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: We can first graph the two solutions to find out if only one equation is needed to find the area. When we graph the given functions, we find that they meet at x = 2 and that we need to break up the equation. For (1,2) we need to use the function 8/x - x^2 and x^2 - 8 / x from (2,4). Once we integrate these expressions according to our integration rules, we end up with 8 ln x - x^3 / 3 and x^3 / 3 - 8 ln x. Now we can solve for our areas. 8 ln (2) – (2)^3 / 3 –(8ln(1)-(1)^3/3) = 2.8785-(-1/3) = 3.2118. (4)^3 / 3 - 8 ln (4) – ((2)^3/3-8ln(2)) = 10.2429 – (-2.8785) = 13.1214. Now to get the total area, we can add our two found areas together. So 3.2118 + 13.1214 = 16.3332.
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): My answer differed from the given solution because the given solution used fractions. Is this is more accurate I should start keeping the bases instead of dividing into decimals.
Whether you work in fractions or decimals depends on the situation, but when in doubt, work in fractions, for the very reasons you state.
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: To find the consumer and producer surpluses, we can first combine our two given equations into one, getting (1000-.4x^2) – 42x. Now we can solve for x, finding that x = 20 is correct. When we graph the two given equations, we find that they meet at (20, 840). Now we find that everything above the line y = 840 is the consumer surplus and everything below it is the producer surplus. We can first find the consumer surplus using the equation 1000 - .4 x^2 – 840. When we simplify the expression we get 160 - .4 x^2, and then we find that its integral is 160x- (.4)/3x^3. Now we can use the Fundamental Theorem of Calculus to find the area of the consumer surplus. So on the interval (0,20), our area is equivalent to 160(20) – (.4)/3(20^3) = 2133.3333. Now we can find the producer surplus by taking the interval of the equation 840 - 42 x, which we find to be 840x-21x^2. Now we can use the Fundamental Theorem of Calculus again to find the area of the producer surplus. So 840(20)-21(20^2) = 8400.
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): I had trouble deciding what equations to use for the consumer surplus and producer surplus, I just need to do some more practice with these.
Self-critique Rating:
&#This looks very good. Let me know if you have any questions. &#