course MTH 272 04/11/10, 1900 assignment #010010. `query 10
......!!!!!!!!...................................
13:47:55 5.5.23 (was 5.5.28 area in region defined by y=8/x, y = x^2, x = 1, x = 4
......!!!!!!!!...................................
RESPONSE --> 8/x = x^2 0=x^2 - (8/x) 0= (1/x) ((x^3) - 8) x^3 - 8 = 0 x^3=8 x=2 x=1,2,4 (x^2 - (8/x)) dx on [1,2] + ((8/x) - x^2) dx on [2,4] ((x^3)/3 - 8 ln x) + (8 ln x - (x^3)/3) [((8/3) - 8 ln 2) - ((1/3) - 8 ln 1)] + [((-64/3) + 8 ln 4) - ((-8/3) + 8 ln 2)] [((7/3) - 8 ln 2) + ((-56/3) + 8 ln 2)] -49/3 confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
13:48:10 These graphs intersect when 8/x = x^2, which we solve to obtain x = 2. For x < 2 we have 8/x > x^2; for x > 2 the inequality is the reverse. So we integrate 8/x - x^2 from x = 1 to x = 2, and x^2 - 8 / x from x = 2 to x = 4. Antiderivative are 8 ln x - x^3 / 3 and x^3 / 3 - 8 ln x. We obtain 8 ln 2 - 8/3 - (8 ln 1 - 1/3) = 8 ln 2 - 7/3 and 64/3 - 8 ln 4 - (8 ln 2 - 8/3) = 56/3 - 8 ln 2. Adding the two results we obtain 49/3. **
......!!!!!!!!...................................
RESPONSE --> Instead of having a positive 49/3, I found a negative 49/3. I switched two terms around in the second function so that its format would mirror that of the first function. It is likely that this is the area where I misplaced a negative sign. ------------------------------------------------ Self-critique rating #$&*ent: 2
.................................................
......!!!!!!!!...................................
13:50:24 5.5.44 (was 5.5.40 demand p1 = 1000-.4x^2, supply p2=42x
......!!!!!!!!...................................
RESPONSE --> p1=p2 1000 - .4x^2 = 42x 0=.4x^2 + 42x - 1000 .4(x^2 + 105x - 2500) .4(x + 125)(x - 20) x=-125,20 x=20 Demand p1=1000 - .4(20^2) p1=840 Supply p2=42(20) p2=840 Consumer Surplus (on the interval [0,20]) integral [demand function - price] dx 1000 - .4x^2 - 840 1000x - (4/10)(1/3)x^3 - 840x 160x - (2/15)x^3 [ 160(20) - (2/15)(20^3) ] - 0 2133.33 Producer Surplus (on the interval [0,20]) integral [price - supply function] dx 840 - 42x 840x - 21x^2 [840(20) - 21(20^2)] - 0 8400 confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
13:51:40 1000-.4x^2 = 42x is a quadratic equation. Rearrange to form -.4 x^2 - 42 x + 1000 = 0 and use the quadratic formula. You get x = 20 At x = 20 demand is 1000 - .4 * 20^2 = 840, supply is 42 * 20 = 840. The demand and supply curves meet at (20, 840). The area of the demand function above the equilibrium line y = 840 is the integral of 1000 - .4 x^2 - 840 = 160 - .4 x^2, from x = 0 to the equlibrium point at x = 20. This is the consumer surplus. The area of the supply function below the equilibrium line is the integral from x = 0 to x = 20 of the function 840 - 42 x. This is the producer surplus. The consumer surplus is therefore integral ( 160 - .4 x^2 , x from 0 to 20) = 2133.33 (antiderivative is 160 x - .4 / 3 * x^3). *&*&
......!!!!!!!!...................................
RESPONSE --> ------------------------------------------------ Self-critique rating #$&*ent: 3
............................................"