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(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/701111260077113873405360.jpg (38a) Express y in terms of x(38b) Express A as a function of x. State the domain of the... 顯示更多 (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/701111260077113873405360.jpg (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/701111260077113873405371.jpg (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
最佳解答:
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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急 Functions 40分連答案發問:
(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/701111260077113873405360.jpg (38a) Express y in terms of x(38b) Express A as a function of x. State the domain of the... 顯示更多 (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/701111260077113873405360.jpg (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/701111260077113873405371.jpg (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
最佳解答:
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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