maths( f.5 )

a) Let the vertical stick be x and the horizontal one will be 76-x.

Area of the kite: 2x〔( 76-x )/ 2 〕
=76x-x^2
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b) 76x - x^2 ≥ 720
x^2 - 76x + 720 ≤ 0
11.1 ≤ x ≤ 64.9 (corr. to 3 sig. fig.)

2010-10-11 22:57:48 補充：
Sorry for texting wrong answer
I think this might be right:

2010-10-11 22:57:53 補充：
Let the vertical stick be x and the horizontal one will be 76-x.

Area of the kite: 2(x〔( 76-x )/ 2 〕)/2
=38x-1/2x^2
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b) 38x-1/2x^2 ≥ 720
x^2-76x+1440 ≤ 0
36 ≤ x ≤ 40

Let the horizontal stick = x cm. ( In fact it does not matter which stick is assumed to be x cm.)
so the vertical stick = (76 - x) cm. long.
Area of kite = 2 x Area of triangle with height (x/2) and base (76 - x)
= 2[(x/2)(76 - x)]/2 = x(76 - x)/2
Now x(76 - x)/2 > 720
76x - x^2 > 1440
x^2 - 76x + 1440 < 0
(x - 40)(x - 36) < 0
so 36 < x < 40.