1 Maths question
2011-08-19 3:26 am
圖片參考：http://imgcld.yimg.com/8/n/HA00656648/o/701108180112713873445490.jpg
In the above figure, P(a,b) is a moving point on line L1, and OBPA(shaded region) is a rectangle. Given that the area of OBPA =
圖片參考：http://imgcld.yimg.com/8/n/HA00656648/o/701108180112713873445491.jpg
Find the maximum and minimum area of the shaded area and the corresponding coordinates of P.
THX a lot!!~
回答 (2)
2011-08-19 5:56 am
✔ 最佳答案
By method of completing square, Area = (12a - 3a^2)/2 = - 3/2(a^2 - 4a) = -3/2[(a - 2)^2 - 4] = - 3/2(a - 2)^2 + 6So the max. area = 6 when a = 2.
Area = ab = 6 = 2b, so b = 3. P is (2,3).
Min. area = 0 when P is (4,0) or ( 0, 6).
2011-08-19 7:22 am
Let A=area of OBPA,P=perimeter of OBPA,
x=length of OBPA, y=height of OBPA
A=xy-------(1)
P=2x+2y------(2)
From(2), y=-x+P/2------(3)
Sub.(3) into (1)
A=-x^2+(P/2)x
A'=-2x+P/ 2
A"=-2
When x is max. or min. , A'=0
-2x+P/2=0
P/2=2x
P=4x
Sub. P=4xinto (2)
4x=2x+2y
x=y
SinceA" is negative, so when x=y, A will be max.
So,sub. x=y into L1
3x+2x-12=0
x=12/5
So,when x=y=12/5, the area of OBPA will be max, and it is (12/5)^2=144/25
Also,we can easily see that when x or y equal to 0, that area will be 0. So, themin. area is 0.
x=length of OBPA, y=height of OBPA
A=xy-------(1)
P=2x+2y------(2)
From(2), y=-x+P/2------(3)
Sub.(3) into (1)
A=-x^2+(P/2)x
A'=-2x+P/ 2
A"=-2
When x is max. or min. , A'=0
-2x+P/2=0
P/2=2x
P=4x
Sub. P=4xinto (2)
4x=2x+2y
x=y
SinceA" is negative, so when x=y, A will be max.
So,sub. x=y into L1
3x+2x-12=0
x=12/5
So,when x=y=12/5, the area of OBPA will be max, and it is (12/5)^2=144/25
Also,we can easily see that when x or y equal to 0, that area will be 0. So, themin. area is 0.
參考: Hope I can help you! ^_^ (From me)
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