1. In a competition, the competitors have to use the minimum area of cardboard to make a square-based cuboid container with no lid. If the volume is 256cm3, find the dimensions of the container.
2. What is the shortest distance between the line 2x-y+1=0 and (3,-1)?
3. A store estimtes that when the temperature is t degree celcius (8 </= t </= 20), the daily number of customers N(t) (in hundreds) buying hot drink can be modelled by the following expression: N(t)=40te^(-0.08t)
a) at what temperature will the number of customers be the greatest? Find also the corresponding number of people and give the answer cor. to nearest integer.
b) at what temperature will the number of customers be the lowest?
4. When the selling price of pork is $x /kg, the daily sales (in kg) is S(x)= (500/(x+20))-1
a) if R(x) (in $) is the revenue when the price is $x, express R(x) in terms of x,
b) which selling price of pork per kg will result in the greatest revenue?
Application of differentiation
2010-04-19 8:36 am
回答 (1)
2010-04-30 9:29 pm
✔ 最佳答案
1. let A cm^3 be the area of the cardboard usedlet x cm be the side of the square base
let h cm be the height of the cuboid
Volume = 256
(x^2)(h) = 256 , h = 256/x^2
A = x^2 + 4hx
= x^2 + 4 ( 256 /x^2)(x)
= (x^3 + 1024) / x
dA/dx = [ x(3x^2) - (x^3+1024) ] /x^2
= (2x^3 - 1024) / x^2
when dA/dx = 0 , x^3 = 512
(x - 8)(x^2 + 8x + 64 ) = 0
x = 8
d^2A/dx^2 = 2 (x^3+1024)/x^3 , at x = 8 , d^2A/dx^2 = 6 > 0
Therefore, at x = 8 , h = 4 , the area of cardboard used is mininum
The dimensions of the cuboid is 8cm * 8cm * 4cm
2. The shortest distance
= | [ ( 2(3) - (-1) + 1 ) / sqrt[ 2^2 + (-1)^2] |
= (8 / sqrt (5))
= (8/5) sqrt (5)
3. (a) N(t) = 40te^(-0.08t)
N'(t) = 40 d/dt [ te^(-0.08t) ]
= 40 [ t d/dt ( e^(-0.08t) ) + e^(-0.08t) ]
= 40 [ t ( e^(-0.08t) ) (-0.08) + e^(-0.08t) ]
= 40e^(-0.08t) (1 - 0.08 t)
Therefore , for N'(t) = 0 , t = 1/0.08 = 12.5
for t < 12.5 slightly, N'(t) > 0,
for t > 12.5 slightly, N'(t) < 0,
Therefore at t= 12.5 N(t) is maximum
The greatest numbers of customers at 12.5°C
= N(12.5)
= 184 ( corr. to the nearest integer )
(b)
for 8≤t<12.5 , N'(t) > 0
for 12.5 < t≤ 20 , N'(t) < 0
Therefore, the lowest number of customers is attained either at t= 8 or 20
N(8) = 169 ( corr. to the nearest integer )
N(20) = 162 ( corr. to the nearest integer )
At 20°C, the number of customers is the lowest
4.(a) R(x) = S(x)‧x
= 500x / (x+20) - x
(b) R'(x) = 10000/(x+20)^2 - 1
For R'(x) = 0 ,
(x+20)^2 = 10000
x = 80 or -120 (rejected)
x < 80 slightly , R'(x) > 0,
x > 80 slightly , R'(x) < 0,
Therefore, at x = 80 , R(x) is the greatest
Selling price of pork is $ 80 / kg to result in the greatest revenue
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