course Mth 272
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RESPONSE --> graphs intersect at point (2,4) int 1 to 2 of x^2 is x^/3 and 2 to 4 of 8/x is 8 ln(x) 1 to 2, 2^3/3 - 1^3/3 = 7/3 2 to 4, 8ln(4) - 8ln(2) = 8ln2 A = 7/3 + 8ln2 = 7.8785 confidence assessment: 2
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19:56:39 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 --> self critique assessment: 2
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19:56:54 5.5.44 (was 5.5.40 demand p1 = 1000-.4x^2, supply p2=42x
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RESPONSE --> -.4x^2 - 42x + 1000 = 0 x = 20 1000 - .4(20)^2 = 840 Consumer surplus = [-.4/3(20^3)+160(20)] - [-.4/3(0^3)+160(0)] = 2133.33 Producer Surplus = [840(20) -21(20)^2] - [840(0) -21(0)^2] = 8400 confidence assessment: 3
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19:56:57 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 --> self critique assessment: 3
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