F.4 Maths---Variations

The cost of a souvenir of surface area Acm(square) is $ C. It is giventhat C is the sum of two parts, one part varies directly as A while theother part varies
direct as A(square) and inversely as n, where n is the number of souvenir
produced. When A=50 and n=500, C=350; when A=20 and n=400, C=100.

(a) Express C in terms of A and n.
Ans:
let C=k1A+k2A^2/n ,where k1 and k2 are constant,

When A=50 and n=500, C=350
350=k1(50)+k2(50)^2 /500
350=50k1+5k2--(1)

when A=20 and n=400, C=100
100=k1(20)+k2(20)^2/400
100=20k1+k2--(2)

(1)-5(2):
350-500=50k1+5k2-100k1-5k2
-150=-50k1
k1=3

put k1=3 into (1),
350=50(3)+5k2
k2=40//

so C=3A+40A^2/n//

(b) The selling price of a souvenir of surface area A cm(square) is $8A and the profit in selling the souvenir is $P .

Ans:
P
=Selling Price - Cost
=8A-(3A+40A^2/n)
=5A-40A^2/n//

To Be Continued......

a. C = Ah + (A^2)(k)/n (h.k is not = 0)
when A=50 and n=500, C=350 ,
350 = 50h + 2500k/500
350 = 50h + 5k
70 = 10h + k -----------------(1)
when A=20 and n=400, C=100 ,
100 = 20h + 400k/400
100 = 20h + k ------------------(2)
(2) - (1)
30 = 10h
h = 3
sub h = 3 into (1) ,
70 = 10(3) + k
k = 40
C = 3A + 40(A^2)/n

b.(1) P = 8A - C
P = 8A - [3A + 40(A^2)/n]
P = 8A - 3A - 40(A^2)/n
P = 5A - 40(A^2)/n
(2) Since P : n = 5:32 , sub P = 5 , n = 32
5 = 5A - 40(A^2)/32
5 = 5A - 5A^2/4
20 = 20A - 5A^2
0 = -5A^2 + 20A - 20
A = 2
A : n = 2 : 32 = 1 : 16
(3) Since A : n = 1 : 16 , when n = 500 , A = 31.25
C = 3A + 40(A^2)/n
C = 3(31.25) + 40(31.25^2)/500
C = 93.75 + 78.125
C = 171.875
The selling price = 8A = 250
The profit of the souvenir = 250 - 171.875
= 78.125
No . The profit of $ 100 cannot be made in selling a souvenir .
(4) when n = 400
P = 8A - 3A - 40A^2/400
P = 5A - A^2/10
P = -1/10 (A^2 - 50A)
P = -1/10 [A^2 - 50A + (50/2)^2 - (50/2)^2]
P = -1/10 (A-25)^2 + [(1/10)(50/2)^2]
P = -1/10 (A-25)^2 + 62.5
Therefore , A = 25 , the greatest profit = $ 62.5 .

lee條好似係past paper
我之前都唔識做
但係後來發覺唔難

a)
照代哂d數
得出 350 = 50k1 + 5k2 同 100 = 20k1 + k2
之後解左 k1 = 3 , k2 = 40
所以 C = 3A + 40A^2 / n

b1)
Profit = 賣價 減 成本
所以 P = 8A - C = 5A -40A^2 / n

b2)
lee part最難
P : n = 5 : 32 即 P = 5n / 32
代入去b1)條式度:
5n / 32 = 5A - 40A^2 / n
之後兩邊都乘32n 目的係消走d分數
得出 5n^2 = 160An - 1280A^2
搬哂去一邊同埋約簡d數
見到 256A^2 - 32An + n^2 = 0
之後用中2學過o既方法factorize
(16A - n)^2 = 0
16A - n =0
A / n = 1 / 16
咁 A : n 咪係 1: 16囉

b3)
佢比左n = 500
用返b2)就計倒A
A = 500 / 16 = 31.25
再計P = 5A -40A^2 / n = 5(31.25) - 40(31.25)^2 / 500 = 78.125
78.125細過100
所以答案係cannot

b4)
P = 5A - 40A^2 /400 = 5A - 1/10 A^2
將 -1/10 搬出去
P = -1/10 ( A^2 - 50A )
= -1/10 ( A^2 - 50A + 25^2 - 25^2 )
= -1/10 ( A - 25 )^2 + 1/10 x 25^2
= -1/10 ( A - 25 )^2 + 62.5
當A = 25 , P係最中
所以答案係 $62.5

C=aA + bA^2/n
When A=50,n=500,C=350. That is 350=50a +5b.........(1)
When A=20,n=400, C=100. That is 100 =20a +b .........(2)
(1) - 5 x (2), we get 350 -500 = -50a, therefore, a =3 and b=40
That is C=3A + 40A^2/n.................(3)
When Cost = Selling price - profit = 8A - P. Substitute into (3), we get
8A - P = 3A + 40A^2/n, that is P =5A - 40A^2/n.........(4)
Now P:n=5:32, that is P=5n/32, substitute this into (4), we get
5n/32 = 5A - 40A^2/n or n/32 = A - 8A^2/n, n=32A - 256A^2/n
that is n^2 = 32nA -256A^2 or n^2 -32nA + 256A^2 = 0, therefore,
(n-16A)^2 = 0. That is n=16A, therefore, A:n = 1:16.
Substitute this relation into (4), we get P = 5n/16 - 5n/32 = 5n/32.
When n=500, P=5 x 500/32 = 78.125, therefore, a profit of $100 cannot be made.

2008-06-11 18:06:40 補充：
For n=400, equation (4) becomes P=5A - A^2/10. For quadratic function y=ax^2 + bx +c , y max. happens when x =-b/2a. Therefore, P is maximum when A = -5/(-2/10) = 25, P maximum = 5(25) - (25)(25)/10 = 125 - 62.5 = 62.5