tfzoofss asked in 科學及數學數學 · 8 years ago

急 Functions 40分連答案

(38) The fig. shows the dimensions of an L-shaped cardboard. The perimeter of the cardboard is 30 cm. Let A cm^2 be the area of the cardboard

圖片參考:http://imgcld.yimg.com/8/n/HA05425562/o/7011112600...

(38a) Express y in terms of x

(38b) Express A as a function of x. State the domain of the function

(38c) Find the maximum area of the cardboard and the cooresponding value of x

(40) In the fig., triangle ABC is an isosceles triangle, where AB = 8 cm and AC = BC = 5 cm. A rectangle PQRS is inscribedin triangle ABC and SC = RC. Suppose SR = x cm

圖片參考:http://imgcld.yimg.com/8/n/HA05425562/o/7011112600...

(40a) Show that the area of PQRS is (3x - (3/8)x^2) cm^2

(40b) Find the area of the largest rectangle that can be inscribed in triangle ABC

Answers:

(38a) y = 11 - x

(38b) A = -x^2 + 12x + 3; the domain is the collectiom of the real number x, where 0 < x < 11

(38c) max. area = 39 cm^2 when x = 6

(40b) 6 cm^2

1 Answer

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    8 years ago
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    38a.

    y + 3 + 1 + (x+3) + (y+1) + x = 30

    2x + 2y + 8 = 30 => x + y = 11

    y = 11 - x

    38b.

    A = x(y+1) + 3(1)

    A = x(11-x+1) + 3

    A = x(12-x) + 3

    A = 12x - x^2 + 3 and 0 < x < 11

    38c.

    by completing the square

    A = -(x^2 - 12x - 3)

    A = -(x^2 - 12x + 36 -36 - 3)

    A = -(x^2 - 12x + 36) + 39

    A = -(x - 6)^2 + 39

    as -(x - 6)^2 always < 0

    Maximum of A = 39 when x = 6

    40a.

    height of C from AB = sqrt[5^2 - (8/2)^2] = 3

    triangle ABC similar to triangle SRC

    CS/AC = SR/AB = RC/BC = height of C from SR / height of C from AB

    x/8 = height of C from SR / 3

    height of C from SR = 3x/8

    height of rectangular PQRS = 3 - (3x/8)

    area of PQRS = x[3 - (3x/8)]

    40b.

    by completing the square

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

    A = -(3/8)[x^2 - 8x + 16 - 16]

    A = -(3/8)[(x-4)^2 - 16]

    A = -(3/8)(x-4)^2 + 6

    as -(x - 4)^2 always < 0

    Maximum of A = 6 when x = 4

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