A) 10, 10
B) 16, 4
C) 8, 12
D) 12, 8
Correct Answer: D
Solution :
Let \[x+y=20\Rightarrow y=20-x\] and \[{{x}^{3}}.{{y}^{2}}=z\Rightarrow z={{x}^{3}}.{{y}^{2}}\] \[z={{x}^{3}}{{(20-x)}^{2}}\]Þ \[z=400{{x}^{3}}+{{x}^{5}}-40{{x}^{4}}\] \[\frac{dz}{dx}=1200{{x}^{2}}+5{{x}^{4}}-160{{x}^{3}}\] Now \[\frac{dz}{dx}=0\], then \[x=12,\ 20\] Now \[\frac{{{d}^{2}}z}{d{{x}^{2}}}=2400x+16{{x}^{3}}-480{{x}^{2}}\]; \[{{\left( \frac{{{d}^{2}}z}{d{{x}^{2}}} \right)}_{x=12}}=-ive\] Hence \[x=12\]is the point of maxima \[\therefore x=12,\ \ y=8\].
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