# 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

• 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%

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