F.4 Maths(HELP!!!)(10 marks)
2010-12-21 9:20 pm
It is given that y/2=2x^2-bx+3b-16, where b is a constant, if the minimum value of y is zero, find the values of b.
回答 (2)
2010-12-21 9:50 pm
2010-12-21 9:41 pm
y/2=2x^2-bx+3b-16
y=4x^2-2bx+6b-32
dy/dx = 8x-2b; d^2 y/dx^2 = 8>0
y attains its absolute minimum if dy/dx=0 --> 8x-2b=0 -->8x=2b-->b=4x
When b=4x,
y=4x^2-2(4x)x+6(4x)-32
=-4x^2+24x-32
=-4(x^2-6x+8)
=-4(x-4)(x-2)
As the absolute minimum value of y is zero, we have
y=-4(x-4)(x-2)=0
x=2 or 4
b=8 or 16 respectively
So, the solution is (1) x=2, b=8 ; (2) x=4, b=16.
y=4x^2-2bx+6b-32
dy/dx = 8x-2b; d^2 y/dx^2 = 8>0
y attains its absolute minimum if dy/dx=0 --> 8x-2b=0 -->8x=2b-->b=4x
When b=4x,
y=4x^2-2(4x)x+6(4x)-32
=-4x^2+24x-32
=-4(x^2-6x+8)
=-4(x-4)(x-2)
As the absolute minimum value of y is zero, we have
y=-4(x-4)(x-2)=0
x=2 or 4
b=8 or 16 respectively
So, the solution is (1) x=2, b=8 ; (2) x=4, b=16.
收錄日期: 2021-04-23 20:42:26
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101221000051KK02241