S.4 MATH (函數與圖像) 超急急急!! 20點

1:

(a)

長方形ABCD面積 = 8 * 12 cm2 = 96 cm2

ΔCDQ面積 = (1/2) * 12 * (8 – x) cm2 = (48 – 6x) cm2

ΔCBP面積 = (1/2) * 8 * (12 – x) cm2 = (48 – 4x) cm2

ΔAPQ面積 = (1/2) * x * x = x2/2 cm2



ΔPCQ面積

=長方形ABCD面積 -ΔCDQ面積 -ΔCBP面積 -ΔAPQ面積

= [96 – (48 – 6x) – (48 – 4x) – x2/2] cm2

= [96 – 48 + 6x – 48 + 4x – x2/2] cm2

= (10x – x2/2) cm2



(b)

ΔPCQ面積

= (10x – x2/2) cm2

= (20x – x2)/2 cm2

= –(x2 – 20x)/2 cm2

= [102/2 – (x2 – 20x + 102)/2] cm2

= [50 – (x – 10)2/2] cm2



無論x是任何實數，–(x – 10)2/2 ≤ 0

所以，50 – (x – 10)2/2 ≤ 50

ΔPCQ的最大面積 = 50 cm2

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2:

(a)(i)

由A作BC的垂直平分線（三角形的高），與BC 相交於 M。

AM = 4 cm

BM = 24/2 cm = 12 cm

BQ = x cm



PQ/AM = BQ/BM

PQ/(4 cm) = (x cm)/(12 cm)

PQ = x/3 cm



(a)(ii)

QR = (24 – 2x) cm

長方形PQRS面積

= PQ * QR

= (x/3)(24 – 2x) cm2

= (8x – 2x2/3) cm2



(b)

長方形PQRS面積

= (8x – 2x2/3) cm2

= -(2/3)(x2 – 12x) cm2

= [(2/3)(6)2 – (2/3)(x2 – 12x + 62)] cm2

= [24 – (2/3)(x – 6)2] cm2



無論x是任何實數，–(2/3)(x – 6)2 ≤ 0

所以 24 – (2/3)(x – 6)2 ≤ 24

長方形PQRS的最大面積 = 24 cm2

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3:

(a)

盒子闊度和長度的總和 = (84 – 3*8)/2 cm = 30 cm

盒子的闊度 = x cm

盒子的長度 = (30 – x) cm

盒子的高度 = 3 cm



盒子的容積

= 長 * 闊 * 高

= (30 – x) * x * 3 cm2

= (90x – 3x2) cm2

(b)

盒子的容積

= (90x – 3x2) cm2

= -3(x2 – 30x) cm2

= [3(15)2 – 3(x2 – 30x + 152)] cm2

= [675 – 3(x – 15)2] cm2



無論x是任何實數，–3(x – 15)2 ≤ 0

所以 675 – 3(x – 15)2 ≤ 675

最大值時，–3(x – 15)2 = 0

x – 15 = 0

x = 15

30 – x = 15

要使容積達致最大，長度 = 15 cm，闊度 = 15 cm。
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