Assignment 10

course Mth 272

ϞIj{yhh䕄assignment #010

010. `query 10

Applied Calculus II

10-05-2008

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22:31:40

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 -->

set y's equal, solve for x

8/x=x^2

x=2

antiderivs for points 1 to 4, x=2

1 - 2

8ln(x )-x^3 / 3

2 - 4

x^3/ 3 - 8ln(x)

= 49/3= 16.3

confidence assessment: 3

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22:31:49

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 -->

ok

self critique assessment: 3

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22:34:17

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

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RESPONSE -->

-0.4 x^2 - 42 x + 1000 = 0

solve for x, x=20

plugin for x demand

1000-0.4* 20^2 =840

supply

42x

42*20= 840

supply and demand intersect at (x=20,y=840)

consumer surplus from 0

1000 -0.4 x^2 -840

= 160 -0.4 x^2, antideriv= 160x -0.4 / 3*x^3

= 2133.3

confidence assessment: 2

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22:34:48

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 -->

Complicated problem but apparently I understand it, or at least on the right track to get the correct answer.

self critique assessment: 2

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Very good.

Note that your test arrived in the mail today. I should have graded within a day or two. Typically it would have been graded on the day of receipt, but activities associated with relocating my lab (which is now in its last phases) and the risk of misplacing documents have made it necessary to be particularly careful with the handling of tests.