M1 Application of Differention

[匿名]

2012-04-30 2:26 am

1.The net profit P(x) in thousand dollars per day of a boutique is governed by P(x)=20(e^0.03x) - x -25 for 0≦ x≦ 100,where x(in thousands) is the number of customers of the boutique on one day.
(a)If no customers visits the boutique on a day,find the loss of the boutique on that day.
(b)Determine the number of customers of the boutique on a day so that the greatest daily profit is attained.

2.At any time t(in hours),the relationship between the number N of tourists at a ski-resort and the air temperature θ°C can be modeled by N=2930-(θ+400)[ln(θ+49)]^2，
(a)Express dN/dt in terms of θ and dθ/dt
(b)At a certain moment,the air temperature is -40°C and it is falling at a rate of 0.5°C per hour.Find, to the nearest integer,the rate of increase of the number of tourists at that moment.
[HKALE M&S 96]

回答 (1)

用戶25975

2012-04-30 4:33 am

✔ 最佳答案

1a) Sub x = 0:

P(0) = 20 - 0 - 25 = -5

So the loss is 5 thousand dollars

b) P'(x) = 0.6e0.03x - 1

P"(x) = 0.018e0.03x

For P'(x) = 0:

e0.03x = 5/3

P"(x) = 0.03 > 0

Hence when e0.03x = 5/3, P(x) attains a min.

So when x = 100, P(x) will be max. since after e0.03x = 5/3, P(x) is increasing.

Max. = P(100) = 20e3 - 100 - 25 = 277 thousand (corr. to nearest thousand)

2a) dN/dt = (dN/dθ) (dθ/dt)

= {(θ + 400) d [ln (θ + 49)]2/dθ + [ln (θ + 49)]2} dθ/dt

= [(θ + 400) x 2ln (θ + 49) (1/θ) + [ln (θ + 49)]2] dθ/dt

= [2 (1 + 400/θ) ln (θ + 49) + [ln (θ + 49)]2] dθ/dt

b) When θ = 40:

dN/dθ = [2 (1 - 400/40) ln (-40 + 49) + [ln (-40 + 49)]2 = -34.72

Hence dN/dt = (-34.72) x (-0.5) = 17.4 per hour

參考: 原創答案

👍 0 👎 0