Problem on minimization
2011-01-22 1:25 am
圖片參考：http://img263.imageshack.us/img263/4448/problemonminmization.png
A particle is moving in a rectangle ABCD with AB = 80 cm and AD = 60 cm. Given that the particle moves at 2 cm/s on the boundary of ABCD and moves at 1 cm/s inside the rectangle. Assume that the particle moves in straight lines. It starts from point A, moves towards the side DC, reaches a point P on DC and then to C. Let DP = x cm.
(a) Find the length of the path travelled by the particle from A to C in terms of x. (Ans: [√(x^2 + 3600) + (80 - x)] cm)
(b) Find the length of the path which takes the minimum time for the particle to travel from A to C. (Ans: 115 cm)
(c) What is the minimum time needed for the particle to travel from A to C? (Ans: 92.0 s)
回答 (2)
2011-01-22 2:41 am
✔ 最佳答案
a)AP + PC= √(x² + 60²) + (80 - x)= √(x² + 3600) + (80 - x) cmThe minimum time = √(x² + 3600) + (80 - x)/2= √(x² + 3600) - x/2 + 40Let y + 40 = √(x² + 3600) - x/2 + 40y = √(x² + 3600) - x/24y² + 4yx + x² = 4x² + 144003x² - 4yx + 14400 - 4y² = 0 ........(*)△ = 16y² - 4*3(14400 - 4y²) ≥ 0 for real x.64y² ≥ 172800y ≥ 30√3
b) Sub y = 30√3 to (*) :3x² - 4*30√3 x + 14400 - 4(30√3)² = 0x² - 40√3 x + 1200 = 0(x - 20√3)² = 0x = 20√3The length of the path which takes the minimum time for the particle to travel from A to C= √((20√3)² + 3600) + (80 - 20√3)= 115.0 cm
c) The minimum time = 30√3 + 40 = 92.0 s
2011-01-21 22:58:31 補充：
Oh sorry , I haven't learnt differentiation so I use such method.
2011-01-22 5:54 am
Inequality can be a method...
but this is an exercise extracted from app. of differentiation...
but this is an exercise extracted from app. of differentiation...
收錄日期: 2021-04-21 22:25:28
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