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a-maths (最佳解答取20分)
發問:
1) A plane is to fly 3000 km at a speed of v km/h. when flying at v km/h, the plane consumes fuel at the rate of (50 + 0.00001v3次方) litres per hour.Find the speed for the greatest fuel economy and the amount of fuel used at this speed.2) a manufacturer's total cost function is given byC =... 顯示更多 1) A plane is to fly 3000 km at a speed of v km/h. when flying at v km/h, the plane consumes fuel at the rate of (50 + 0.00001v3次方) litres per hour. Find the speed for the greatest fuel economy and the amount of fuel used at this speed. 2) a manufacturer's total cost function is given by C = (q^2)/4 + 3q + 400 where c is the total cost of producing q units. at what level of output will average cost per unit be a minimum? What is this minimum?
1. Let P be the amount of fuel consumed for the whole journey, t be the time taken for it. t = 3000/v P = (50 + 0.00001*v3)*t = 3000/v*(50 + 0.00001*v3) = 150000/v + 0.03v2 To consume the minimum amount of fuel, dP/dv = 0 and d2P/dv2 > 0 dP/dv = -150000/v2 + 0.06v = 0 -150000 + 0.06v3 = 0 (for v not equal to 0) v3 = 150000/0.06 = 2500000 v = 135.72 km/h d2P/dv2 = 300000/v3 + 0.06 When v = 135.72, d2P/dv2 = 0.18 > 0 So the fuel consumption is minimum if the speed is 135.72 km/h. The corresponding amount of fuel consumed = 150000/135.72 + 0.03*135.722 = 1657.81 litres. 2. C = q2/4 + 3q + 400 Let A be the average cost per unit, A = C/q = q/4 + 3 + 400/q For the average cost to be minimum, dA/dq = 0 and d2A/dq2 > 0 dA/dq = 1/4 - 400/q2 = 0 400 / q2 = 1/4 q2 = 400*4 = 1600 q = 40, since q >= 0 d2A/dq2 = 800/q3 when q = 40, d2A/dq2 = 800/403 = 0.0125 > 0 So A is min. if q = 40. When 40 units are produced, the average cost per unit will be the minimum. Minimum average cost per unit = q/4 + 3 + 400/q = 40/4 + 3 + 400/40 = 10 + 3 + 10 = $23
其他解答:
1) First of all, find out the expression of the total amount of fuel consumed in the 3000km journey in terms of v. Suppose the plane fly with a constant velocity throughout the whole journey, then the rate of fuel consumption is (50 + 0.00001v^3) litres per hour which is given by the question. And the time for the 3000km journey is given by: 3000/v hours Therefore, total amount of fuel consumed in the 3000km journey, C, is: C = (50 + 0.00001v^3)(3000/v) C = 150000/v + 0.03v^2 Now, dC/dv = -150000/v^2 + 0.06v d^2C/dv^2 = 300000/v^3 + 0.06 For dC/dv = 0: -150000/v^2 + 0.06v = 0 0.06v = 150000/v^2 v^3 = 2500000 v = 135.7 km/h (correct to 1 d.p.) Checking for d^2C/dv^2 at v = 135.7: d^2C/dv^2 (v = 135.7) = 300000/v^3 + 0.06 = 300000/250000 + 0.06 > 0 Thus, v = 135.7 is a minumum for C (greatest fuel economy). So amount of fuel used at this speed is: 150000/135.7 + 0.03(135.7)^2 = 1658 litres (correct to the nearest litres) 2) First of all, average cost per unit is C/q, i.e. R = q/4 + 3 + 400/q (denote the average cost per unit by R) To minimize R, dR/dq = 1/4 - 400/q^2 d^2R/dq^2 = 800/q^3 Now, dR/dq = 0 1/4 - 400/q^2 = 0 1/4 = 400/q^2 q^2 = 1600 q = 40 Checking with d^2R/dq^2, it is positive for any positive value of q, so q = 40 is a minumum point for R. Hence, 40 units should be produced and the total cost will be: 40/4 + 3 + 400/40 = 23 And the minumum cost per unit will be 23/40 = 0.575 per unit.0D7DAC4E7B8CAAC5
a-maths (最佳解答取20分)
發問:
1) A plane is to fly 3000 km at a speed of v km/h. when flying at v km/h, the plane consumes fuel at the rate of (50 + 0.00001v3次方) litres per hour.Find the speed for the greatest fuel economy and the amount of fuel used at this speed.2) a manufacturer's total cost function is given byC =... 顯示更多 1) A plane is to fly 3000 km at a speed of v km/h. when flying at v km/h, the plane consumes fuel at the rate of (50 + 0.00001v3次方) litres per hour. Find the speed for the greatest fuel economy and the amount of fuel used at this speed. 2) a manufacturer's total cost function is given by C = (q^2)/4 + 3q + 400 where c is the total cost of producing q units. at what level of output will average cost per unit be a minimum? What is this minimum?
