Maxima and MinimaHard
Question
The greatest value of x3 - 18x2 + 96x in the interval (0 , 9) is-
Options
A.128
B.60
C.160
D.160
Solution
f(x) = x3 - 18x2 + 96x
f′(x) = 3x2 - 36x + 96
f′(x) = x2 - 12x + 32 = 0
= x2 - 8x - 4x + 32 = 0
= x(x - 8) - 4 (x - 8)
f′(x) = (x - 4) (x - 8) = 0
x = 4, 8
f″(x) = 2x - 12
f″ (4) = -4 < 0
x = 4 point of maxima
f(4) = 64 - 16 × 18 + 96 × 4
= 448 - 288 = 160
f′(x) = 3x2 - 36x + 96
f′(x) = x2 - 12x + 32 = 0
= x2 - 8x - 4x + 32 = 0
= x(x - 8) - 4 (x - 8)
f′(x) = (x - 4) (x - 8) = 0
x = 4, 8
f″(x) = 2x - 12
f″ (4) = -4 < 0
x = 4 point of maxima
f(4) = 64 - 16 × 18 + 96 × 4
= 448 - 288 = 160
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