標題:
f.4 amathz..
發問:
The maximum value of the function f(x)=4k+18x-kx^2 (k is a positive constant) is 45. Find k.
最佳解答:
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
其他解答:
f.4 amathz..
發問:
The maximum value of the function f(x)=4k+18x-kx^2 (k is a positive constant) is 45. Find k.
最佳解答:
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
其他解答:
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