A) 12
B) 9
C) 8
D) 6
Correct Answer: A
Solution :
\[f(x)=2x+3y\] when \[xy=6\] \[f(x)=2x+3y=2x+\frac{18}{x}\] \[{f}'(x)=2-\frac{18}{{{x}^{2}}}=0\] Þ \[x=\pm 3\] and \[{f}''(x)=\frac{36}{{{x}^{3}}}\Rightarrow {f}''(3)>0\] Putting \[x=+3\], we get the minimum value to be 12.You need to login to perform this action.
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