assignment 29 

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Applied Calculus II

Asst # 28

08-30-2006

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16:22:48

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Applied Calculus II

Asst # 29

08-01-2006

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13:53:05

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**** 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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13:58:02

x direction: -4

y direction: -2

Good. The details, for your reference:

** The x derivative is 2x; at (-2,1,3) we have x = -2 so the slope is 2 * -2 = -4.

The slope in the y direction is the y partial derivaitve -2y; at y = 1 this is -2. **

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13:58:03

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

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13:58:34

-4

i found the partial derivative equaling to 2x and then used the point -2 and multiplied giving -4

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13:58:35

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

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13:59:00

-2

i calculated the partial derivative to equal -2y and used the point 1 giving -2

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13:59:01

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

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14:08:01

fx = 1/x

fy= -1/y

fxx=1/-x^2

fyy=1/y^2

fxy=-1/x^2

fyx=1/y^2

You are not applying the Chain Rule correctly:

** 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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14:08:02

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**** What is fxx at the given point?

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14:08:11

1/-x^2

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14:08:12

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**** What is fyx at the given point?

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14:08:18

1/y^2

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14:08:18

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**** What is fxy at the given point?

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14:08:25

-1/x^2

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14:08:26

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**** What is fyy at the given point?

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14:08:32

1/y^2

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14:08:32

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**** What is fx at the given point?

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14:08:36

1/x

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14:08:37

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**** What is fy at the given point?

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14:08:42

-1/y

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14:08:43

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**** 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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14:12:13

plant 1 and plant 2 at 72

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14:12:14

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**** What is the marginal revenue for plant 1?

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14:12:54

i got the partial derivative to get 200-8x1 - 8x2 to equal 72

** 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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14:12:55

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**** What is the marginal revenue for plant 2?

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14:13:09

i did the same and got the same expression and 72

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14:13:10

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

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14:14:22

marginal means to take the derivative of the total in this case revenue. both plant 1 and 2 are involved in the total revenue hence using partial derivatives helps separate each plant

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14:14:22

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

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14:16:27

they both are selling the same over the counter medicine. so in competition they would want to sell equal amounts in the long run even to have a profit of 0

*&*& The marginal revenues for each plant may depend on each 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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14:16:28

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**** 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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14:16:46

200-8x1-8x2

** 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. **

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See my notes and let me know if you have questions.