1)
The temperature on a particular day can be modeled by the function
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Where t is the time elapsed, in hours, after 5am and C is the temperatures in degrees Celsius.
Find the maximum temperature and the time when this temperature is reached.


2)
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.


3)
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?

繳交日期係12/1

[ 本帖最後由 bbq55 於 2007-1-6 08:27 PM 編輯 ]

第1題死圖,,幫你唔到..SOR

2. 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.

3. 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

THANK YOU SO MUCH!!!
第1條個圖係
c = -4cos(πt)/12 +14
希望幫一幫我啦

問題要你找出curve的最高點
即係 cos(πt)= -1

c= 43/3
t=1,3,5,7,9...