Raymond 發問於 科學及數學數學 · 6 年 前

Question of Variations

1. The average cost of producing a toy ball $C consists of two parts. One part varies directly as the radius r cm of the ball and the second part varies directly as the square of the radius of the ball and inversely as the total number of balls n produced. When r=6 and n=60, C=42; when r=2.4 and n=48, C=12.

(a) Express C in terms of r and n.

(b) Find the radius of the toy ball when 50 balls are produced and the cost of a ball is $24.8

(c) The selling price of a toy ball of radius r cm is $9.6r and the profit from selling the ball is $P.

(i) Express P in terms of r and n.

(ii) Suppose 80 balls are produced. Using the method of completing the square, find the greatest profit from selling a toy ball and the corresponding radius of the ball.

(Give the answer correct to 1 decimal place if necessary.)

2. The cost of making a wooden sphere varies directly as its volume and the cost of painting varies directly as its surface area. The total cost $C of making and painting a wooden sphere of radius 1 m is $200 and that of radius 2 m is $1120.

(a) Express C in terms of r.

(b) A company orders a wooden sphere of radius 2.5 m for decoration. Find the total cost of making and painting the sphere.

(c) The manufacturer of the wooden spheres reduces the cost of making by 10% and the cost of painting by 5%. What is the overall percentage change in the total cost of making and painting a sphere of radius 2.5 m?

1 個解答

評分
  • 6 年 前
    最佳解答

    1. a) C = kr + (m)r^2(1/n) (where k,m are non-zero variable constant)

    Sub r=6, n=60, C=42

    42 = 6k + (36m)(1/60)

    42 = 6k + 3m/5 ------- (i)

    Sub r=2.4, n=48, C=12

    12 = 2.4k + (5.76m)(1/48)

    12 = 2.4k + 3m/25 -------- (ii)

    (ii) x 5 - (i) :

    18 = 6k

    k = 3

    m = 40

    Therefore C = 3r + (r^2)(40/n)

    b) Sub C=24.8, n=50

    24.8 = 3r + (r^2)(40/50)

    24.8 = 3r + 0.8(r^2)

    124 = 15r + 4(r^2)

    r = 4

    c) i) P = 9.6r - 3r - (r^2)(40/n)

    = 6.6r - (r^2)(40/m)

    ii) P = 6.6r - (r^2)(40/80)

    = -1/2 (r^2) + 6.6r

    = -1/2 [(r^2) - 13.2r)]

    = -1/2 [(r^2) - 6.6]^2 + 21.78

    Therefore greatest profit is $21.8 when the redius is 6.6

    2. a) C = k(Volume) + m(Surface area)

    (where k,m are non-zero variable constant)

    C = k(4/3)(π)r^3 + m(4)(π)r^2

    Sub C=200, r=1

    200 = (4π/3)k + 4πm --------- (i)

    Sub C=1120, r=2

    1120 = (32π/3)k + 16πm ---------- (ii)

    (ii) - (i) x 4 :

    320 = (16π/3)k

    k = 60/π = 19.1

    m = 30/π = 9.5

    Therefore C = 80(r^3) + 120(r^2)

    b) Sub r=2.5

    C = $2000

    c) C(original) = $2000

    C(reduced) = $1837.5

    Percentage change = (2000-1837.5)/2000 x 100%

    =-8.125%

    資料來源: 在下
還有問題嗎?立即提問即可得到解答。