A) \[4\]
B) \[1\]
C) \[3\]
D) \[0\]
Correct Answer: A
Solution :
Let two numbers are \[x\] and\[(3-x)\]. \[\therefore \]Product, \[P=x{{(3-x)}^{2}}\] \[\Rightarrow \] \[\frac{dP}{dx}=-2x(3-x)+{{(3-x)}^{2}}\] \[\Rightarrow \] \[\frac{dP}{dx}=(3-x)(3-3x)\] and \[\frac{{{d}^{2}}P}{d{{x}^{2}}}=6x-12\] For maxima or minima,\[\frac{dP}{dx}=0\] \[\Rightarrow \] \[(3-x)(3-3x)=0\] \[\Rightarrow \] \[x=3,\,\,1\] At\[(x=3)\], \[\frac{{{d}^{2}}P}{d{{x}^{2}}}=18-12=6>0\] (minima) At\[(x=1)\], \[\frac{{{d}^{2}}P}{d{{x}^{2}}}=-6<0\] So, \[P\]is maximum at\[x=1\]. Maximum value of \[P=1{{(3-1)}^{2}}=4\]
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