A) 4
B) 3
C) 2
D) 1
Correct Answer: A
Solution :
| Let x, y be two numbers such that |
| \[x+y=3\Rightarrow y=3-x\].and let product \[\operatorname{P}=x{{y}^{2}}\]thus \[\operatorname{P}=x{{(3-x)}^{2}}={{x}^{3}}-6{{x}^{2}}+9x\] |
| For a maxima or minima \[\frac{dP}{dx}=0\] |
| Thus \[\frac{\operatorname{dP}}{dx}=3{{x}^{2}}-12x+9\]and\[\frac{{{d}^{2}}P}{d{{x}^{2}}}=6x-12\] |
| Now,\[\frac{\operatorname{dP}}{dx}=0\Rightarrow 3{{x}^{2}}-12x+9=0\Rightarrow x=1,3.\] |
| Thus, \[{{\left( \frac{{{d}^{2}}P}{d{{x}^{2}}} \right)}_{x=1}}=-6\]and\[{{\left( \frac{{{d}^{2}}P}{d{{x}^{2}}} \right)}_{x=3}}=6\] |
| Thus P is maximum when \[x=1\text{ }\Rightarrow y=\text{ }2\] |
| So, \[\operatorname{P}=1.{{2}^{2}}=4.\] |
You need to login to perform this action.
You will be redirected in
3 sec
