2題f.4 maths,,急

lsp_student，F.4math係唔會學微分架  人地可能唔知你做緊乜野

應該用completing square method.
1ai)

XA/AB = 8 /12 (ΔXAB ~ΔXYZ)

(8-x)/AB = 8 /12 = 2/3

AB = 3(8-x)/2

ii)

Area of rectangle = BA x AY

= [3(8-x)/2](x)

= -3x²/2 + 12x

= -3/2(x²-8x)

= -3/2(x²-8x+4²)-(-3/2)(4²)

= -3/2(x-4)²-(-3/2)(4²)

= -3/2(x-4)² + 24

So, the largest possible area is 24 cm².
2)

Let the smaller number be x and the larger number be 10-x.

x² + (10-x)²

= 2x²-20x+100

= x²-10x+50

= (x²-10x +5²) + 50 - 5²

= (x-5)² +25

When x = 5,

the sum of their squares is a minimum.

So, the two positive number is 5 and 5.

i) angle AXB = angle YXZ
angle XAB = angle XYZ
so triangle ABC is similar to triangle XAB (AAA similarity)
hence, AB/YZ = AX/XY, i.e. AB = AX*YZ/XY = (8-x)*12/8 = 3(8-x)/2 cm

ii) area of rectangle ABCY = AB * AY = x * (24-3x)/2 = 3x(8-x)/2 cm^2

b) set f(x) = 3x(8-x)/2 = (24x - 3x^2)/2
f'(x) = (24 - 6x) / 2
set f'(x) = 0, we have (24-6x) /2 = 0, hence 6x = 24 i.e. x = 4
as f'(x) > 0 for x < 4 and f'(x) < 0 for x > 4
so we have x = 4 provided the maximum value of f(x)
The largest possible area of rectangle ABCY = f(4) = 3*4(8-4)/2 = 24cm^2

希望幫到你

2007-12-17 14:03:26 補充：
tommylautszki
多謝你個reminding.我都剩係記得F.4 有學過微分. (原來係 A.MATH)應該用completing square method 先岩一時口快~ 答深左 Sorry