A) 160, 0
B) 60, 0
C) 160, 128
D) 120, 28
Correct Answer: C
Solution :
Let \[y={{x}^{3}}-18{{x}^{2}}+96x\]Þ \[\frac{dy}{dx}=3{{x}^{2}}-36x+96=0\] \ \[{{x}^{2}}-12x+32=0\Rightarrow (x-4)\,(x-8)=0\], \[x=4,\,8\] Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=6x-36\] At \[x=4\,\,\,\frac{{{d}^{2}}y}{d{{x}^{2}}}=24-36=-12<0\] \ At \[x=4\] function will be maximum and \[{{[f(4)]}_{\text{max}\text{.}}}=64-288+384=160\] At \[x=8\], \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=48-36=12>0\] \At\[x=8\], function will be minimum and \[{{[f(8)]}_{\text{min}\text{.}}}=128.\]
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