A) 4
B) 3
C) 2
D) 1
Correct Answer: A
Solution :
Let the first number be \[3-x\], then second will be \[x\]. Accordingly, we have to maximize \[(3-x){{x}^{2}}\] Let \[f(x)=(3-x){{x}^{2}}=3{{x}^{2}}-{{x}^{3}}\Rightarrow f'(x)=6x-3{{x}^{2}}\] \[\therefore f'(x)=0\Rightarrow x=0,\ 2\] Also \[{f}''\,(x)=6-6x\]. Obviously, \[{f}''\,(2)=-6<0\]. Therefore, required max. value \[=(3-2).\,{{2}^{2}}=4.\]You need to login to perform this action.
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