1. The selling price of each copy of a weekly magazine is $x, and the total profit on selling the magazine is $y. It is given that one part of y varies directly as x, and the other part varies directly as the square of x. When x=42, y=79800. When x=50, y=75000.
(a)Express y in term of x.
(b)Find the value of y when x=45.
(c)Use the method of completing the square to write y in the form y=a+b(x+c)^2, where a, b, c are constants.
I know how to calculate(a) and (b) but I don't know how to calaulate(c), so just show the calculation of (c) is enough. thx~
Maths problem of variation[急]
2009-08-04 5:16 am
回答 (2)
2009-08-04 5:42 am
✔ 最佳答案
(a) Let y=Ax+Bx^2Sub. x=42, y=79800. When x=50, y=75000
79800=42A+1764B
75000=50A+2500B
So from (2) 1500=A+50B=>A=1500-50B
Sub, into (1) 79800=42(1500-50B)+1764B
=>79800=63000-2100B+1764B
=>B=-50
A=4000
So y=4000x-50x^2
(b) Sub. x=45, we have y=78750
(c)
y
=4000x-50x^2
=-50(x-40)^2+80000
2009-08-04 5:46 am
As a demonstration, let y = 4x^2 + 6x + 5 [actual value should be those in part (a)].
y = 4x^2 + 6x + 5
= 4(x^2 + 6x/4) + 5
= 4(x^2 + 3x/2) + 5
= 4[(x + 3/4)^2 - (3/4)^2] + 5 ( 3/4 is HALF of 3/2 and having same sign.)
= 4[(x + 3/4)^2 - 9/16] + 5
= 4(x + 3/4)^2 - 9/4 + 5
= 4(x + 3/4)^2 + 11/4.
so in this case, a = 11/4, b = 4 and c = 3/4.
same method can be applied to any quadratic function y.
y = 4x^2 + 6x + 5
= 4(x^2 + 6x/4) + 5
= 4(x^2 + 3x/2) + 5
= 4[(x + 3/4)^2 - (3/4)^2] + 5 ( 3/4 is HALF of 3/2 and having same sign.)
= 4[(x + 3/4)^2 - 9/16] + 5
= 4(x + 3/4)^2 - 9/4 + 5
= 4(x + 3/4)^2 + 11/4.
so in this case, a = 11/4, b = 4 and c = 3/4.
same method can be applied to any quadratic function y.
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