A) 75
B) 91
C) 84
D) 96
Correct Answer: A
Solution :
| [a] Let 20 be divided in two parts such that first part = x |
| \[\therefore \] Second part = 20-x |
| Now, assume that |
| \[P={{x}^{3}}(20-x)=20{{x}^{3}}-{{x}^{4}}\] |
| Now, \[\frac{dP}{dx}=60{{x}^{2}}-4{{x}^{3}};\] and \[\frac{{{d}^{2}}P}{d{{x}^{2}}}=120x-12{{x}^{2}}\] |
| Put \[\frac{dP}{dx}=0\] for maxima or minima |
| \[\Rightarrow \frac{dP}{dx}=0\] |
| \[\Rightarrow \,\,4{{x}^{2}}(15-x)=0\Rightarrow x=0,x=15\] |
| \[\therefore {{\left( \frac{{{d}^{2}}P}{d{{x}^{2}}} \right)}_{x=15}}=120\times 15-12\times (225)\] |
| \[=1800-2700=-900<0\] |
| \[\therefore \] P is a maximum at \[x=15\]. |
| \[\therefore \] First part = 15 |
| and second part \[=20-15=15\] |
| Required product \[=15\times 5=75\] |
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