Assignment 10

course mth 272

ñg~Ÿœÿœ¢{ÒݴۄæXý÷á~assignment #010

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

á×{ƒ¯h׌ݼ—ê}ŠÇË­ßÇšï…

Applied Calculus I

06-21-2006

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21:20:49

5.5.23 (was 5.5.28 area in region defined by y=8/x, y = x^2, x = 1, x = 4

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RESPONSE -->

antiderivative: 8 ln x - x^3/3 and then the inverse of that (x^3/3 - 8 ln x )

8 ln 2 - 8/3 - (8 ln 1 - 1/3) = 8 ln 2 - 7/3

64/3 - 8 ln 4 - (8 ln 2 - 8/3) = 56/3 - 8 ln 2

8 ln 2 - 7/3+ 56/3 - 8 ln 2 = 49/3 or 16 and 1/3

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21:21:27

** These graphs intersect when 8/x = x^2, which we solve to obtain x = 2.

For x < 2 we have 8/x > x^2; for x > 2 the inequality is the reverse.

So we integrate 8/x - x^2 from x = 1 to x = 2, and x^2 - 8 / x from x = 2 to x = 4.

Antiderivative are 8 ln x - x^3 / 3 and x^3 / 3 - 8 ln x. We obtain

8 ln 2 - 8/3 - (8 ln 1 - 1/3) = 8 ln 2 - 7/3 and

64/3 - 8 ln 4 - (8 ln 2 - 8/3) = 56/3 - 8 ln 2.

Adding the two results we obtain 49/3. **

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RESPONSE -->

the graph intersects giving x=2

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21:24:21

5.5.44 (was 5.5.40 demand p1 = 1000-.4x^2, supply p2=42x

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RESPONSE -->

-.4 x^2 - 42 x + 1000 = 0 so x = 20 and supply with the demand is 840 ( 20*42)

C surplus: integral giving you 160 -.4x^2

P surplus: 840-42x

so the C surplus = 2133 from 0 to 20 = x

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21:24:24

5.5.44 (was 5.5.40 demand p1 = 1000-.4x^2, supply p2=42x

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RESPONSE -->

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21:24:29

17:42:46

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RESPONSE -->

?

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21:24:44

** 1000-.4x^2 = 42x is a quadratic equation. Rearrange to form

-.4 x^2 - 42 x + 1000 = 0 and use the quadratic formula.

You get x = 20

At x = 20 demand is 1000 - .4 * 20^2 = 840, supply is 42 * 20 = 840.

The demand and supply curves meet at (20, 840).

The area of the demand function above the equilibrium line y = 840 is the integral of 1000 - .4 x^2 - 840 = 160 - .4 x^2, from x = 0 to the equlibrium point at x = 20. This is the consumer surplus.

The area of the supply function below the equilibrium line is the integral from x = 0 to x = 20 of the function 840 - 42 x. This is the producer surplus.

The consumer surplus is therefore integral ( 160 - .4 x^2 , x from 0 to 20) = 2133.33 (antiderivative is 160 x - .4 / 3 * x^3). *&*&

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RESPONSE -->

antiderivative is 160 x - .4 / 3 * x^3

.................................................

"

You did well on these questions. Let me know if you have questions.

Assignment 10

course mth 272

ñg~Ÿœÿœ¢{ÒݴۄæXý÷á~assignment #010

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

á×{ƒ¯h׌ݼ—ê}ŠÇË­ßÇšï…

Applied Calculus I

06-21-2006

......!!!!!!!!...................................

21:20:49

5.5.23 (was 5.5.28 area in region defined by y=8/x, y = x^2, x = 1, x = 4

......!!!!!!!!...................................

RESPONSE -->

antiderivative: 8 ln x - x^3/3 and then the inverse of that (x^3/3 - 8 ln x )

8 ln 2 - 8/3 - (8 ln 1 - 1/3) = 8 ln 2 - 7/3

64/3 - 8 ln 4 - (8 ln 2 - 8/3) = 56/3 - 8 ln 2

8 ln 2 - 7/3+ 56/3 - 8 ln 2 = 49/3 or 16 and 1/3

.................................................

......!!!!!!!!...................................

21:21:27

** These graphs intersect when 8/x = x^2, which we solve to obtain x = 2.

For x < 2 we have 8/x > x^2; for x > 2 the inequality is the reverse.

So we integrate 8/x - x^2 from x = 1 to x = 2, and x^2 - 8 / x from x = 2 to x = 4.

Antiderivative are 8 ln x - x^3 / 3 and x^3 / 3 - 8 ln x. We obtain

8 ln 2 - 8/3 - (8 ln 1 - 1/3) = 8 ln 2 - 7/3 and

64/3 - 8 ln 4 - (8 ln 2 - 8/3) = 56/3 - 8 ln 2.

Adding the two results we obtain 49/3. **

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RESPONSE -->

the graph intersects giving x=2

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......!!!!!!!!...................................

21:24:21

5.5.44 (was 5.5.40 demand p1 = 1000-.4x^2, supply p2=42x

......!!!!!!!!...................................

RESPONSE -->

-.4 x^2 - 42 x + 1000 = 0 so x = 20 and supply with the demand is 840 ( 20*42)

C surplus: integral giving you 160 -.4x^2

P surplus: 840-42x

so the C surplus = 2133 from 0 to 20 = x

.................................................

......!!!!!!!!...................................

21:24:24

5.5.44 (was 5.5.40 demand p1 = 1000-.4x^2, supply p2=42x

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

21:24:29

17:42:46

......!!!!!!!!...................................

RESPONSE -->

?

.................................................

......!!!!!!!!...................................

21:24:44

** 1000-.4x^2 = 42x is a quadratic equation. Rearrange to form

-.4 x^2 - 42 x + 1000 = 0 and use the quadratic formula.

You get x = 20

At x = 20 demand is 1000 - .4 * 20^2 = 840, supply is 42 * 20 = 840.

The demand and supply curves meet at (20, 840).

The area of the demand function above the equilibrium line y = 840 is the integral of 1000 - .4 x^2 - 840 = 160 - .4 x^2, from x = 0 to the equlibrium point at x = 20. This is the consumer surplus.

The area of the supply function below the equilibrium line is the integral from x = 0 to x = 20 of the function 840 - 42 x. This is the producer surplus.

The consumer surplus is therefore integral ( 160 - .4 x^2 , x from 0 to 20) = 2133.33 (antiderivative is 160 x - .4 / 3 * x^3). *&*&

......!!!!!!!!...................................

RESPONSE -->

antiderivative is 160 x - .4 / 3 * x^3

.................................................

"

You did well on these questions. Let me know if you have questions.