assignment 10

course Mth 272

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:

(8/x)=x^2

x(8/x)=(x^2)x Times both sides by X

8=x^3

x=2

From 1 to 2 use (8/x)-x^2 formula

AND

From 2 to 4 use x^2 - (8/x) formula

Now, anti-derivatives

(8/x)-x^2= 8lnx- x^3/3

x^2-(8/x)= x^3/3-8lnx

Then, by plugging the boundaries in, we get.........

5.545-2.6666666= 2.878 (1 to 2)

10.24- (-2.878)= 13.118 (2 to 4)

2.878+13.118= 16.0

Confidence rating:

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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):

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:

First, set equations equal to each other

1000-.4x^2=42x

42x+.4x^2-1000

SIMPLIFY= x^2+105x-2500 (Divide all numbers by .4)

(x-20) (x+125)= 0

(x-20)=0

x=20

Now, plug 20 into either equation to get an answer for Y

42(20)=y

Y=840

Now, find consumer surplus.......

1000-.4x^2-840

=160-.4x^2

Now, find the anti-derivative

160x-.4x^3/3

=160x-(2/15)x^3

Now, plug 20 into this formula for consumer surplus

160(20)-(2/15)*(20)^3

=3200-1066.67

=2,133.33

Confidence rating:

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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):

Self-critique Rating:

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&#Very good work. Let me know if you have questions. &#