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最佳解答:1. Let P be the amount of fuel consumed for the whole journey, t be the time taken for it. t = 3000/v P = (50 + 0.00001*v3)*t = 3000/v*(50 + 0.00001*v3) = 150000/v + 0.03v2 To consume the minimum amount of fuel, dP/dv = 0 and d2P/dv2 > 0 dP/dv = -150000/v2 + 0.06v = 0 -150000 + 0.06v3 = 0 (for v not equal to 0) v3 = 150000/0.06 = 2500000 v = 135.72 km/h d2P/dv2 = 300000/v3 + 0.06 When v = 135.72, d2P/dv2 = 0.18 > 0 So the fuel consumption is minimum if the speed is 135.72 km/h. The corresponding amount of fuel consumed = 150000/135.72 + 0.03*135.722 = 1657.81 litres. 2. C = q2/4 + 3q + 400 Let A be the average cost per unit, A = C/q = q/4 + 3 + 400/q For the average cost to be minimum, dA/dq = 0 and d2A/dq2 > 0 dA/dq = 1/4 - 400/q2 = 0 400 / q2 = 1/4 q2 = 400*4 = 1600 q = 40, since q >= 0 d2A/dq2 = 800/q3 when q = 40, d2A/dq2 = 800/403 = 0.0125 > 0 So A is min. if q = 40. When 40 units are produced, the average cost per unit will be the minimum. Minimum average cost per unit = q/4 + 3 + 400/q = 40/4 + 3 + 400/40 = 10 + 3 + 10 = $23
其他解答:
1) First of all, find out the expression of the total amount of fuel consumed in the 3000km journey in terms of v. Suppose the plane fly with a constant velocity throughout the whole journey, then the rate of fuel consumption is (50 + 0.00001v^3) litres per hour which is given by the question. And the time for the 3000km journey is given by: 3000/v hours Therefore, total amount of fuel consumed in the 3000km journey, C, is: C = (50 + 0.00001v^3)(3000/v) C = 150000/v + 0.03v^2 Now, dC/dv = -150000/v^2 + 0.06v d^2C/dv^2 = 300000/v^3 + 0.06 For dC/dv = 0: -150000/v^2 + 0.06v = 0 0.06v = 150000/v^2 v^3 = 2500000 v = 135.7 km/h (correct to 1 d.p.) Checking for d^2C/dv^2 at v = 135.7: d^2C/dv^2 (v = 135.7) = 300000/v^3 + 0.06 = 300000/250000 + 0.06 > 0 Thus, v = 135.7 is a minumum for C (greatest fuel economy). So amount of fuel used at this speed is: 150000/135.7 + 0.03(135.7)^2 = 1658 litres (correct to the nearest litres) 2) First of all, average cost per unit is C/q, i.e. R = q/4 + 3 + 400/q (denote the average cost per unit by R) To minimize R, dR/dq = 1/4 - 400/q^2 d^2R/dq^2 = 800/q^3 Now, dR/dq = 0 1/4 - 400/q^2 = 0 1/4 = 400/q^2 q^2 = 1600 q = 40 Checking with d^2R/dq^2, it is positive for any positive value of q, so q = 40 is a minumum point for R. Hence, 40 units should be produced and the total cost will be: 40/4 + 3 + 400/40 = 23 And the minumum cost per unit will be 23/40 = 0.575 per unit.0D7DAC4E7B8CAAC5
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