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?
One more question:
- 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%