a.maths.....

2008-03-28 12:59 am

1).A piece of wire of length 8cm is cut into two pieces and each piece is bent

into a square.If the total area enclosed is minimum,find the length of the sides

of each square.

2).ABCD is a rectangle.AB = 8cm,AD = 6cm and AP=BQ=CR=DS= x cm.

Find the value of x such that the area of the parallelogram PQRS is minimum.

thx......

回答 (1)

NCY

2008-03-28 1:18 am

✔ 最佳答案

1)Let the one of the part be x cm,and the other part be (8 – x) cm,
Also,let A be the area .
A = (x/4)2 + [(8 –x)/4]2
= x2/16 + (64 – 16x + x2)/16
= (64 – 16x + 2x2)/16
= (x2 – 8x + 32)/8
= (x2 – 8x + 16 + 16)/8
= [(x – 4)2 + 16]/8

∴When x = 4,A gains a minimum.
∴each side of the squares = 4/4 = 1cm

2)Let A be the area,
The area of ABCD = AB AD = 8 6 = 48cm2
A = 48 ﹣[2x(6 ﹣x) / 2] ﹣[2x(8 ﹣x) / 2]
 = 48 ﹣x(6 ﹣x) ﹣x(8 ﹣x)
 = 48 ﹣6x + x2 ﹣8x + x2
 = 2x2 ﹣14x + 48
 = 2(x2 ﹣7x) + 48
 = 2[x2 ﹣7x + (7/2)2 ﹣(7/2)2 ] + 48
 = 2(x2 ﹣7/2)2 ﹣2(49/4) + 48
 = 2(x2 ﹣7/2)2 + 47/2
∴when the value of x = 7/2,A gains the minimum

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