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The maximum and minimum values of x^3-18...

The maximum and minimum values of `x^3-18x^2+96x` in interval `(0,9)` are

A

160,0

B

128,160

C

160,128

D

0,160

Text Solution

Verified by Experts

The correct Answer is:
C

Let `y=x^(3)-18x^(2)+96ximplies(dy)/(dx)=3x^(2)-36x+96=0`
`becausex^(2)-12x+32=0implies(x-4)(x-8)=0,x=4,8`
Now `(d^(2)Y)/(dx^(2))=6x-36`
At `x=4(d^(2)y)/(dx^(2))=24-36=-12lt0`
At `x=4` function will be maximum
And `[f(4)]_(max)=64-288+384=160`
At `x=8,(d^(2)y)/(dx^(2))=48-36=12gt0`
At `x=8`, function will be minimum and `[f(8)]_(min)=128`.
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