QUERY29

course MTH 272

05/12,2300

assignment #029029.

Applied Calculus II

05-12-2010

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20:45:20

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 -->

z=x^2-y^2 at (-2,1,3)

fx=2x

slope in x-direction: fx=2(-2)=-4

fy=-2y

slope in y-direction: fy=-2(1)=-2

confidence rating #$&* 2

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20:46:24

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 at (-2,1,3)

fx=2x

slope in x-direction: fx=2(-2)=-4

Slope in the x-direction is found by keeping y constant and differenting with respect to x.

confidence rating #$&* 2

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20:47:01

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

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RESPONSE -->

z=x^2-y^2 at (-2,1,3)

fy=-2y

slope in y-direction: fy=-2(1)=-2

Slope in the y-direction is found by keeping x constant and differentiating with respect to y.

confidence rating #$&* 2

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20:54:52

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) at (2,1)

fx= 1/ (x-y)

=(x - y)^-1

fy=1/ (x-y)

=(x-y)^-1

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

= -1/ (x-y)^2

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

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

=-1/(x-y)^2

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

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

= 1/ (x-y)^2

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

confidence rating #$&* 2

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20:55:18

What is fxx at the given point?

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RESPONSE -->

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

= -1/ (x-y)^2

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

confidence rating #$&* 2

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20:55:35

What is fyx at the given point?

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RESPONSE -->

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

= 1/ (x-y)^2

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

confidence rating #$&* 2

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20:55:47

What is fxy at the given point?

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RESPONSE -->

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

= 1/ (x-y)^2

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

confidence rating #$&* 2

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20:56:01

What is fyy at the given point?

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RESPONSE -->

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

=-1/(x-y)^2

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

confidence rating #$&* 2

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20:56:39

What is fx at the given point?

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RESPONSE -->

fx= 1/ (x-y)

=(x - y)^-1

1/ (2-1) =1

confidence rating #$&* 2

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20:57:13

What is fy at the given point?

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RESPONSE -->

fy=1/ (x-y)

=(x-y)^-1

fy=1/(2-1) =1

confidence rating #$&* 2

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21:00:49

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 -->

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

dR/dx1=200 - 8x1 - 8x2

dR/dx2=200 - 8x1 - 8x2

x1=4,x2=12

dR/dx1=200 - 8(4) - 8(12)=72

dR/dx2=200 - 8(4) - 8(12)=72

confidence rating #$&* 2

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21:01:14

What is the marginal revenue for plant 1?

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RESPONSE -->

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

dR/dx1=200 - 8x1 - 8x2

x1=4,x2=12

dR/dx1=200 - 8(4) - 8(12)=72

confidence rating #$&* 2

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21:01:28

What is the marginal revenue for plant 2?

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RESPONSE -->

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

dR/dx2=200 - 8x1 - 8x2

x1=4,x2=12

dR/dx2=200 - 8(4) - 8(12)=72

confidence rating #$&* 2

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21:01:55

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

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RESPONSE -->

If R is representative of the revenue, the marginal revenue is a derivative of this function. X1 is the variable that bears an impact on the revenue of plant 1.

confidence rating #$&* 2

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21:04:16

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 -->

The revenue of one plant might certainly impact the production of the other plant. If the products were complementary, as one plant has increased success, so too might the other plant. If the the products were substitutes, the relationship would be more of a competitive one. Increased success of one plant would potentially decrease sales from the other.

confidence rating #$&* 2

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21:04:59

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 -->

Each marginal revenue function contains both x1 and x2.

confidence rating #$&* 2

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