AMATHS一條f.4,考試啦,快快快,40分

Q .1
Total area S = 3600 (given)
A = S - 上下兩條bar - 左右兩條bar
__= S - (12+13)*x - (8+8)*(y-12-13)
__= S - 25*x - 16*y + 400
__= 4000 - 25*x - 16*(S/x)
__= 4000 - 25*x - 57600/x

Q.2
(i) x must be greater than or equal to 16
(ii) y must be greater than or equal to 25,
___ thus x must be smaller than or equal to 3600/25 = 144
therefor, 16 小於或等於 x 小於或等於 144
A = 4000 - 25*x - 57600/x
dA/dx = -25 + 57600/x^2
For dA/dx = 0, 57600/x^2 = 25
____________________x = sqrt(2304) = 48
Thus, a turning point is at 48,
as x passes through 48, dA/dx drops from positive to negative,
therefore, it is a maximum point.
put x = 48 into A, we have maximum value of A = 1600

Q.3 For x larger than or equal 50,
dA/dx is negative, that means A is always decreasing,
the maximum value is at x = 50
A = 4000-25x-57600/x = 1598

Q.4 x / y = x / (3600/x) = x^2 / 3600
thus 4/9 , x/y , 9/16 即 1600 , x^2 , 2025 即 40, x, 45
dA/dx positive within the range, which means A is increasing,
largest value occurs at x= 45
A = 4000-25x-57600/x = 1595

2008-06-19 08:47:07 補充：
我用逗號係因為打唔到 細過或等於個箭嘴

一開始俾左 x*y = 3600, 所以 y = 3600/x
4/9 細過或等於 x/y 細過或等於 9/16
= 4/9 細過或等於 x / (3600/x) 細過或等於 9/16
= 4/9 細過或等於 x 二次 除 3600 細過或等於 9/16
= 3600*4/9 細過或等於 x二次 細過或等於 3600*9/16
= 1600 細過或等於 x二次 細過或等於 2025
= 40 細過或等於 x 細過或等於 45

2008-06-19 19:19:25 補充：
http://www.sendspace.com/file/k9h69n

(1)Width of A = (x - 16), length of A = (y - 25).
Therefore, Area of A = (x - 16)(y - 25) = xy -16y -25x + 400.
Let xy=k, where k is a constant, therefore, y=k/x. Substitute into A, we get
A = k - 16k/x -25x +400.
Put k = 3600, we get A = 3600 -16(3600)/x -25x +400 = 4000 -25x -57600/x....(1)
(2) dA/dx = -25 +57600/x^2. Put dA/dx = 0, we get x^2 = 57600/25 = 48 (-48 is rejected since x is positive).
Put x = 48 into (1), we get A = 4000 -25(48) - 57600/48 = 1600.
(3) Since A is largest when x =48 and decreases when x increases, therefore, for x greater or equal to 50, A is largest when x = 50 , that is A= 4000 -(25)(50) -57600/50 = 4000 - 1250 - 1152 = 1598.