a-maths (最佳解答取20分)

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

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.