course mth 272 DWwy`㏿a}Eassignment #010
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12:15:57 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 --> im looking through my examples and I don't see one like this. I will give a detailed critique when I see how you've done this one. confidence assessment: 0
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12:25:32 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 --> oh yeah. You do not use all of the coordinates. You graph and find which ones are intersecting. Which in this case is where 8/x = x^2 x = 2. Now at x < 2 we have 8/x > x^2; for x > 2 the inequality is the reverse. Integrate 8/x - x^2 from x = 1 to x = 2 antiderivative = 8 ln x - x^3 / 3 8 ln 2 - 8/3 - (8 ln 1 - 1/3) = 8 ln 2 - 7/3 x^2 - 8 / x from x = 2 to x = 4. antiderivative = x^3 / 3 - 8 ln x 64/3 - 8 ln 4 - (8 ln 2 - 8/3) = 56/3 - 8 ln 2 add together to get A = 49/3 self critique assessment: 2
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12:35:30 5.5.44 (was 5.5.40 demand p1 = 1000-.4x^2, supply p2=42x
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RESPONSE --> 1000 - .4x^2 = 42x x= -2605 that can't be right. then you would find price per unit by plugging in x into p2 then take the integral from 0 to x with (price -supply function)dx take the antiderivative then apply fundamental theorem and add together. confidence assessment: 1
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12:51:53 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 --> wow, ok I forgot all about the quadratic formula. so you rearrange to form -.4 x^2 - 42 x + 1000 = 0 then use the quadratic formula x = 20 at x=20 demand is 1000 - .4 * 20^2 = 840 supply is 42 * 20 = 840 demand and supply curves meet at (20,840) For consumer surplus: use the integral of 1000 - .4 x^2 - 840 = 160 - .4 x^2, from x = 0 to x = 20 antiderivative is 160 x - .4 / 3 * x^3 =2122.33 For producer surplus: use the integral of 840 - 42 x from x = 0 to x = 20 antiderivative = 840x - 21x^2 solve using fundamental theorem self critique assessment: 2
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