confidence assessment: 3

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21:44:13

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 -->

ok

confidence assessment: 3

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21:44:40

What is the marginal revenue for plant 1?

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RESPONSE -->

fx1= 200-8x1-8x2

confidence assessment: 3

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21:45:09

What is the marginal revenue for plant 2?

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RESPONSE -->

fx2= 200-16x1x2- 8x2

If the function is

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

then f_x2 = 200 - 8 x1- 8 x2

confidence assessment: 3

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21:45:52

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

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RESPONSE -->

Because we're trying to find the change in marginal revenue for x1 and how it is affected by x2.

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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21:47:25

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 -->

This deals with monopolistic competition. When other firms enter an industry, the already present firms must account for their presence by becoming more competitive. In some cases they will lose and in some they will gain.

confidence assessment: 3

The marginal revenues for each plant may depend on the each other for a variety of reasons; for example 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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21:50:19

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 is multivariable, so partial derivation is needed. This means that they are interdependent, because ones progress affects the other's directly.

The specific reason is that both derivatives contain x1 and x2 terms, so both marginal revenues depend on both the production of plant 1 and of plant 2.

confidence assessment: 3

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