A) 3, 13
B) 4, 12
C) 6, 10
D) 8, 8
Correct Answer: D
Solution :
\[x+y=16\Rightarrow y=16-x\]Þ \[{{x}^{2}}+{{y}^{2}}={{x}^{2}}+{{(16-x)}^{2}}\] Let \[z={{x}^{2}}+{{(16-x)}^{2}}\Rightarrow z'=4x-32\] To be minimum of \[z,\,\,z''>0\], and it is. Therefore \[4x-32=0\Rightarrow x=8\Rightarrow y=8\]You need to login to perform this action.
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