course mth 272 ñg~Ÿœÿœ¢{ÒݴۄæXý÷á~assignment #010
......!!!!!!!!...................................
21:20:49 5.5.23 (was 5.5.28 area in region defined by y=8/x, y = x^2, x = 1, x = 4
......!!!!!!!!...................................
RESPONSE --> antiderivative: 8 ln x - x^3/3 and then the inverse of that (x^3/3 - 8 ln x ) 8 ln 2 - 8/3 - (8 ln 1 - 1/3) = 8 ln 2 - 7/3 64/3 - 8 ln 4 - (8 ln 2 - 8/3) = 56/3 - 8 ln 2 8 ln 2 - 7/3+ 56/3 - 8 ln 2 = 49/3 or 16 and 1/3
.................................................
......!!!!!!!!...................................
21:21:27 ** 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. **
......!!!!!!!!...................................
RESPONSE --> the graph intersects giving x=2
.................................................
......!!!!!!!!...................................
21:24:21 5.5.44 (was 5.5.40 demand p1 = 1000-.4x^2, supply p2=42x
......!!!!!!!!...................................
RESPONSE --> -.4 x^2 - 42 x + 1000 = 0 so x = 20 and supply with the demand is 840 ( 20*42) C surplus: integral giving you 160 -.4x^2 P surplus: 840-42x so the C surplus = 2133 from 0 to 20 = x
.................................................
......!!!!!!!!...................................
21:24:24 5.5.44 (was 5.5.40 demand p1 = 1000-.4x^2, supply p2=42x
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:24:29 17:42:46
......!!!!!!!!...................................
RESPONSE --> ?
.................................................
......!!!!!!!!...................................
21:24:44 ** 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). *&*&
......!!!!!!!!...................................
RESPONSE --> antiderivative is 160 x - .4 / 3 * x^3
.................................................
"
course mth 272 ñg~Ÿœÿœ¢{ÒݴۄæXý÷á~assignment #010
......!!!!!!!!...................................
21:20:49 5.5.23 (was 5.5.28 area in region defined by y=8/x, y = x^2, x = 1, x = 4
......!!!!!!!!...................................
RESPONSE --> antiderivative: 8 ln x - x^3/3 and then the inverse of that (x^3/3 - 8 ln x ) 8 ln 2 - 8/3 - (8 ln 1 - 1/3) = 8 ln 2 - 7/3 64/3 - 8 ln 4 - (8 ln 2 - 8/3) = 56/3 - 8 ln 2 8 ln 2 - 7/3+ 56/3 - 8 ln 2 = 49/3 or 16 and 1/3
.................................................
......!!!!!!!!...................................
21:21:27 ** 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. **
......!!!!!!!!...................................
RESPONSE --> the graph intersects giving x=2
.................................................
......!!!!!!!!...................................
21:24:21 5.5.44 (was 5.5.40 demand p1 = 1000-.4x^2, supply p2=42x
......!!!!!!!!...................................
RESPONSE --> -.4 x^2 - 42 x + 1000 = 0 so x = 20 and supply with the demand is 840 ( 20*42) C surplus: integral giving you 160 -.4x^2 P surplus: 840-42x so the C surplus = 2133 from 0 to 20 = x
.................................................
......!!!!!!!!...................................
21:24:24 5.5.44 (was 5.5.40 demand p1 = 1000-.4x^2, supply p2=42x
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:24:29 17:42:46
......!!!!!!!!...................................
RESPONSE --> ?
.................................................
......!!!!!!!!...................................
21:24:44 ** 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). *&*&
......!!!!!!!!...................................
RESPONSE --> antiderivative is 160 x - .4 / 3 * x^3
.................................................
"