assignment 10

course MTH 272

7/5 4 pm

If your solution to a stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

010. `query 10

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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:

The graphs cross at 8/x = x^2, which solves as x = 2.

x < 2= 8/x > x^2

x>2 = x^2<8/x

integrate:

8/x - x^2 (from x = 1 to x = 2) and x^2 - 8 / x (from x = 2 to x = 4)

8 ln x - x^3 / 3 and x^3 / 3 - 8 ln x.

This gives you 8 ln 2 - 8/3 - (8 ln 1 - 1/3) = 8 ln 2 - 7/3

Also, the other integral is 64/3 - 8 ln 4 - (8 ln 2 - 8/3) = 56/3 - 8 ln 2.

You have to add these to get the answer. This= 49/3

Confidence rating: 2.5

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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): I know this type of problem fairly well, but there are still some parts I need practice on.

Self-critique Rating:

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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:

Demand p1=1000-.4x^2

Supply p2=42x

The demand is a quadratic equation.

a=-.4x^2

b=-42x

c=1000

solve through quad formula. Final answer=20

20=demand where 1000-.4*400

This gets 840

There is a supply curve too obviously. It touches the demand curve at x=20 and y=840

Confidence rating:1.5

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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): I get most of the first part of the problem, but the part that gets me is the part where I have to find the surplus of the produced and consumer. The solution helped me get it a little better. But that is the part of the problem I am yet to understand fully.

You won't have a question on a test that requires you to know the definitions of consumer surplus and producer surplus. However you should be able to apply those definitions if they are given.

In other words you might not remember that

Consumer surplus is the area of the demand function above the equilibrium line.

Producer surplus is the area of the supply function below the equilibrium line.

But you should be able to find the area of the demand function above the equilibrium line.

and the area of the supply function below the equilibrium line.

Self-critique Rating:

3"

&#Good responses. See my notes and let me know if you have questions. &#