Assignment 10

course MTH272

02-13-10 4:32pm

010. `query 10*********************************************

Question: `q5.5.1 (previously 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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Your solution: 8/x=x^2 x=2

8lnx - x^3/3

8ln2 - 8/3

2.878

confidence rating #$&*:2

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Given Solution:

`a 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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Self-critique (if necessary):ok I missed the x and 1 and x and 2 and then 2 and 4 and putting it all together.

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Self-critique rating #$&*:2

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Question: `q5.5.4 (previously 5.5.44 (was 5.5.40) ) demand p1 = 1000-.4x^2, supply p2=42x

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Your solution: -.4x^2-42x+1000= 0

x=20

demand of p1 = 1000-.4 (20^2) = 840

p2 = 42(20) = 840

(20, 840)

confidence rating #$&*:2

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Given Solution:

`a 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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Self-critique (if necessary):well i forgot about doing the integral. Have to do that for a test.

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Self-critique rating #$&*:1"

You could have done that last integral. Be sure you understand about splitting the region when the graphs cross.

&#Please let me know if you have questions. &#