f.4 amathz..
2008-10-21 2:05 am
The maximum value of the function f(x)=4k+18x-kx^2 (k is a positive constant) is 45. Find k.
回答 (1)
2008-10-21 2:17 am
✔ 最佳答案
f(x) = -kx^2 + 18x + 4k= -k (x^2 - (18/k) x - 4 )
= -k (x^2 - (18/k) x + 81/k^2 - 81/k^2 - 4 )
= -k (x^2 - (18/k) x + 81/k^2) + 81/k + 4k
= -k (x - 9/k)^2 + 81/k + 4k
So max of f(x) = 81/k + 4k = 45
=> 81 + 4k^2 = 45k
=> 4k^2 -45k + 81 = 0
=> k = (45 + (45^2 - 4*4*81)^0.5)/(2*4) or (45 - (45^2 - 4*4*81)^0.5)/(2*4)
=> k = 9 or 2.25
收錄日期: 2021-04-13 16:10:39
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https://hk.answers.yahoo.com/question/index?qid=20081020000051KK01243